Otras dos asignaturas

5 files changed, 11264 insertions, 0 deletions

diff --git a/aalg/n.lyx b/aalg/n.lyx
new file mode 100644
index 0000000..d86d8f5
--- /dev/null
+++ b/aalg/n.lyx
@@ -0,0 +1,199 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\usepackage{tikz}
+\input{../defs}
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize 10
+\spacing single
+\use_hyperref false
+\papersize a5paper
+\use_geometry true
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 0.2cm
+\topmargin 0.7cm
+\rightmargin 0.2cm
+\bottommargin 0.7cm
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle empty
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Ampliación de Álgebra y Geometría
+\end_layout
+
+\begin_layout Date
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+def
+\backslash
+cryear{2018}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "../license.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Bibliografía:
+\end_layout
+
+\begin_layout Itemize
+Ampliación de Álgebra Lineal y Geometría 2016–17, Claudi Busqué Roca, Departamen
+to de Matemáticas, Universidad de Murcia.
+\end_layout
+
+\begin_layout Chapter
+Introducción a las cónicas
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n1.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Estudio métrico de las cónicas
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n2.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Cónicas proyectivas
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n3.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Formas bilineales y cuadráticas
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n4.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/aalg/n1.lyx b/aalg/n1.lyx
new file mode 100644
index 0000000..613776c
--- /dev/null
+++ b/aalg/n1.lyx
@@ -0,0 +1,3205 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\usepackage{tikz}
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style french
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Un 
+\series bold
+doble cono recto
+\series default
+ es la figura obtenida al girar una recta 
+\begin_inset Formula $g$
+\end_inset
+
+ alrededor de una recta 
+\begin_inset Formula $h$
+\end_inset
+
+, llamada 
+\series bold
+eje
+\series default
+, que la corta en un solo punto, el 
+\series bold
+vértice
+\series default
+.
+ La recta 
+\begin_inset Formula $g$
+\end_inset
+
+ y las que se obtienen al girar 
+\begin_inset Formula $g$
+\end_inset
+
+ alrededor del eje se llaman 
+\series bold
+generatrices
+\series default
+.
+ Una (
+\series bold
+sección
+\series default
+)
+\series bold
+ cónica
+\series default
+ es la intersección de un doble cono recto con un plano que lo corta.
+ Secciones cónicas 
+\series bold
+no degeneradas
+\series default
+:
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Circunferencia
+\series default
+: El plano es perpendicular al eje y no pasa por el vértice.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Elipse
+\series default
+: El plano forma un ángulo con el eje mayor al que este forma con una generatriz
+, sin ser perpendicular, y no pasa por el vértice.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Parábola
+\series default
+: El plano es paralelo a una generatriz y no pasa por el vértice.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Hipérbola
+\series default
+: El plano forma un ángulo con el eje menor al que este forma con una ge
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+ne
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+ra
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+triz, y no pasa por el vértice.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{center}
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{tikzpicture}
+\end_layout
+
+\begin_layout Plain Layout
+
+% Cone
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw (-2,-4) -- (3,6) (2,-4) -- (-3,6);
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[domain=-3:3] plot (
+\backslash
+x, {6+0.2*sqrt(9-
+\backslash
+x*
+\backslash
+x)});
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[domain=-3:3] plot (
+\backslash
+x, {6-0.2*sqrt(9-
+\backslash
+x*
+\backslash
+x)});
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[domain=-2:2] plot (
+\backslash
+x, {-4+0.2*sqrt(4-
+\backslash
+x*
+\backslash
+x)});
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[domain=-2:2] plot (
+\backslash
+x, {-4-0.2*sqrt(4-
+\backslash
+x*
+\backslash
+x)});
+\end_layout
+
+\begin_layout Plain Layout
+
+% Circumference
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw (-0.75,1) -- (0.75,1);
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[->] (3,1) node[right]{Circunferencia} -- (0.875,1);
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[orange,domain=-0.5:0.5] plot (
+\backslash
+x, {1+0.2*sqrt(0.25-
+\backslash
+x*
+\backslash
+x)});
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[orange,domain=-0.5:0.5] plot (
+\backslash
+x, {1-0.2*sqrt(0.25-
+\backslash
+x*
+\backslash
+x)});
+\end_layout
+
+\begin_layout Plain Layout
+
+% Ellipse
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw (-0.9736067977499789,1.38819660112501) -- (1.473606797749979,2.61180339887499);
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[->] (3,2) node[right]{Elipse} -- (1,2);
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[green,domain=-0.75:1.25] plot (
+\backslash
+x, {1.875+0.5*
+\backslash
+x+0.2*sqrt(1-(
+\backslash
+x-0.25)*(
+\backslash
+x-0.25))});
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[green,domain=-0.75:1.25] plot (
+\backslash
+x, {1.875+0.5*
+\backslash
+x-0.2*sqrt(1-(
+\backslash
+x-0.25)*(
+\backslash
+x-0.25))});
+\end_layout
+
+\begin_layout Plain Layout
+
+% Parabola
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw (-0.6118033988749895,-0.7763932022500211) -- (1,-4);
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[->] (3,-2.5) node[right]{Parábola} -- (1.125,-2.5);
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[red,domain=-0.5:1] plot (
+\backslash
+x, {-2*
+\backslash
+x-2+0.2*sqrt(2*
+\backslash
+x+1)});%1.183493515728975
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[red,domain=-0.5:1] plot (
+\backslash
+x, {-2*
+\backslash
+x-2-0.2*sqrt(2*
+\backslash
+x+1)});%0.8365064842710254
+\end_layout
+
+\begin_layout Plain Layout
+
+% Hyperbola
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw (-1, -4) -- (-2, 6);
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[->] (3,4.75) node[right]{Hipérbola} -- (-1.3125,4.75);
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[purple,domain=-2:-1.75] plot (
+\backslash
+x, {-10*
+\backslash
+x-14+0.2*sqrt(24*
+\backslash
+x*
+\backslash
+x+70*
+\backslash
+x+49)});%-1.960007140870476
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[purple,domain=-2:-1.75] plot (
+\backslash
+x, {-10*
+\backslash
+x-14-0.2*sqrt(24*
+\backslash
+x*
+\backslash
+x+70*
+\backslash
+x+49)});%-2.050493666883967
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[purple,domain=-1.166:-1] plot (
+\backslash
+x, {-10*
+\backslash
+x-14+0.2*sqrt(24*
+\backslash
+x*
+\backslash
+x+70*
+\backslash
+x+49)});%-0.9604664856861941
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw[purple,domain=-1.166:-1] plot (
+\backslash
+x, {-10*
+\backslash
+x-14-0.2*sqrt(24*
+\backslash
+x*
+\backslash
+x+70*
+\backslash
+x+49)});%-1.030648215444662
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{tikzpicture}
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{center}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Secciones cónicas no degeneradas.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Secciones cónicas 
+\series bold
+degeneradas
+\series default
+: Cuando el plano pasa por el vértice, obtenemos un punto si el ángulo del
+ plano con el eje es mayor al del eje con la generatriz, una recta si es
+ igual y un par de rectas que se cortan si es menor.
+\end_layout
+
+\begin_layout Section
+Circunferencia
+\end_layout
+
+\begin_layout Standard
+Una circunferencia es el lugar geométrico de los puntos del plano a la misma
+ distancia, llamada 
+\series bold
+radio
+\series default
+, a un punto fijo, el 
+\series bold
+centro
+\series default
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $h$
+\end_inset
+
+ el eje del cono, 
+\begin_inset Formula $g$
+\end_inset
+
+ la generatriz 
+\begin_inset Formula $V$
+\end_inset
+
+ el vértice y 
+\begin_inset Formula $O\neq V$
+\end_inset
+
+ el punto de corte de 
+\begin_inset Formula $h$
+\end_inset
+
+ con el plano perpendicular.
+ Si 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ están en la circunferencia, se corresponden con un giro de centro 
+\begin_inset Formula $V$
+\end_inset
+
+, luego 
+\begin_inset Formula $\Vert\overrightarrow{VA}\Vert=\Vert\overrightarrow{VB}\Vert$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\Vert\overrightarrow{OA}\Vert^{2}=\Vert\overrightarrow{VA}\Vert^{2}-\Vert\overrightarrow{VO}\Vert^{2}=\Vert\overrightarrow{VB}\Vert^{2}-\Vert\overrightarrow{VO}\Vert^{2}=\Vert\overrightarrow{OB}\Vert^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Fijado un sistema de referencia ortonormal, la ecuación de la circunferencia
+ 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ de centro 
+\begin_inset Formula $O=(a,b)$
+\end_inset
+
+ y radio 
+\begin_inset Formula $r$
+\end_inset
+
+, que denotamos 
+\begin_inset Formula ${\cal C}(O,r)$
+\end_inset
+
+, es 
+\begin_inset Formula $(x-a)^{2}+(y-b)^{2}=r^{2}$
+\end_inset
+
+, que podemos desarrollar como 
+\begin_inset Formula $x^{2}+y^{2}-2ax-2by+a^{2}+b^{2}-r^{2}=0$
+\end_inset
+
+.
+ Situando el origen de coordenadas en 
+\begin_inset Formula $O$
+\end_inset
+
+, obtenemos la 
+\series bold
+ecuación reducida de la circunferencia
+\series default
+: 
+\begin_inset Formula 
+\[
+x^{2}+y^{2}=r^{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Las 
+\series bold
+ecuaciones paramétricas
+\series default
+ de un cierto objeto 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ son las componentes de una aplicación biyectiva 
+\begin_inset Formula $p:I\rightarrow{\cal C}$
+\end_inset
+
+ donde 
+\begin_inset Formula $I$
+\end_inset
+
+ es un intervalo de 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Para las circunferencias, tenemos 
+\begin_inset Formula 
+\begin{eqnarray*}
+\left\{ \begin{array}{rcl}
+x & = & r\cos t\\
+y & = & r\sin t
+\end{array}\right. & \text{ con } & t\in[0,2\pi)
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Dadas dos circunferencias 
+\begin_inset Formula ${\cal C}(O,r)$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal C}(O',r')$
+\end_inset
+
+ con 
+\begin_inset Formula $r<r'$
+\end_inset
+
+, sea 
+\begin_inset Formula $d:=\Vert\overrightarrow{OO'}\Vert\neq0$
+\end_inset
+
+, estas se cortan en dos puntos si 
+\begin_inset Formula $r-r'<d<r+r'$
+\end_inset
+
+ y en uno si 
+\begin_inset Formula $d=r-r'$
+\end_inset
+
+ ó 
+\begin_inset Formula $d=r+r'$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula ${\cal C}:={\cal C}(O,r)$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal D}:={\cal C}(O',r')$
+\end_inset
+
+, la ecuación de 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ en un cierto referencial ortonormal es 
+\begin_inset Formula $x^{2}+y^{2}=r^{2}$
+\end_inset
+
+, y rotando este referencial, obtenemos uno con 
+\begin_inset Formula ${\cal C}\equiv x^{2}+y^{2}=r^{2}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal D}\equiv(x-d)^{2}+y^{2}=r'^{2}$
+\end_inset
+
+.
+ Así, si 
+\begin_inset Formula $P=(x,y)\in{\cal {\cal C}}\cap{\cal D}$
+\end_inset
+
+, restando la ecuación de 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ a la de 
+\begin_inset Formula ${\cal D}$
+\end_inset
+
+ obtenemos que 
+\begin_inset Formula $-2dx+d^{2}=r'^{2}-r^{2}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $x=\frac{d^{2}+r^{2}-r'^{2}}{2d}$
+\end_inset
+
+, luego 
+\begin_inset Formula $y=\pm\sqrt{r^{2}-x^{2}}=\pm\sqrt{r^{2}-\left(\frac{d^{2}+r^{2}-r'^{2}}{2d}\right)^{2}}$
+\end_inset
+
+.
+ Esta última ecuación tiene solución cuando
+\begin_inset Formula 
+\begin{multline*}
+|r|\geq\left|\frac{d^{2}+r^{2}-r'^{2}}{2d}\right|\overset{r\geq r'}{\iff}r\geq\frac{d^{2}+r^{2}-r'^{2}}{2d}\iff2dr\geq d^{2}+r^{2}-r'^{2}\iff\\
+\iff r'^{2}\geq(d-r)^{2}\iff r'\geq|d-r|\iff r'\geq d-r,r-d\iff r-r'\leq d\leq r+r'
+\end{multline*}
+
+\end_inset
+
+Vemos de forma análoga que esta solución es única cuando 
+\begin_inset Formula $d\in\{r-r',r+r'\}$
+\end_inset
+
+, y de lo contrario es doble.
+\end_layout
+
+\begin_layout Standard
+Un punto 
+\begin_inset Formula $P$
+\end_inset
+
+ pertenece a la circunferencia en que 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ son diametralmente opuestos si y sólo si 
+\begin_inset Formula $\overrightarrow{AP}\bot\overrightarrow{BP}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula ${\cal C}(O,r)$
+\end_inset
+
+ esta circunferencia,
+\begin_inset Formula 
+\begin{multline*}
+\Vert\overrightarrow{OP}\Vert^{2}=(\overrightarrow{OA}+\overrightarrow{AP})\cdot(\overrightarrow{OB}+\overrightarrow{BP})=\overrightarrow{OA}\cdot\overrightarrow{OB}+\overrightarrow{OA}\cdot\overrightarrow{BP}+\overrightarrow{AP}\cdot\overrightarrow{OB}+\overrightarrow{AP}\cdot\overrightarrow{BP}=\\
+=-r^{2}+\overrightarrow{OA}\cdot\overrightarrow{BP}-\overrightarrow{AP}\cdot\overrightarrow{OA}+\overrightarrow{AP}\cdot\overrightarrow{BP}=-r^{2}+\overrightarrow{OA}\cdot(\overrightarrow{BP}-\overrightarrow{AP})+\overrightarrow{AP}\cdot\overrightarrow{BP}=\\
+=-r^{2}+\overrightarrow{OA}\cdot(\overrightarrow{BO}+\overrightarrow{OA})+\overrightarrow{AP}\cdot\overrightarrow{BP}=-r^{2}+2r^{2}+\overrightarrow{AP}\cdot\overrightarrow{BP}=r^{2}+\overrightarrow{AP}\cdot\overrightarrow{BP}
+\end{multline*}
+
+\end_inset
+
+luego 
+\begin_inset Formula $P\in{\cal C}(O,r)\iff\Vert\overrightarrow{OP}\Vert^{2}=r^{2}\iff\overrightarrow{AP}\cdot\overrightarrow{BP}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una recta es 
+\series bold
+tangente
+\series default
+ a una circunferencia si la corta en un único punto, y 
+\series bold
+secante
+\series default
+ si la corta en dos puntos.
+ Sea 
+\begin_inset Formula $P$
+\end_inset
+
+ un punto exterior a la circunferencia 
+\begin_inset Formula ${\cal C}:={\cal C}(O,r)$
+\end_inset
+
+ (
+\begin_inset Formula $\Vert\overrightarrow{OP}\Vert>r$
+\end_inset
+
+), existen dos y solo dos tangentes a 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ por 
+\begin_inset Formula $P$
+\end_inset
+
+, y si 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ son los puntos de tangencia, 
+\begin_inset Formula $\Vert\overrightarrow{PA}\Vert=\Vert\overrightarrow{PB}\Vert$
+\end_inset
+
+ y 
+\begin_inset Formula $\overrightarrow{OA}\bot\overrightarrow{PA}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $M:=\frac{O+P}{2}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal D}:={\cal C}(M,\Vert\overrightarrow{MO}\Vert)$
+\end_inset
+
+, sabemos que 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal D}$
+\end_inset
+
+ se cortan en dos puntos 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $O,P,A\in{\cal D}$
+\end_inset
+
+ siendo 
+\begin_inset Formula $O$
+\end_inset
+
+ y 
+\begin_inset Formula $P$
+\end_inset
+
+ diametralmente opuestos, tenemos 
+\begin_inset Formula $\overrightarrow{OA}\bot\overrightarrow{PA}$
+\end_inset
+
+, luego 
+\begin_inset Formula $PA$
+\end_inset
+
+ es tangente a 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+, porque cualquier otro punto 
+\begin_inset Formula $A'\in PA$
+\end_inset
+
+ cumple 
+\begin_inset Formula $\Vert\overrightarrow{OA'}\Vert=\sqrt{\Vert\overrightarrow{OA}\Vert^{2}+\Vert\overrightarrow{AA'}\Vert^{2}}>\Vert\overrightarrow{OA}\Vert$
+\end_inset
+
+, y por tanto 
+\begin_inset Formula $A'\notin{\cal C}$
+\end_inset
+
+.
+ Por el mismo argumento, 
+\begin_inset Formula $PB$
+\end_inset
+
+ es tangente a 
+\begin_inset Formula $B$
+\end_inset
+
+.
+ Además, por ser 
+\begin_inset Formula $\overrightarrow{OA}\bot\overrightarrow{PA}$
+\end_inset
+
+ y 
+\begin_inset Formula $\overrightarrow{OB}\bot\overrightarrow{PB}$
+\end_inset
+
+, 
+\begin_inset Formula $\Vert\overrightarrow{PA}\Vert^{2}=\Vert\overrightarrow{OP}\Vert^{2}-\Vert\overrightarrow{OA}\Vert^{2}=\Vert\overrightarrow{OP}\Vert^{2}-r^{2}=\Vert\overrightarrow{OP}\Vert^{2}-\Vert\overrightarrow{OB}\Vert^{2}=\Vert\overrightarrow{PB}\Vert^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Finalmente, supongamos que existe una tercera recta que pasa por 
+\begin_inset Formula $P$
+\end_inset
+
+ y es tangente a 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ en un punto 
+\begin_inset Formula $D$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $Q$
+\end_inset
+
+ el punto de 
+\begin_inset Formula $PD$
+\end_inset
+
+ tal que 
+\begin_inset Formula $QM\bot PD$
+\end_inset
+
+, tenemos que 
+\begin_inset Formula $\min\{\Vert\overrightarrow{P'M}\Vert\}_{P'\in PD}=\Vert\overrightarrow{QM}\Vert$
+\end_inset
+
+, y para cualquier otro punto 
+\begin_inset Formula $Q'\in PD$
+\end_inset
+
+ se tiene 
+\begin_inset Formula $\Vert\overrightarrow{Q'M}\Vert>\Vert\overrightarrow{QM}\Vert$
+\end_inset
+
+, luego 
+\begin_inset Formula $PM\bot PD\iff\min\{\Vert\overrightarrow{P'M}\Vert\}_{P'\in PD}=\Vert\overrightarrow{PM}\Vert$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $PD$
+\end_inset
+
+ es perpendicular a 
+\begin_inset Formula $PM$
+\end_inset
+
+, también lo es a 
+\begin_inset Formula $PO$
+\end_inset
+
+, luego para un punto 
+\begin_inset Formula $P'\in PD$
+\end_inset
+
+, 
+\begin_inset Formula $\Vert\overrightarrow{P'O}\Vert>\Vert\overrightarrow{PO}\Vert>r$
+\end_inset
+
+ y 
+\begin_inset Formula $PD$
+\end_inset
+
+ no corta a 
+\begin_inset Formula ${\cal C}\#$
+\end_inset
+
+.
+ Si por el contrario 
+\begin_inset Formula $Q\neq P$
+\end_inset
+
+, tomando 
+\begin_inset Formula $P'\neq P$
+\end_inset
+
+ como la simetría de 
+\begin_inset Formula $P$
+\end_inset
+
+ sobre la recta 
+\begin_inset Formula $QM$
+\end_inset
+
+ es fácil ver que 
+\begin_inset Formula $P'\in PD\cap{\cal D}$
+\end_inset
+
+, y como 
+\begin_inset Formula $P$
+\end_inset
+
+ y 
+\begin_inset Formula $O$
+\end_inset
+
+ son diametralmente opuestos en 
+\begin_inset Formula ${\cal D}$
+\end_inset
+
+, 
+\begin_inset Formula $PP'\bot OP'$
+\end_inset
+
+.
+ Así, si 
+\begin_inset Formula $\Vert\overrightarrow{OP'}\Vert>r$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\Vert\overrightarrow{OD}\Vert\geq\Vert\overrightarrow{OP'}\Vert>r\#$
+\end_inset
+
+, y si 
+\begin_inset Formula $\Vert\overrightarrow{OP'}\Vert<r$
+\end_inset
+
+, tomando la simetría de 
+\begin_inset Formula $D\in{\cal C}$
+\end_inset
+
+ sobre la recta 
+\begin_inset Formula $OP'$
+\end_inset
+
+ obtenemos un punto 
+\begin_inset Formula $D'\in{\cal C}\cap PD$
+\end_inset
+
+, luego 
+\begin_inset Formula $PD$
+\end_inset
+
+ es secante
+\begin_inset Formula $\#$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Si alguien tiene una demostración más corta o procesable de que no hay tercera
+ recta, que me lo diga, por favor.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Por tres puntos no alineados 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $B$
+\end_inset
+
+ y 
+\begin_inset Formula $C$
+\end_inset
+
+ pasa una única circunferencia.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ una circunferencia que pasa por 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $B$
+\end_inset
+
+ y 
+\begin_inset Formula $C$
+\end_inset
+
+.
+ Necesariamente el centro, 
+\begin_inset Formula $O$
+\end_inset
+
+, debe estar en las mediatrices de 
+\begin_inset Formula $AC$
+\end_inset
+
+ y de 
+\begin_inset Formula $AB$
+\end_inset
+
+, que llamaremos 
+\begin_inset Formula $m$
+\end_inset
+
+ y 
+\begin_inset Formula $m'$
+\end_inset
+
+ respectivamente.
+ La intersección de estas es un único punto, pues de lo contrario estas
+ serían paralelas y por tanto 
+\begin_inset Formula $AB$
+\end_inset
+
+ y 
+\begin_inset Formula $AC$
+\end_inset
+
+ lo serían también entre sí, pero como tienen un punto 
+\begin_inset Formula $A$
+\end_inset
+
+ en común, los tres puntos estarían alineados.
+ Así, podemos tomar 
+\begin_inset Formula $\{O\}:=m\cap m'$
+\end_inset
+
+ y entonces 
+\begin_inset Formula ${\cal C}(O,\overrightarrow{OA})$
+\end_inset
+
+ sería la única circunferencia que pasa por los tres puntos.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula ${\cal C}\equiv ax^{2}+bxy+cy^{2}+dx+ey+f=0$
+\end_inset
+
+ (por ejemplo, una circunferencia) y 
+\begin_inset Formula $\ell\nsubseteq{\cal C}$
+\end_inset
+
+ una recta, entonces 
+\begin_inset Formula $|\ell\cap{\cal C}|\leq2$
+\end_inset
+
+.
+ 
+\series bold
+Demostración
+\series default
+: Tras un cambio de coordenadas ortonormal tal que 
+\begin_inset Formula $\ell\equiv y=0$
+\end_inset
+
+, tenemos 
+\begin_inset Formula 
+\[
+\left\{ \begin{array}{rrcl}
+{\cal C}: & a'x^{2}+b'xy+c'y^{2}+d'x+e'y+f' & = & 0\\
+\ell: & y & = & 0
+\end{array}\right.\iff a'x^{2}+d'x+f'=0
+\]
+
+\end_inset
+
+que al ser una ecuación de segundo grado, tiene a lo sumo 2 soluciones salvo
+ si 
+\begin_inset Formula $a'=d'=f'=0$
+\end_inset
+
+, pero entonces sería 
+\begin_inset Formula ${\cal C}\equiv b'xy+c'y^{2}+e'y=0$
+\end_inset
+
+ y 
+\begin_inset Formula $\ell\subseteq{\cal C}\#$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+La elipse
+\end_layout
+
+\begin_layout Standard
+Una elipse es el lugar geométrico de los puntos del plano cuya suma de distancia
+s a dos puntos fijos, llamados 
+\series bold
+focos
+\series default
+, es constante.
+ 
+\begin_inset Note Comment
+status open
+
+\begin_layout Plain Layout
+
+\series bold
+Demostración
+\series default
+: Sean dos esferas inscritas en el cono y tangentes a la sección del cono
+ que nos da la elipse
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Queda demostrar que estas son únicas.
+\end_layout
+
+\end_inset
+
+, si 
+\begin_inset Formula $F$
+\end_inset
+
+ y 
+\begin_inset Formula $F'$
+\end_inset
+
+ son los puntos de tangencia, dado un punto 
+\begin_inset Formula $P$
+\end_inset
+
+ arbitrario de la elipse, si 
+\begin_inset Formula $A_{P}$
+\end_inset
+
+ y 
+\begin_inset Formula $B_{P}$
+\end_inset
+
+ son puntos de las esferas (huecas) que contienen respectivamente a 
+\begin_inset Formula $F$
+\end_inset
+
+ y 
+\begin_inset Formula $F'$
+\end_inset
+
+ y que están alineados con 
+\begin_inset Formula $P$
+\end_inset
+
+ y el vértice del cono, entonces 
+\begin_inset Formula $\Vert\overrightarrow{PF'}\Vert=\Vert\overrightarrow{PB_{P}}\Vert$
+\end_inset
+
+ y 
+\begin_inset Formula $\Vert\overrightarrow{PF}\Vert=\Vert\overrightarrow{PA_{P}}\Vert$
+\end_inset
+
+, pues todas las tangentes a una esfera dese un mismo punto (
+\begin_inset Formula $P$
+\end_inset
+
+) tienen la misma longitud
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+También hay que probar esto último.
+\end_layout
+
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\Vert\overrightarrow{PF}\Vert+\Vert\overrightarrow{PF'}\Vert=\Vert\overrightarrow{A_{P}B_{P}}\Vert$
+\end_inset
+
+, pero 
+\begin_inset Formula $\Vert\overrightarrow{A_{P}B_{P}}\Vert$
+\end_inset
+
+ no depende del punto 
+\begin_inset Formula $P$
+\end_inset
+
+, pues 
+\begin_inset Formula $A_{P}$
+\end_inset
+
+ y 
+\begin_inset Formula $B_{P}$
+\end_inset
+
+ mantienen distancia constante con el vértice del cono por estar en la intersecc
+ión del cono y la esfera que es una circunferencia perpendicular al eje
+ del cono
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Lo cual hay que demostrar también.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $F$
+\end_inset
+
+ y 
+\begin_inset Formula $F'$
+\end_inset
+
+ son los focos de la elipse, la recta 
+\begin_inset Formula $FF'$
+\end_inset
+
+ es el 
+\series bold
+eje principal
+\series default
+, la mediatriz del segmento 
+\begin_inset Formula $FF'$
+\end_inset
+
+, el 
+\series bold
+eje secundario
+\series default
+, y su intersección es el 
+\series bold
+centro
+\series default
+ 
+\begin_inset Formula $O$
+\end_inset
+
+.
+ Los 
+\series bold
+vértices
+\series default
+ son los puntos de la elipse que intersecan con el eje principal (
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $A'$
+\end_inset
+
+) o el secundario (
+\begin_inset Formula $B$
+\end_inset
+
+ y 
+\begin_inset Formula $B'$
+\end_inset
+
+).
+ Llamamos 
+\series bold
+semidistancia focal
+\series default
+ a 
+\begin_inset Formula $c:=\Vert\overrightarrow{OF}\Vert$
+\end_inset
+
+, 
+\series bold
+distancia focal
+\series default
+ a 
+\begin_inset Formula $2c$
+\end_inset
+
+, 
+\series bold
+semieje principal
+\series default
+ a 
+\begin_inset Formula $a:=\Vert\overrightarrow{OA}\Vert$
+\end_inset
+
+ y 
+\series bold
+semieje secundario
+\series default
+ a 
+\begin_inset Formula $b:=\Vert\overrightarrow{OB}\Vert$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Para todo punto 
+\begin_inset Formula $P$
+\end_inset
+
+ de la elipse, 
+\begin_inset Formula $\Vert\overrightarrow{PF}\Vert+\Vert\overrightarrow{PF'}\Vert=2a$
+\end_inset
+
+, pues 
+\begin_inset Formula $\Vert\overrightarrow{AF}\Vert+\Vert\overrightarrow{AF}\Vert+\Vert\overrightarrow{FF'}\Vert=\Vert\overrightarrow{AF}\Vert+\Vert\overrightarrow{AF'}\Vert=\Vert\overrightarrow{A'F}\Vert+\Vert\overrightarrow{A'F'}\Vert=\Vert\overrightarrow{A'F'}\Vert+\Vert\overrightarrow{F'F}\Vert+\Vert\overrightarrow{A'F'}\Vert$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $\Vert\overrightarrow{AF}\Vert=\Vert\overrightarrow{A'F'}\Vert$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $\Vert\overrightarrow{PF}\Vert+\Vert\overrightarrow{PF'}\Vert=\Vert\overrightarrow{AF}\Vert+\Vert\overrightarrow{AF'}\Vert=\Vert\overrightarrow{A'F'}\Vert+\Vert\overrightarrow{AF'}\Vert=\Vert\overrightarrow{AA'}\Vert=2a$
+\end_inset
+
+.
+ De aquí que 
+\begin_inset Formula $\Vert\overrightarrow{BF}\Vert=a$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $a^{2}=\Vert\overrightarrow{BF}\Vert^{2}=\Vert\overrightarrow{BO}\Vert^{2}+\Vert\overrightarrow{OF}\Vert^{2}=b^{2}+c^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+excentricidad
+\series default
+ de la elipse a 
+\begin_inset Formula $\epsilon:=\frac{c}{a}$
+\end_inset
+
+, y tenemos que 
+\begin_inset Formula $b=a\sqrt{1-\epsilon^{2}}$
+\end_inset
+
+ y 
+\begin_inset Formula $0\leq\epsilon<1$
+\end_inset
+
+, si bien cuando 
+\begin_inset Formula $\epsilon=0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $F=F'$
+\end_inset
+
+ y tenemos una circunferencia.
+\end_layout
+
+\begin_layout Standard
+Una 
+\series bold
+ecuación reducida de la elipse
+\series default
+ es 
+\begin_inset Formula 
+\[
+\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
+\]
+
+\end_inset
+
+siendo 
+\begin_inset Formula $a$
+\end_inset
+
+ el semieje mayor y 
+\begin_inset Formula $b$
+\end_inset
+
+ el menor.
+ En efecto, en cierto referencial ortonormal la elipse tiene focos 
+\begin_inset Formula $F=(c,0)$
+\end_inset
+
+ y 
+\begin_inset Formula $F'=(-c,0)$
+\end_inset
+
+, y para un punto 
+\begin_inset Formula $P(x,y)$
+\end_inset
+
+ en la elipse, 
+\begin_inset Formula $2a=\Vert\overrightarrow{PF}\Vert+\Vert\overrightarrow{PF'}\Vert=\sqrt{(x-c)^{2}+y^{2}}+\sqrt{(x+c)^{2}+y^{2}}$
+\end_inset
+
+, luego 
+\begin_inset Formula $4a^{2}+x^{2}+2cx+c^{2}+y^{2}-4a\sqrt{(x+c)^{2}+y^{2}}=x^{2}-2cx+c^{2}+y^{2}$
+\end_inset
+
+, y simplificando, 
+\begin_inset Formula $a^{2}+cx=a\sqrt{(x+c)^{2}+y^{2}}$
+\end_inset
+
+ y de aquí, elevando al cuadrado y simplificando, 
+\begin_inset Formula $a^{2}y^{2}+b^{2}x^{2}=a^{2}b^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Unas 
+\series bold
+ecuaciones paramétricas
+\series default
+ de esta elipse son 
+\begin_inset Formula 
+\begin{eqnarray*}
+\left\{ \begin{array}{rcl}
+x & = & a\cos t\\
+y & = & b\sin t
+\end{array}\right. & \text{ con } & t\in[0,2\pi)
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Decimos que una recta es 
+\series bold
+tangente
+\series default
+ a una elipse si la corta en un único punto y 
+\series bold
+secante
+\series default
+ si la corta en dos puntos.
+ La 
+\series bold
+propiedad focal de la elipse
+\series default
+ dice que, dado un punto 
+\begin_inset Formula $P$
+\end_inset
+
+ de una elipse de focos 
+\begin_inset Formula $F$
+\end_inset
+
+ y 
+\begin_inset Formula $F'$
+\end_inset
+
+, la recta bisectriz del ángulo entre 
+\begin_inset Formula $-\overrightarrow{PF}$
+\end_inset
+
+ y 
+\begin_inset Formula $\overrightarrow{PF'}$
+\end_inset
+
+ es tangente a la elipse.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $\ell$
+\end_inset
+
+ dicha recta, basta ver que cualquier otro punto 
+\begin_inset Formula $P'\in\ell$
+\end_inset
+
+ no está en la elipse.
+ Sea 
+\begin_inset Formula $G:=s_{\ell}(F)$
+\end_inset
+
+ (el simétrico), entonces 
+\begin_inset Formula $P$
+\end_inset
+
+ está en el segmento 
+\begin_inset Formula $F'G$
+\end_inset
+
+ y 
+\begin_inset Formula $\Vert\overrightarrow{F'P}\Vert+\Vert\overrightarrow{FP}\Vert=\Vert\overrightarrow{F'P}\Vert+\Vert\overrightarrow{GP}\Vert=\Vert\overrightarrow{F'G}\Vert<\Vert\overrightarrow{F'P'}\Vert+\Vert\overrightarrow{P'G}\Vert=\Vert\overrightarrow{F'P'}\Vert+\Vert\overrightarrow{P'F}\Vert$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+La recta tangente a la elipse 
+\begin_inset Formula ${\cal C}\equiv\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
+\end_inset
+
+ por el punto 
+\begin_inset Formula $P(x_{0},y_{0})$
+\end_inset
+
+ es 
+\begin_inset Formula $\ell\equiv\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^{2}}=1$
+\end_inset
+
+.
+ En particular, la tangente a la circunferencia 
+\begin_inset Formula ${\cal C}\equiv x^{2}+y^{2}=r^{2}$
+\end_inset
+
+ por 
+\begin_inset Formula $P\in{\cal C}$
+\end_inset
+
+ es 
+\begin_inset Formula $\ell\equiv x_{0}x+y_{0}y=r^{2}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración
+\series default
+: Sea 
+\begin_inset Formula 
+\[
+\ell\equiv\left\{ \begin{array}{rcl}
+x & = & x_{0}+ut\\
+y & = & y_{0}+vt
+\end{array}\right.
+\]
+
+\end_inset
+
+Los puntos de 
+\begin_inset Formula $\ell$
+\end_inset
+
+ que están en la elipse satisfacen 
+\begin_inset Formula $\frac{(x_{0}+ut)^{2}}{a^{2}}+\frac{(y_{0}+vt)^{2}}{b^{2}}=1$
+\end_inset
+
+, y operando obtenemos 
+\begin_inset Formula $\left(\frac{2ux_{0}}{a^{2}}+\frac{2vy_{0}}{b^{2}}\right)t+\left(\frac{u^{2}}{a^{2}}+\frac{v^{2}}{b^{2}}\right)t^{2}=1-\frac{x_{0}^{2}}{a^{2}}-\frac{y_{0}^{2}}{b^{2}}=0$
+\end_inset
+
+, que se cumple para 
+\begin_inset Formula $t\in\left\{ 0,-\frac{2\left(\frac{ux_{0}}{a^{2}}+\frac{vy_{0}}{b^{2}}\right)}{\frac{u^{2}}{a^{2}}+\frac{v^{2}}{b^{2}}}\right\} $
+\end_inset
+
+.
+ Estos dos valores son iguales si y sólo si 
+\begin_inset Formula $\frac{ux_{0}}{a^{2}}+\frac{vy_{0}}{b^{2}}=0$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $\frac{x_{0}}{a^{2}}(x-x_{0})+\frac{y_{0}}{b^{2}}(y-y_{0})=0$
+\end_inset
+
+ o, equivalentemente, 
+\begin_inset Formula $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^{2}}=\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+La hipérbola
+\end_layout
+
+\begin_layout Standard
+Una hipérbola es el lugar geométrico de los puntos del plano cuya diferencia
+ de distancias a dos puntos fijos distintos, llamados 
+\series bold
+focos
+\series default
+, es constante en valor absoluto.
+ Si los focos son 
+\begin_inset Formula $F$
+\end_inset
+
+ y 
+\begin_inset Formula $F'$
+\end_inset
+
+, llamamos 
+\series bold
+eje principal
+\series default
+ a la recta 
+\begin_inset Formula $FF'$
+\end_inset
+
+, 
+\series bold
+eje secundario
+\series default
+ a la mediatriz del segmento 
+\begin_inset Formula $FF'$
+\end_inset
+
+ y 
+\series bold
+centro
+\series default
+ a donde intersecan ambos ejes.
+ Los 
+\series bold
+vértices
+\series default
+ de la hipérbola son los sus puntos de corte con el eje principal.
+ Llamamos 
+\series bold
+semidistancia focal
+\series default
+ a 
+\begin_inset Formula $c:=\Vert\overrightarrow{OF}\Vert$
+\end_inset
+
+, 
+\series bold
+distancia focal
+\series default
+ a 
+\begin_inset Formula $2c$
+\end_inset
+
+, 
+\series bold
+semieje principal
+\series default
+ a 
+\begin_inset Formula $a:=\Vert\overrightarrow{OA}\Vert$
+\end_inset
+
+ y 
+\series bold
+semieje secundario
+\series default
+ a 
+\begin_inset Formula $b:=\sqrt{c^{2}-a^{2}}$
+\end_inset
+
+.
+ Una hipérbola es 
+\series bold
+equilátera
+\series default
+ si 
+\begin_inset Formula $a=b$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+excentricidad
+\series default
+ a 
+\begin_inset Formula $\epsilon:=\frac{c}{a}>1$
+\end_inset
+
+, y tenemos que 
+\begin_inset Formula $b=a\sqrt{\epsilon^{2}-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Todo punto 
+\begin_inset Formula $P$
+\end_inset
+
+ de la hipérbola 
+\begin_inset Formula ${\cal H}$
+\end_inset
+
+ cumple 
+\begin_inset Formula $\Vert\overrightarrow{PF}\Vert-\Vert\overrightarrow{PF'}\Vert=\pm2a$
+\end_inset
+
+.
+ 
+\series bold
+Demostración
+\series default
+: Sabemos que 
+\begin_inset Formula $\Vert\overrightarrow{AA'}\Vert=\Vert\overrightarrow{A'F}\Vert-\Vert\overrightarrow{AF}\Vert=\Vert\overrightarrow{AF'}\Vert-\Vert\overrightarrow{A'F'}\Vert$
+\end_inset
+
+, y que 
+\begin_inset Formula $\Vert\overrightarrow{AF'}\Vert-\Vert\overrightarrow{AF}\Vert=\Vert\overrightarrow{A'F}\Vert-\Vert\overrightarrow{A'F'}\Vert$
+\end_inset
+
+.
+ Sustituyendo 
+\begin_inset Formula $\Vert\overrightarrow{AF'}\Vert$
+\end_inset
+
+ en la segunda ecuación, nos queda 
+\begin_inset Formula $\Vert\overrightarrow{A'F'}\Vert+\Vert\overrightarrow{A'F}\Vert-2\Vert\overrightarrow{AF}\Vert=\Vert\overrightarrow{A'F}\Vert-\Vert\overrightarrow{A'F'}\Vert$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\Vert\overrightarrow{A'F'}\Vert=\Vert\overrightarrow{AF}\Vert$
+\end_inset
+
+, lo que significa que 
+\begin_inset Formula $\Vert\overrightarrow{OA'}\Vert=\Vert\overrightarrow{OF'}\Vert-\Vert\overrightarrow{A'F'}\Vert=\Vert\overrightarrow{OF}\Vert-\Vert\overrightarrow{AF}\Vert=\Vert\overrightarrow{OA}\Vert=a$
+\end_inset
+
+ y 
+\begin_inset Formula $\Vert\overrightarrow{AA'}\Vert=2a$
+\end_inset
+
+.
+ Así, dado un punto 
+\begin_inset Formula $P\in{\cal H}$
+\end_inset
+
+ arbitrario, se tiene 
+\begin_inset Formula $|\Vert\overrightarrow{PF}\Vert-\Vert\overrightarrow{PF'}\Vert|=\Vert\overrightarrow{AF'}\Vert-\Vert\overrightarrow{AF}\Vert=\Vert\overrightarrow{AF'}\Vert-\Vert\overrightarrow{A'F'}\Vert=\Vert\overrightarrow{AA'}\Vert=2a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una 
+\series bold
+ecuación reducida de la hipérbola
+\series default
+ de semieje principal 
+\begin_inset Formula $a$
+\end_inset
+
+ y secundario 
+\begin_inset Formula $b$
+\end_inset
+
+ es
+\begin_inset Formula 
+\[
+\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
+\]
+
+\end_inset
+
+En efecto, si tomamos el referencial ortonormal en el que 
+\begin_inset Formula $F=(c,0)$
+\end_inset
+
+ y 
+\begin_inset Formula $F'=(-c,0)$
+\end_inset
+
+, tenemos 
+\begin_inset Formula $\pm2a=\Vert\overrightarrow{PF}\Vert-\Vert\overrightarrow{PF'}\Vert=\sqrt{(x-c)^{2}+y^{2}}-\sqrt{(x+c)^{2}+y^{2}}$
+\end_inset
+
+, es decir, 
+\begin_inset Formula $4a^{2}+(x+c)^{2}+y^{2}\pm4a\sqrt{(x+c)^{2}+y^{2}}=(x-c)^{2}+y^{2}$
+\end_inset
+
+, y simplificando, 
+\begin_inset Formula $a^{2}+cx=\pm a\sqrt{(x+c)^{2}+y^{2}}$
+\end_inset
+
+.
+ Elevando al cuadrado y simplificando, nos queda que 
+\begin_inset Formula $b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Unas ecuaciones paramétricas para esta hipérbola son
+\begin_inset Formula 
+\begin{eqnarray*}
+\left\{ \begin{array}{rcl}
+x & = & a\cosh t\\
+y & = & b\sinh t
+\end{array}\right. & \text{ con } & t\in\mathbb{R}
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+sremember{FUVR1}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\begin{eqnarray*}
+\cosh(x)=\frac{e^{x}+e^{-x}}{2} & \sinh(x)=\frac{e^{x}-e^{-x}}{2} & \cosh^{2}(x)-\sinh^{2}(x)=1
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+eremember
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Una recta 
+\begin_inset Formula $\ell$
+\end_inset
+
+ es una 
+\series bold
+asíntota
+\series default
+ de la hipérbola 
+\begin_inset Formula ${\cal H}$
+\end_inset
+
+ si 
+\begin_inset Formula $\ell\cap{\cal H}=\emptyset$
+\end_inset
+
+ y 
+\begin_inset Formula $d(\ell,{\cal H})=0$
+\end_inset
+
+, es
+\series bold
+ asintótica
+\series default
+ si es paralela a una asíntota, y es 
+\series bold
+tangente
+\series default
+ si corta a la hipérbola en un único punto sin ser asintótica.
+ Las rectas 
+\begin_inset Formula $y=\pm\frac{b}{a}x$
+\end_inset
+
+ son las (únicas) asíntotas de la hipérbola dada por la ecuación reducida.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $\ell\equiv y=\pm\frac{b}{a}x$
+\end_inset
+
+, 
+\begin_inset Formula $(x,y)\in{\cal H}\cap\ell\iff\left(\frac{x}{a}\right)^{2}-\left(\frac{\pm\frac{b}{a}x}{b}\right)^{2}=1$
+\end_inset
+
+, pero 
+\begin_inset Formula $\left(\frac{x}{a}\right)^{2}-\left(\frac{\pm\frac{b}{a}x}{b}\right)^{2}=\left(\frac{x}{a}\right)^{2}-\left(\frac{x}{a}\right)^{2}=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $\ell\cap{\cal H}=\emptyset$
+\end_inset
+
+.
+ Ahora bien, dado 
+\begin_inset Formula $t\in\mathbb{R}$
+\end_inset
+
+, el punto 
+\begin_inset Formula $P:=(a\cosh t,b\sinh t)\in{\cal H}$
+\end_inset
+
+ está en la misma abscisa que 
+\begin_inset Formula $Q:=(a\cosh t,b\cosh t)\in\ell$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $d(P,Q)=b(\cosh t-\sinh t)=be^{-t}$
+\end_inset
+
+, que tiende a 0 cuando 
+\begin_inset Formula $t$
+\end_inset
+
+ tiende a 
+\begin_inset Formula $+\infty$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $d(\ell,{\cal H})=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\iff y=\pm b\sqrt{\frac{x^{2}}{a^{2}}-1}=\pm\frac{b}{a}\sqrt{x^{2}-a^{2}}$
+\end_inset
+
+.
+ Así, una recta de la forma 
+\begin_inset Formula $\ell\equiv x=r$
+\end_inset
+
+ intersecará con 
+\begin_inset Formula ${\cal H}$
+\end_inset
+
+ en 
+\begin_inset Formula $(r,\pm\sqrt{r^{2}-a^{2}})$
+\end_inset
+
+ si 
+\begin_inset Formula $|r|\geq|a|$
+\end_inset
+
+.
+ De lo contrario, observamos que todo punto 
+\begin_inset Formula $P(x,y)\in{\cal H}$
+\end_inset
+
+ cumple 
+\begin_inset Formula $|x|\geq a$
+\end_inset
+
+, luego 
+\begin_inset Formula $d(P,\ell)^{2}=(x-r)^{2}$
+\end_inset
+
+, pero como 
+\begin_inset Formula $|x|\geq|a|>|r|$
+\end_inset
+
+, entonces 
+\begin_inset Formula $|x|\neq|r|$
+\end_inset
+
+, 
+\begin_inset Formula $x\neq r$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $(x-r)^{2}>0$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Si 
+\begin_inset Formula $\ell\equiv y=mx+n$
+\end_inset
+
+ para ciertos 
+\begin_inset Formula $m,n\in\mathbb{R}$
+\end_inset
+
+, vemos que para que sea 
+\begin_inset Formula $d(\ell,{\cal H})=0$
+\end_inset
+
+ pero 
+\begin_inset Formula $\ell\cap{\cal H}=\emptyset$
+\end_inset
+
+, la distancia 0 debe tenerse como un límite.
+ De lo contrario, dada la función 
+\begin_inset Formula $h:\mathbb{R}\rightarrow\mathbb{R}$
+\end_inset
+
+ definida por 
+\begin_inset Formula $h(t):=d((a\cosh t,b\sinh t),\ell)$
+\end_inset
+
+, debería haber un 
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+ con 
+\begin_inset Formula $\lim_{t\rightarrow c}h(c)=0$
+\end_inset
+
+, pero por ser 
+\begin_inset Formula $h$
+\end_inset
+
+ continua se tendría 
+\begin_inset Formula $h(c)=0$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $d(c,\ell)=0$
+\end_inset
+
+ y si ahora definimos 
+\begin_inset Formula $g_{c}:\mathbb{R}\rightarrow\mathbb{R}$
+\end_inset
+
+ como 
+\begin_inset Formula $g_{c}(t):=d((a\cosh c,b\sinh c),mt+n)$
+\end_inset
+
+, por el mismo argumento existiría un 
+\begin_inset Formula $d\in\mathbb{R}$
+\end_inset
+
+ con 
+\begin_inset Formula $d((\cosh c,\sinh c),md+n)=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\ell\cap{\cal H}\neq\emptyset\#$
+\end_inset
+
+.
+ 
+\begin_inset Newline newline
+\end_inset
+
+Centrémonos ahora en el 
+\begin_inset Quotes fld
+\end_inset
+
+hemisferio norte
+\begin_inset Quotes frd
+\end_inset
+
+ de la hipérbola (
+\begin_inset Formula $\{(x,y)\in{\cal H}:y\geq0\}$
+\end_inset
+
+), dado por 
+\begin_inset Formula $y=\frac{b}{a}\sqrt{x^{2}-a^{2}}$
+\end_inset
+
+.
+ Si definimos la función 
+\begin_inset Formula $f:(-\infty,-a]\cup[a,+\infty)\rightarrow\mathbb{R}$
+\end_inset
+
+ como 
+\begin_inset Formula $f(x):=mx+n-\frac{b}{a}\sqrt{x^{2}-a^{2}}$
+\end_inset
+
+, tenemos que 
+\begin_inset Formula $\lim_{x\rightarrow+\infty}f(x)$
+\end_inset
+
+ ó 
+\begin_inset Formula $\lim_{x\rightarrow-\infty}f(x)$
+\end_inset
+
+ debe ser 0 para que 
+\begin_inset Formula $\ell$
+\end_inset
+
+ sea una asíntota en el hemisferio norte de 
+\begin_inset Formula ${\cal H}$
+\end_inset
+
+.
+ Ahora bien, 
+\begin_inset Formula $\lim_{x\rightarrow+\infty}mx+n-\frac{b}{a}\sqrt{x^{2}-a^{2}}$
+\end_inset
+
+ converge si y sólo si 
+\begin_inset Formula $m=\frac{b}{a}$
+\end_inset
+
+, y en este caso converge a 
+\begin_inset Formula $n$
+\end_inset
+
+, por lo que debe ser 
+\begin_inset Formula $m=\frac{b}{a}$
+\end_inset
+
+ y 
+\begin_inset Formula $n=0$
+\end_inset
+
+.
+ Para el 
+\begin_inset Formula $\lim_{x\rightarrow-\infty}f(x)$
+\end_inset
+
+ nos encontramos con lo mismo pero con 
+\begin_inset Formula $m=-\frac{b}{a}$
+\end_inset
+
+.
+ El hemisferio sur se hace de forma análoga, tomando 
+\begin_inset Formula $\hat{f}(x):=mx+n+\frac{b}{a}\sqrt{x^{2}-a^{2}}$
+\end_inset
+
+, y las condiciones que deben cumplir 
+\begin_inset Formula $m$
+\end_inset
+
+ y 
+\begin_inset Formula $n$
+\end_inset
+
+ son las mismas.
+\end_layout
+
+\begin_layout Standard
+La 
+\series bold
+propiedad focal de la hipérbola
+\series default
+ afirma que dado un punto 
+\begin_inset Formula $P$
+\end_inset
+
+ de una hipérbola de focos 
+\begin_inset Formula $F$
+\end_inset
+
+ y 
+\begin_inset Formula $F'$
+\end_inset
+
+, la recta bisectriz del ángulo entre 
+\begin_inset Formula $-\overrightarrow{PF}$
+\end_inset
+
+ y 
+\begin_inset Formula $\overrightarrow{PF'}$
+\end_inset
+
+ es tangente a la elipse en 
+\begin_inset Formula $P$
+\end_inset
+
+.
+ 
+\series bold
+Demostración
+\series default
+: Sea 
+\begin_inset Formula $\ell$
+\end_inset
+
+ dicha recta y 
+\begin_inset Formula $E:=s_{\ell}(F)$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $E\in PF'$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\Vert\overrightarrow{EF'}\Vert=|\Vert\overrightarrow{PF'}\Vert-\Vert\overrightarrow{PE}\Vert|=|\Vert\overrightarrow{PF'}\Vert-\Vert\overrightarrow{PF}\Vert|=2a$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $P\neq P'\in\ell$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\Vert\overrightarrow{P'E}\Vert<\Vert\overrightarrow{P'F'}\Vert+\Vert\overrightarrow{EF'}\Vert<\Vert\overrightarrow{P'E}\Vert+2\Vert\overrightarrow{EF'}\Vert$
+\end_inset
+
+, por lo que restando 
+\begin_inset Formula $\Vert\overrightarrow{P'E}\Vert+\Vert\overrightarrow{EF'}\Vert$
+\end_inset
+
+, nos queda 
+\begin_inset Formula $-\Vert\overrightarrow{EF'}\Vert<\Vert\overrightarrow{P'F'}\Vert-\Vert\overrightarrow{P'E}\Vert<\Vert\overrightarrow{EF'}\Vert$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $|\Vert\overrightarrow{P'F'}\Vert-\Vert\overrightarrow{P'E}\Vert|<\Vert\overrightarrow{EF'}\Vert=2a$
+\end_inset
+
+ y 
+\begin_inset Formula $P'$
+\end_inset
+
+ no está en la hipérbola.
+ Queda ver que 
+\begin_inset Formula $\ell$
+\end_inset
+
+ no es asintótica.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Comment
+status open
+
+\begin_layout Plain Layout
+ Para ello vemos que la hipérbola divide al plano en 3 regiones abiertas
+ conexas y definimos 
+\begin_inset Formula $f(Q):=\Vert\overrightarrow{QF}\Vert-\Vert\overrightarrow{QF'}\Vert$
+\end_inset
+
+.
+ Se tiene que 
+\begin_inset Formula $f(Q)=\pm2a$
+\end_inset
+
+ en los puntos de la hipérbola, 
+\begin_inset Formula $f(Q)<-2a$
+\end_inset
+
+ en la región que contiene un foco, 
+\begin_inset Formula $f(Q)>2a$
+\end_inset
+
+ en la región del otro y 
+\begin_inset Formula $|f(Q)|<2a$
+\end_inset
+
+ en el medio.
+ Hemos visto que en 
+\begin_inset Formula $\ell$
+\end_inset
+
+ se tiene 
+\begin_inset Formula $-2a\leq f\leq2a$
+\end_inset
+
+, luego 
+\begin_inset Formula $\ell$
+\end_inset
+
+ nunca cruza a ninguna de las regiones que contiene un foco y por tanto
+ no puede ser una recta asintótica
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Habría que demostrar esto último.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+La recta tangente a la hipérbola 
+\begin_inset Formula ${\cal H}\equiv\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
+\end_inset
+
+ en el punto 
+\begin_inset Formula $P(x_{0},y_{0})$
+\end_inset
+
+ es 
+\begin_inset Formula $\ell\equiv\frac{x_{0}x}{a^{2}}-\frac{y_{0}y}{b^{2}}=1$
+\end_inset
+
+.
+ 
+\series bold
+Demostración
+\series default
+: Sea
+\begin_inset Formula 
+\[
+\ell\equiv\left\{ \begin{array}{rcl}
+x & = & x_{0}+ut\\
+y & = & y_{0}+vt
+\end{array}\right.
+\]
+
+\end_inset
+
+Los puntos de 
+\begin_inset Formula $\ell$
+\end_inset
+
+ en la hipérbola satisfacen 
+\begin_inset Formula $\frac{(x_{0}+ut)^{2}}{a^{2}}-\frac{(y_{0}+vt)^{2}}{b^{2}}=1$
+\end_inset
+
+, y operando, 
+\begin_inset Formula $\left(\frac{2ux_{0}}{a^{2}}-\frac{2vy_{0}}{b^{2}}\right)t+\left(\frac{u^{2}}{a^{2}}-\frac{v^{2}}{b^{2}}\right)t^{2}=1-\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=0$
+\end_inset
+
+, lo que se cumple para 
+\begin_inset Formula $t\in\left\{ 0,-\frac{2\left(\frac{ux_{0}}{a^{2}}-\frac{vy_{0}}{b^{2}}\right)}{\frac{u^{2}}{a^{2}}-\frac{v^{2}}{b^{2}}}\right\} $
+\end_inset
+
+.
+ Estos dos valores son iguales si y sólo si 
+\begin_inset Formula $\frac{ux_{0}}{a^{2}}=\frac{vy_{0}}{b^{2}}$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $\frac{x_{0}}{a^{2}}(x-x_{0})=\frac{y_{0}}{b^{2}}(y-y_{0})$
+\end_inset
+
+ y 
+\begin_inset Formula $\frac{x_{0}x}{a^{2}}-\frac{y_{0}y}{b^{2}}=\frac{x_{0}^{2}}{a^{2}}-\frac{y_{0}^{2}}{b^{2}}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+La parábola
+\end_layout
+
+\begin_layout Standard
+Una parábola es el lugar de los puntos del plano que equidistan de un punto
+ llamado 
+\series bold
+foco
+\series default
+ (
+\begin_inset Formula $F$
+\end_inset
+
+), y una recta llamada 
+\series bold
+directriz
+\series default
+ (
+\begin_inset Formula $l$
+\end_inset
+
+).
+ La perpendicular a 
+\begin_inset Formula $l$
+\end_inset
+
+ por 
+\begin_inset Formula $F$
+\end_inset
+
+ es el 
+\series bold
+eje
+\series default
+ (
+\series bold
+principal
+\series default
+) de la parábola y el punto en que la parábola interseca con el eje es el
+ 
+\series bold
+vértice
+\series default
+.
+\end_layout
+
+\begin_layout Standard
+Una ecuación reducida de la parábola con 
+\begin_inset Formula $d(F,l)=:p$
+\end_inset
+
+ es
+\begin_inset Formula 
+\[
+y^{2}=2px
+\]
+
+\end_inset
+
+En efecto, si tomamos un referencial ortonormal en el que 
+\begin_inset Formula $x=0$
+\end_inset
+
+ sea el eje de la parábola, el origen sea el vértice, 
+\begin_inset Formula $F=(\frac{p}{2},0)$
+\end_inset
+
+ y 
+\begin_inset Formula $l\equiv x=-\frac{p}{2}$
+\end_inset
+
+, sea 
+\begin_inset Formula $P(x,y)$
+\end_inset
+
+ un punto 
+\begin_inset Quotes fld
+\end_inset
+
+genérico
+\begin_inset Quotes frd
+\end_inset
+
+ de la parábola, entonces 
+\begin_inset Formula $\sqrt{(x-\frac{p}{2})^{2}+y^{2}}=\Vert\overrightarrow{PF}\Vert=d(P,l)=x+\frac{p}{2}$
+\end_inset
+
+, luego 
+\begin_inset Formula $x^{2}-px+\frac{p^{2}}{4}+y^{2}=x^{2}+px+\frac{p^{2}}{4}$
+\end_inset
+
+ e 
+\begin_inset Formula $y^{2}=2px$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Unas ecuaciones paramétricas son
+\begin_inset Formula 
+\begin{eqnarray*}
+\left\{ \begin{array}{rcl}
+x & = & \frac{t^{2}}{2p}\\
+y & = & t
+\end{array}\right. & \text{ con } & t\in\mathbb{R}
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Una recta es 
+\series bold
+tangente
+\series default
+ a una parábola si la corta en un solo punto sin ser paralela al eje.
+ La 
+\series bold
+propiedad focal de la parábola
+\series default
+ afirma que si 
+\begin_inset Formula $P$
+\end_inset
+
+ es un punto de la parábola de directriz 
+\begin_inset Formula $l$
+\end_inset
+
+ y foco 
+\begin_inset Formula $F$
+\end_inset
+
+ y 
+\begin_inset Formula $A$
+\end_inset
+
+ es la intersección de 
+\begin_inset Formula $l$
+\end_inset
+
+ con su perpendicular por 
+\begin_inset Formula $P$
+\end_inset
+
+ entonces la recta bisectriz del ángulo entre 
+\begin_inset Formula $\overrightarrow{PF}$
+\end_inset
+
+ y 
+\begin_inset Formula $\overrightarrow{PA}$
+\end_inset
+
+ es tangente a la parábola en 
+\begin_inset Formula $P$
+\end_inset
+
+.
+ 
+\series bold
+Demostración
+\series default
+: Sea 
+\begin_inset Formula $r$
+\end_inset
+
+ la bisectriz y 
+\begin_inset Formula $P\neq P'\in r$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\Vert\overrightarrow{P'F}\Vert=\Vert\overrightarrow{P'A}\Vert>d(P',l)$
+\end_inset
+
+, luego 
+\begin_inset Formula $P'$
+\end_inset
+
+ no está en la parábola.
+ Queda ver que 
+\begin_inset Formula $r$
+\end_inset
+
+ no es paralela al eje.
+ Si lo fuese, el ángulo entre 
+\begin_inset Formula $\overrightarrow{PA}$
+\end_inset
+
+ y 
+\begin_inset Formula $r$
+\end_inset
+
+ sería 0 y por tanto también lo sería aquel entre 
+\begin_inset Formula $\overrightarrow{PA}$
+\end_inset
+
+ y 
+\begin_inset Formula $\overrightarrow{PF}$
+\end_inset
+
+ con lo que 
+\begin_inset Formula $\overrightarrow{PA}=\lambda\overrightarrow{PF}$
+\end_inset
+
+ para cierto 
+\begin_inset Formula $\lambda>0$
+\end_inset
+
+, que debe ser 1 porque 
+\begin_inset Formula $\Vert\overrightarrow{PF}\Vert=\Vert\overrightarrow{PA}\Vert=|\lambda|\Vert\overrightarrow{PF}\Vert$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $F=A\in l\#$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+La recta tangente a la parábola 
+\begin_inset Formula $y^{2}=2px$
+\end_inset
+
+ en 
+\begin_inset Formula $P(x_{0},y_{0})$
+\end_inset
+
+ es 
+\begin_inset Formula $\ell\equiv y_{0}y-px=px_{0}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración
+\series default
+: Sea 
+\begin_inset Formula 
+\[
+\ell\equiv\left\{ \begin{array}{rcl}
+x & = & x_{0}+ut\\
+y & = & y_{0}+vt
+\end{array}\right.
+\]
+
+\end_inset
+
+Los puntos de 
+\begin_inset Formula $\ell$
+\end_inset
+
+ en la parábola satisfacen 
+\begin_inset Formula $(y_{0}+vt)^{2}=2p(x_{0}+ut)$
+\end_inset
+
+, y operando, 
+\begin_inset Formula $(2y_{0}v-2pu)t+v^{2}t^{2}=2px_{0}-y_{0}^{2}=0$
+\end_inset
+
+, lo que se cumple para 
+\begin_inset Formula $t\in\left\{ 0,\frac{2(pu-y_{0}v)}{v^{2}}\right\} $
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $v=0$
+\end_inset
+
+, la recta es paralela al eje y no tangente; de lo contrario los dos valores
+ son iguales si y sólo si 
+\begin_inset Formula $pu=y_{0}v$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $p(x-x_{0})=y_{0}(y-y_{0})$
+\end_inset
+
+ e 
+\begin_inset Formula $y_{0}y-px=y_{0}^{2}-px_{0}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Definición alternativa de las cónicas
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ es una cónica no degenerada distinta de una circunferencia si y sólo si
+ existen 
+\begin_inset Formula $\epsilon>0$
+\end_inset
+
+, una recta 
+\begin_inset Formula $\ell$
+\end_inset
+
+ y un punto 
+\begin_inset Formula $F\notin\ell$
+\end_inset
+
+ tales que 
+\begin_inset Formula $\frac{d(P,F)}{d(P,\ell)}=\epsilon$
+\end_inset
+
+, en cuyo caso 
+\begin_inset Formula $\epsilon<1$
+\end_inset
+
+ para una elipse, 
+\begin_inset Formula $\epsilon=1$
+\end_inset
+
+ para una parábola y 
+\begin_inset Formula $\epsilon>1$
+\end_inset
+
+ para una hipérbola.
+ Llamamos a 
+\begin_inset Formula $\ell$
+\end_inset
+
+ la 
+\series bold
+directriz del foco
+\series default
+ 
+\begin_inset Formula $F$
+\end_inset
+
+ y a 
+\begin_inset Formula $p:=d(F,\ell)$
+\end_inset
+
+ el 
+\series bold
+parámetro focal
+\series default
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $O$
+\end_inset
+
+ el punto de intersección entre 
+\begin_inset Formula $\ell$
+\end_inset
+
+ y su perpendicular por 
+\begin_inset Formula $F$
+\end_inset
+
+, y tomamos un referencial ortonormal con origen 
+\begin_inset Formula $O$
+\end_inset
+
+, 
+\begin_inset Formula $\ell\equiv x=0$
+\end_inset
+
+ y 
+\begin_inset Formula $F=(p,0)$
+\end_inset
+
+ siendo 
+\begin_inset Formula $p>0$
+\end_inset
+
+ la distancia focal.
+ Vemos que 
+\begin_inset Formula $\frac{d(P,F)}{d(P,\ell)}=\epsilon\iff(x-p)^{2}+y^{2}=\epsilon^{2}x^{2}\iff(1-\epsilon^{2})x^{2}+y^{2}-2px+p^{2}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Si 
+\begin_inset Formula $\epsilon<1$
+\end_inset
+
+, entonces 
+\begin_inset Formula $1-\epsilon^{2}>0$
+\end_inset
+
+ y
+\begin_inset Formula 
+\begin{multline*}
+0=(1-\epsilon)^{2}\left(x^{2}-\frac{2p}{1-\epsilon^{2}}x\right)+y^{2}+p^{2}=\\
+=(1-\epsilon)^{2}\left(\left(x-\frac{p}{1-\epsilon^{2}}\right)^{2}-\frac{p^{2}}{(1-\epsilon^{2})^{2}}\right)+y^{2}+p^{2}\implies\\
+\implies(1-\epsilon^{2})\left(x-\frac{p}{1-\epsilon^{2}}\right)^{2}+y^{2}=\frac{p^{2}}{1-\epsilon^{2}}-p^{2}=\frac{p^{2}\epsilon^{2}}{1-\epsilon^{2}}\implies\\
+\left.\stackrel[y':=y]{x':=x-\frac{p}{1-\epsilon^{2}}}{}\right\} \implies(1-\epsilon^{2})x'^{2}+y'^{2}=\frac{p^{2}\epsilon^{2}}{1-\epsilon^{2}}\implies\frac{(1-\epsilon^{2})^{2}}{\epsilon^{2}p^{2}}x'^{2}+\frac{1-\epsilon^{2}}{\epsilon^{2}p^{2}}y'^{2}=1\implies\\
+\left.\stackrel[b:=\frac{\epsilon p}{\sqrt{1-\epsilon^{2}}}]{a:=\frac{\epsilon p}{1-\epsilon^{2}}}{}\right\} \implies\frac{x'^{2}}{a^{2}}+\frac{y'^{2}}{b^{2}}=1
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $\epsilon=1$
+\end_inset
+
+, nos queda 
+\begin_inset Formula $y^{2}-2px+p^{2}=0\iff y^{2}=2p(x-\frac{p}{2})\stackrel[y':=y]{x':=x-\frac{p}{2}}{\implies}y'^{2}=2px'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $\epsilon>1$
+\end_inset
+
+, cambiando el signo a la ecuación de arriba nos queda 
+\begin_inset Formula $(\epsilon^{2}-1)x^{2}-y^{2}+2px-p^{2}=0$
+\end_inset
+
+ con 
+\begin_inset Formula $\epsilon^{2}-1>0$
+\end_inset
+
+, luego
+\begin_inset Formula 
+\begin{multline*}
+(\epsilon^{2}-1)\left(x+\frac{p}{\epsilon^{2}-1}\right)^{2}-y^{2}=p^{2}-\frac{p^{2}}{\epsilon^{2}-1}=\frac{p^{2}\epsilon^{2}}{\epsilon^{2}-1}\implies\\
+\left.\stackrel[y':=y]{x':=x+\frac{p}{\epsilon^{2}-1}}{}\right\} \implies(\epsilon^{2}-1)x'^{2}-y'^{2}=\frac{p^{2}\epsilon^{2}}{\epsilon^{2}-1}\implies\frac{(\epsilon^{2}-1)^{2}}{p^{2}\epsilon^{2}}x'^{2}-\frac{\epsilon^{2}-1}{p^{2}\epsilon^{2}}y'^{2}=1\implies\\
+\left.\stackrel[b:=\frac{\epsilon p}{\sqrt{\epsilon^{2}-1}}]{a:=\frac{\epsilon p}{\epsilon^{2}-1}}{}\right\} \implies\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
+\end{multline*}
+
+\end_inset
+
+ 
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Las cuentas son aproximadamente las de la otra implicación pero al revés.
+ Así, para una elipse 
+\begin_inset Formula $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
+\end_inset
+
+ o una hipérbola 
+\begin_inset Formula $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
+\end_inset
+
+, tomamos 
+\begin_inset Formula $\epsilon=\frac{c}{a}$
+\end_inset
+
+, 
+\begin_inset Formula $F=(c,0)$
+\end_inset
+
+ y 
+\begin_inset Formula $\ell\equiv x=\frac{a^{2}}{c}$
+\end_inset
+
+, mientras que para una parábola 
+\begin_inset Formula $y^{2}=2px$
+\end_inset
+
+, tomamos 
+\begin_inset Formula $\epsilon=1$
+\end_inset
+
+, 
+\begin_inset Formula $F=(\frac{p}{2},0)$
+\end_inset
+
+ y 
+\begin_inset Formula $\ell\equiv x=-\frac{p}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $F=(p,0)$
+\end_inset
+
+ y 
+\begin_inset Formula $\ell\equiv x=0$
+\end_inset
+
+, la distancia entre 
+\begin_inset Formula $F$
+\end_inset
+
+ y 
+\begin_inset Formula $Q:=(p,p\epsilon)$
+\end_inset
+
+, 
+\begin_inset Formula $\lambda:=d(F,Q)=p\epsilon$
+\end_inset
+
+, se llama 
+\series bold
+semilado recto
+\series default
+ de la cónica.
+ La ecuación de una cónica de excentricidad 
+\begin_inset Formula $\epsilon$
+\end_inset
+
+, foco 
+\begin_inset Formula $F=(s,t)$
+\end_inset
+
+ y directriz 
+\begin_inset Formula $\ell\equiv ux+vy+w=0$
+\end_inset
+
+ se puede escribir como
+\begin_inset Formula 
+\[
+(x-s)^{2}+(y-t)^{2}=(lx+my+n)^{2}
+\]
+
+\end_inset
+
+ con 
+\begin_inset Formula $k:=\frac{\epsilon}{\sqrt{u^{2}+v^{2}}}$
+\end_inset
+
+, 
+\begin_inset Formula $l:=ku$
+\end_inset
+
+, 
+\begin_inset Formula $m:=kv$
+\end_inset
+
+ y 
+\begin_inset Formula $n:=kw$
+\end_inset
+
+, la 
+\series bold
+ecuación focal de la cónica
+\series default
+, pues 
+\begin_inset Formula 
+\begin{multline*}
+\frac{d(P,F)}{d(P,\ell)}=\epsilon\iff d(P,F)^{2}=\epsilon^{2}d(P,\ell)^{2}\iff\\
+\iff(x-s)^{2}+(y-t)^{2}=\epsilon^{2}\frac{(ux+vy+w)^{2}}{u^{2}+v^{2}}=(lx+my+n)^{2}
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+sremember{GAE}
+\end_layout
+
+\end_inset
+
+La distancia de un punto 
+\begin_inset Formula $Q=(q_{1},\dots,q_{n})$
+\end_inset
+
+ a un hiperplano 
+\begin_inset Formula ${\cal H}$
+\end_inset
+
+ de ecuación 
+\begin_inset Formula $a_{1}x_{1}+\dots+a_{n}x_{n}+b=0$
+\end_inset
+
+ es 
+\begin_inset Formula $d(Q,{\cal H})=\frac{|a_{1}q_{1}+\dots+a_{n}q_{n}+b|}{\Vert(a_{1},\dots,a_{n})\Vert}$
+\end_inset
+
+.
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+eremember
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/aalg/n2.lyx b/aalg/n2.lyx
new file mode 100644
index 0000000..51dd686
--- /dev/null
+++ b/aalg/n2.lyx
@@ -0,0 +1,1570 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\usepackage{tikz}
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style french
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Fijado un referencial ortonormal en un plano afín euclídeo, llamamos 
+\series bold
+cónica
+\series default
+ al conjunto de puntos 
+\begin_inset Formula $(x,y)$
+\end_inset
+
+ con ecuación 
+\begin_inset Formula 
+\[
+ax^{2}+2bxy+cy^{2}+2ex+2fy+d=0
+\]
+
+\end_inset
+
+ donde al menos uno de los valores 
+\begin_inset Formula $a$
+\end_inset
+
+, 
+\begin_inset Formula $b$
+\end_inset
+
+ o 
+\begin_inset Formula $c$
+\end_inset
+
+ no es nulo.
+ Distintas ecuaciones de este tipo pueden definir la misma cónica, como
+ múltiplos de esta por 
+\begin_inset Formula $\lambda\neq0$
+\end_inset
+
+, o las que dan lugar a la cónica vacía.
+ Esta ecuación se puede expresar como
+\begin_inset Formula 
+\[
+\left(\begin{array}{ccc}
+x & y & 1\end{array}\right)\left(\begin{array}{cc|c}
+a & b & e\\
+b & c & f\\
+\hline e & f & d
+\end{array}\right)\left(\begin{array}{c}
+x\\
+y\\
+1
+\end{array}\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+matriz 
+\series default
+(
+\series bold
+proyectiva
+\series default
+)
+\series bold
+ de la cónica
+\series default
+ a 
+\begin_inset Formula 
+\[
+\overline{A}=\left(\begin{array}{ccc}
+a & b & e\\
+b & c & f\\
+e & f & d
+\end{array}\right)
+\]
+
+\end_inset
+
+y 
+\series bold
+matriz principal de la cónica
+\series default
+ a 
+\begin_inset Formula 
+\[
+A=\left(\begin{array}{cc}
+a & b\\
+b & c
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula 
+\begin{align*}
+\overline{A} & =\left(\begin{array}{c|c}
+A & B\\
+\hline B^{t} & d
+\end{array}\right) & B & =\left(\begin{array}{c}
+e\\
+f
+\end{array}\right) & X & =\left(\begin{array}{c}
+x\\
+y
+\end{array}\right) & \overline{X} & =\left(\begin{array}{c}
+X\\
+\hline 1
+\end{array}\right)
+\end{align*}
+
+\end_inset
+
+podemos expresar la ecuación como 
+\begin_inset Formula $\overline{X}^{t}\overline{A}\overline{X}=0$
+\end_inset
+
+ o como 
+\begin_inset Formula $X^{t}AX+2B^{t}X+d=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Forma reducida
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+sremember{GAE}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Para cambiar coordenadas entre dos referenciales 
+\begin_inset Formula $\Re=(O,{\cal B})$
+\end_inset
+
+ y 
+\begin_inset Formula $\Re'=(O',{\cal B}')$
+\end_inset
+
+ de 
+\begin_inset Formula ${\cal E}$
+\end_inset
+
+, si llamamos 
+\begin_inset Formula $X_{0}:=[O]_{\Re'}=[\overrightarrow{O'O}]_{{\cal B}'}$
+\end_inset
+
+ y 
+\begin_inset Formula $M:=M_{{\cal B}'{\cal B}}$
+\end_inset
+
+, se tiene que:
+\begin_inset Formula 
+\[
+\text{[...]}X'=\text{[...]}=X_{0}+MX
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+eremember
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Podemos emplear la expresión matricial equivalente:
+\begin_inset Formula 
+\[
+\left(\begin{array}{c}
+X'\\
+\hline 1
+\end{array}\right)=\left(\begin{array}{c|c}
+M & X_{0}\\
+\hline 0 & 1
+\end{array}\right)\left(\begin{array}{c}
+X\\
+\hline 1
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+sremember{AlgL}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Los vectores propios de 
+\begin_inset Formula $f$
+\end_inset
+
+ asociados a 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ son todos los vectores no nulos de 
+\begin_inset Formula $\text{Nuc}(f-\lambda Id)$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $V_{\lambda}=\text{Nuc}(f-\lambda Id)=\{v\in V:(f-\lambda Id)(v)=0\}=\{v\in V:f(v)=\lambda v\}$
+\end_inset
+
+ es el 
+\series bold
+subespacio propio
+\series default
+ o 
+\series bold
+característico
+\series default
+ correspondiente al valor propio 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $\lambda\in K$
+\end_inset
+
+ es un valor propio de 
+\begin_inset Formula $f$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\det(f-\lambda Id)=0$
+\end_inset
+
+.
+ [...]
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $P_{f}(x):=\det(xId-f)$
+\end_inset
+
+ es el 
+\series bold
+polinomio característico
+\series default
+ de 
+\series bold
+
+\begin_inset Formula $f$
+\end_inset
+
+
+\series default
+, y 
+\begin_inset Formula $P_{A}(x):=\det(xI_{n}-A)$
+\end_inset
+
+ es el polinomio característico de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Podemos comprobar que
+\begin_inset Formula 
+\[
+P_{A}(x)=x^{n}-\text{tr}(A)x^{n-1}+\dots+(-1)^{n}\det(A)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de diagonalización:
+\series default
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ es diagonalizable si y sólo si
+\begin_inset Formula 
+\[
+P_{f}(x)=(x-\lambda_{1})^{d_{1}}\cdots(x-\lambda_{r})^{d_{r}}
+\]
+
+\end_inset
+
+con 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{r}\in K$
+\end_inset
+
+ distintos dos a dos, y 
+\begin_inset Formula $d_{i}=\dim(\text{Nuc}(\lambda_{i}Id-f))$
+\end_inset
+
+.
+ [...]
+\end_layout
+
+\begin_layout Standard
+Así, para diagonalizar una matriz 
+\begin_inset Formula $A\in M_{n}(K)$
+\end_inset
+
+ en matrices 
+\begin_inset Formula $A=M_{{\cal CB}}DM_{{\cal BC}}$
+\end_inset
+
+, con 
+\begin_inset Formula $D$
+\end_inset
+
+ diagonal, obtenemos su polinomio característico, hallamos sus raíces, que
+ serán los autovalores de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Si la suma de sus multiplicidades da 
+\begin_inset Formula $n$
+\end_inset
+
+, resolvemos cada ecuación 
+\begin_inset Formula $(\lambda Id-f)X=0$
+\end_inset
+
+ para obtener las bases de los subespacios propios, cuya dimensión debería
+ coincidir con la multiplicidad del autovalor si 
+\begin_inset Formula $A$
+\end_inset
+
+ es diagonalizable.
+ Entonces añadimos cada raíz en 
+\begin_inset Formula $D$
+\end_inset
+
+ tantas veces como sea su multiplicidad y razonamos que los vectores correspondi
+entes de la base 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+, y por tanto las correspondientes columnas de 
+\begin_inset Formula $M_{{\cal CB}}$
+\end_inset
+
+, son los de la base de dicho subespacio propio.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+eremember
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $A\in M_{2}(\mathbb{R})$
+\end_inset
+
+ simétrica, existe una matriz ortogonal 
+\begin_inset Formula $Q$
+\end_inset
+
+ de determinante 1 tal que 
+\begin_inset Formula $Q^{t}AQ$
+\end_inset
+
+ es diagonal.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula 
+\[
+A=\left(\begin{array}{cc}
+a & b\\
+b & c
+\end{array}\right)\implies P_{A}(x)=\left|\begin{array}{cc}
+a-x & b\\
+b & c-x
+\end{array}\right|=x^{2}-(a+c)x+(ac-b^{2})
+\]
+
+\end_inset
+
+y el discriminante de 
+\begin_inset Formula $P_{A}(x)=0$
+\end_inset
+
+, 
+\begin_inset Formula $(a-c)^{2}+4b^{2}$
+\end_inset
+
+, es siempre mayor que 0 salvo que 
+\begin_inset Formula $A$
+\end_inset
+
+ ya sea diagonal con 
+\begin_inset Formula $a=c$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $A$
+\end_inset
+
+ tiene dos valores propios distintos y por tanto diagonaliza.
+ Si 
+\begin_inset Formula $u$
+\end_inset
+
+ y 
+\begin_inset Formula $v$
+\end_inset
+
+ son vectores propios de valores propios respectivos 
+\begin_inset Formula $\alpha\neq\beta$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\alpha(u\cdot v)=(\alpha u)\cdot v=f_{A}(u)\cdot v=(Au)^{t}v=u^{t}A^{t}v=u^{t}Av=u\cdot f_{A}(v)=u\cdot\beta v=\beta(u\cdot v)$
+\end_inset
+
+, luego 
+\begin_inset Formula $(\alpha-\beta)(u\cdot v)=0$
+\end_inset
+
+ y como 
+\begin_inset Formula $\alpha\neq\beta$
+\end_inset
+
+ se tiene 
+\begin_inset Formula $u\bot v$
+\end_inset
+
+, luego la base en que diagonaliza 
+\begin_inset Formula $A$
+\end_inset
+
+ se puede escoger ortonormal.
+ Finalmente, si 
+\begin_inset Formula $Q$
+\end_inset
+
+ es la matriz cuyas columnas son estos vectores propios y su determinante
+ es 
+\begin_inset Formula $-1$
+\end_inset
+
+, podemos cambiar el signo de una de las columnas para que el determinante
+ sea 1.
+\end_layout
+
+\begin_layout Standard
+Con esto podemos hacer dos reducciones a cualquier cónica 
+\begin_inset Formula ${\cal G}$
+\end_inset
+
+ y encontrar un referencial ortonormal en que esta tenga ecuación reducida.
+\end_layout
+
+\begin_layout Standard
+Para la primera reducción, sea 
+\begin_inset Formula 
+\[
+\overline{A}=\left(\begin{array}{c|c}
+A & B\\
+\hline B^{t} & d
+\end{array}\right)
+\]
+
+\end_inset
+
+ la matriz de 
+\begin_inset Formula ${\cal G}$
+\end_inset
+
+ en un referencial ortonormal 
+\begin_inset Formula $\Re$
+\end_inset
+
+ y 
+\begin_inset Formula $Q$
+\end_inset
+
+ una matriz ortogonal con 
+\begin_inset Formula $|Q|=1$
+\end_inset
+
+ tal que 
+\begin_inset Formula $Q^{t}AQ=Q^{-1}AQ$
+\end_inset
+
+ sea diagonal.
+ Entonces, si consideramos el referencial 
+\begin_inset Formula $\Re'$
+\end_inset
+
+ tal que
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\begin{eqnarray*}
+\left(\begin{array}{c}
+X\\
+\hline 1
+\end{array}\right)=N\left(\begin{array}{c}
+X'\\
+\hline 1
+\end{array}\right) & \text{con} & N:=\left(\begin{array}{c|c}
+Q & 0\\
+\hline 0 & 1
+\end{array}\right)
+\end{eqnarray*}
+
+\end_inset
+
+la ecuación de 
+\begin_inset Formula ${\cal G}$
+\end_inset
+
+ queda como
+\begin_inset Formula 
+\[
+\left(N\left(\begin{array}{c}
+X'\\
+\hline 1
+\end{array}\right)\right)^{t}\overline{A}N\left(\begin{array}{c}
+X'\\
+\hline 1
+\end{array}\right)=\left(\begin{array}{c|c}
+X'^{t} & 1\end{array}\right)N^{t}\overline{A}N\left(\begin{array}{c}
+X'\\
+\hline 1
+\end{array}\right)=0
+\]
+
+\end_inset
+
+ y la matriz de 
+\begin_inset Formula ${\cal G}$
+\end_inset
+
+ en 
+\begin_inset Formula $\Re'$
+\end_inset
+
+ es
+\begin_inset Formula 
+\[
+\left(\begin{array}{c|c}
+A' & B'\\
+\hline B'^{t} & d'
+\end{array}\right)=N^{t}\overline{A}N=\left(\begin{array}{c|c}
+Q^{t} & 0\\
+\hline 0 & 1
+\end{array}\right)\left(\begin{array}{c|c}
+A & B\\
+\hline B^{t} & d
+\end{array}\right)\left(\begin{array}{c|c}
+Q & 0\\
+\hline 0 & 1
+\end{array}\right)=\left(\begin{array}{c|c}
+Q^{t}AQ & Q^{t}B\\
+\hline B^{t}Q & d
+\end{array}\right)
+\]
+
+\end_inset
+
+luego 
+\begin_inset Formula $A'=Q^{t}AQ$
+\end_inset
+
+, 
+\begin_inset Formula $B'=Q^{t}B$
+\end_inset
+
+ y 
+\begin_inset Formula $d'=d$
+\end_inset
+
+ y el término 
+\begin_inset Formula $xy$
+\end_inset
+
+ se anula en la ecuación de 
+\begin_inset Formula ${\cal G}$
+\end_inset
+
+, lo que nos deja con 
+\begin_inset Formula 
+\[
+\lambda_{1}x'^{2}+\lambda_{2}y'^{2}+2mx'+2ny'+d=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Este cambio es solo vectorial, pues no modifica el origen de coordenadas,
+ y como 
+\begin_inset Formula $|Q|=1$
+\end_inset
+
+, se trata de un giro.
+ Para la segunda reducción, sea 
+\begin_inset Formula $\delta:=\lambda_{1}\lambda_{2}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $\delta>0$
+\end_inset
+
+, podemos suponer 
+\begin_inset Formula $\lambda_{1},\lambda_{2}>0$
+\end_inset
+
+ (de lo contrario cambiamos de signo la ecuación), y completando cuadrados
+ tenemos que 
+\begin_inset Formula $\lambda_{1}x'^{2}+2mx'=\lambda_{1}(x'+\frac{m}{\lambda_{1}})^{2}-\frac{m^{2}}{\lambda_{1}}$
+\end_inset
+
+ y 
+\begin_inset Formula $\lambda_{2}y'^{2}+2ny=\lambda_{2}(y+\frac{n}{\lambda_{2}})^{2}-\frac{n^{2}}{\lambda_{2}}$
+\end_inset
+
+.
+ Nos queda entonces 
+\begin_inset Formula $\lambda_{1}(x'+\frac{m}{\lambda_{1}})^{2}+\lambda_{2}(y'+\frac{n}{\lambda_{2}})^{2}-\frac{m^{2}}{\lambda_{1}}-\frac{n^{2}}{\lambda_{2}}+d=0$
+\end_inset
+
+ y, haciendo la traslación de vector 
+\begin_inset Formula $(\frac{m}{\lambda_{1}},\frac{n}{\lambda_{2}})$
+\end_inset
+
+, nos queda 
+\begin_inset Formula $\lambda_{1}x''^{2}+\lambda_{2}y''^{2}=q$
+\end_inset
+
+, lo que nos deja con una cónica de 
+\series bold
+tipo elíptico
+\series default
+.
+ Si 
+\begin_inset Formula $q>0$
+\end_inset
+
+ es una 
+\series bold
+elipse real
+\series default
+, si 
+\begin_inset Formula $q=0$
+\end_inset
+
+ es un 
+\series bold
+punto
+\series default
+ y si 
+\begin_inset Formula $q<0$
+\end_inset
+
+ es una 
+\series bold
+elipse imaginaria
+\series default
+.
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $\delta<0$
+\end_inset
+
+, por el mismo procedimiento llegamos a que 
+\begin_inset Formula $\lambda_{1}x''^{2}+\lambda_{2}y''^{2}=:q$
+\end_inset
+
+ y, como 
+\begin_inset Formula $\lambda_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $\lambda_{2}$
+\end_inset
+
+ tienen signos opuestos, la ecuación es de 
+\series bold
+tipo hiperbólico
+\series default
+.
+ Si 
+\begin_inset Formula $q=0$
+\end_inset
+
+ tenemos un 
+\series bold
+par de rectas que se cortan
+\series default
+, dadas por 
+\begin_inset Formula $y''=\pm\sqrt{-\frac{\lambda_{1}}{\lambda_{2}}}x''$
+\end_inset
+
+; de lo contrario es una 
+\series bold
+hipérbola
+\series default
+.
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $\delta=0$
+\end_inset
+
+, podemos suponer 
+\begin_inset Formula $\lambda_{1}=0$
+\end_inset
+
+ y 
+\begin_inset Formula $\lambda_{2}\neq0$
+\end_inset
+
+ (no pueden ser ambos 0 porque entonces sería 
+\begin_inset Formula $A=0$
+\end_inset
+
+).
+ Nos queda entonces que 
+\begin_inset Formula $\lambda_{2}y'^{2}+2mx'+2ny'+d=0$
+\end_inset
+
+ y, completando cuadrados, que 
+\begin_inset Formula $\lambda_{2}(y'+\frac{n}{\lambda_{2}})^{2}-\frac{n^{2}}{\lambda_{2}}+2mx'+d=0$
+\end_inset
+
+, una ecuación de 
+\series bold
+tipo parabólico
+\series default
+.
+ Si 
+\begin_inset Formula $m\neq0$
+\end_inset
+
+ podemos escribir la ecuación como 
+\begin_inset Formula $\lambda_{2}(y'+\frac{n}{\lambda_{2}})^{2}+2m(x'-\frac{n^{2}}{2m\lambda_{2}}+\frac{d}{2m})=0$
+\end_inset
+
+, y la traslación de vector 
+\begin_inset Formula $(\frac{d}{2m}-\frac{n^{2}}{2m\lambda_{2}},\frac{n}{\lambda_{2}})$
+\end_inset
+
+ nos lleva la ecuación a 
+\begin_inset Formula $\lambda_{2}y''^{2}+2mx''=0$
+\end_inset
+
+, y tenemos una parábola.
+ Si 
+\begin_inset Formula $m=0$
+\end_inset
+
+, nos queda 
+\begin_inset Formula $\lambda_{2}(y'+\frac{n}{\lambda_{2}})^{2}-\frac{n^{2}}{\lambda_{2}}+d=0$
+\end_inset
+
+ y la traslación de vector 
+\begin_inset Formula $(0,\frac{n}{\lambda_{2}})$
+\end_inset
+
+ nos lleva la ecuación a 
+\begin_inset Formula $\lambda_{2}y''^{2}=q$
+\end_inset
+
+, con lo que tenemos 
+\series bold
+dos rectas paralelas
+\series default
+ si 
+\begin_inset Formula $\frac{q}{\lambda_{2}}>0$
+\end_inset
+
+, una 
+\series bold
+recta doble
+\series default
+ si 
+\begin_inset Formula $q=0$
+\end_inset
+
+ o 
+\series bold
+dos rectas paralelas imaginarias
+\series default
+ si 
+\begin_inset Formula $\frac{q}{\lambda_{2}}<0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Nótese que la ecuación reducida obtenida no es exactamente como las que
+ vimos en el tema anterior para las cónicas no degeneradas.
+ Para obtener estas dividiríamos entre 
+\begin_inset Formula $q$
+\end_inset
+
+ para 
+\begin_inset Formula $\delta\neq0$
+\end_inset
+
+ o entre 
+\begin_inset Formula $\lambda_{2}$
+\end_inset
+
+ para 
+\begin_inset Formula $\delta=0$
+\end_inset
+
+, intercambiaríamos coordenadas si fuera necesario (negando una de las dos
+ para que el cambio sea ortonormal) y, para el caso de la parábola, la giraríamo
+s 
+\begin_inset Formula $\unit[180]{\mathring{}}$
+\end_inset
+
+ en su caso.
+\end_layout
+
+\begin_layout Section
+Invariantes métricos
+\end_layout
+
+\begin_layout Standard
+Dada una cónica con matriz proyectiva 
+\begin_inset Formula $\overline{A}$
+\end_inset
+
+ y matriz principal 
+\begin_inset Formula $A$
+\end_inset
+
+, las cantidades 
+\begin_inset Formula $\Delta:=|\overline{A}|$
+\end_inset
+
+, 
+\begin_inset Formula $\delta:=|A|$
+\end_inset
+
+ y 
+\begin_inset Formula $s:=\text{tr}(A)$
+\end_inset
+
+, llamadas 
+\series bold
+invariantes métricos de la cónica
+\series default
+, se mantienen invariantes al cambiar a otro referencial ortonormal.
+ 
+\series bold
+Demostración:
+\series default
+ Consideremos el cambio de referencial dado por
+\begin_inset Formula 
+\begin{eqnarray*}
+\left(\begin{array}{c}
+X\\
+\hline 1
+\end{array}\right)=N\left(\begin{array}{c}
+X'\\
+\hline 1
+\end{array}\right) & \text{con} & N:=\left(\begin{array}{c|c}
+Q & R\\
+\hline 0 & 1
+\end{array}\right)
+\end{eqnarray*}
+
+\end_inset
+
+con 
+\begin_inset Formula $Q$
+\end_inset
+
+ ortogonal.
+ Entonces la matriz de la cónica en la nueva referencia es 
+\begin_inset Formula $N^{t}\overline{A}N$
+\end_inset
+
+ y la matriz principal es 
+\begin_inset Formula $Q^{t}AQ$
+\end_inset
+
+, y como 
+\begin_inset Formula $|N|=|Q|$
+\end_inset
+
+ y 
+\begin_inset Formula $Q^{t}=Q^{-1}$
+\end_inset
+
+, se tiene 
+\begin_inset Formula $|N^{t}\overline{A}N|=|Q^{t}\overline{A}Q|=|\overline{A}|$
+\end_inset
+
+, 
+\begin_inset Formula $|Q^{t}AQ|=|A|$
+\end_inset
+
+ y 
+\begin_inset Formula $\text{tr}(Q^{t}AQ)=\text{tr}(A)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\align center
+\begin_inset Tabular
+<lyxtabular version="3" rows="4" columns="3">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="middle" width="24text%">
+<column alignment="center" valignment="middle" width="33text%">
+<column alignment="center" valignment="middle" width="33text%">
+<row>
+<cell alignment="center" valignment="top" bottomline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\Delta\neq0$
+\end_inset
+
+: No degenerada
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\Delta=0$
+\end_inset
+
+: Degenerada
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\delta>0$
+\end_inset
+
+: Ecuación elíptica
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Elipse, imaginaria si 
+\begin_inset Formula $s\Delta>0$
+\end_inset
+
+ o real si 
+\begin_inset Formula $s\Delta<0$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Punto
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\delta<0$
+\end_inset
+
+: Ecuación hiperbólica
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Hipérbola
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Dos rectas secantes
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\delta=0$
+\end_inset
+
+: Ecuación parabólica
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Parábola
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Recta doble, o dos rectas paralelas reales o imaginarias
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Además, si 
+\begin_inset Formula $\delta\neq0$
+\end_inset
+
+, la ecuación reducida es 
+\begin_inset Formula $\lambda_{1}x^{2}+\lambda_{2}y^{2}=-\frac{\Delta}{\delta}$
+\end_inset
+
+, mientras que si 
+\begin_inset Formula $\delta=0$
+\end_inset
+
+ y 
+\begin_inset Formula $\Delta\neq0$
+\end_inset
+
+ la ecuación reducida es 
+\begin_inset Formula $y^{2}+2\sqrt{-\frac{\Delta}{s^{3}}}x=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Consideremos 
+\begin_inset Formula $\delta\neq0$
+\end_inset
+
+.
+ Entonces tenemos una cónica de tipo elíptica o hiperbólica que tras la
+ doble reducción es 
+\begin_inset Formula $\lambda_{1}x^{2}+\lambda_{2}y^{2}=q$
+\end_inset
+
+, con lo que
+\begin_inset Formula 
+\[
+\Delta=\left|\begin{array}{ccc}
+\lambda_{1} &  & 0\\
+ & \lambda_{2}\\
+0 &  & -q
+\end{array}\right|=-\lambda_{1}\lambda_{2}q=-\delta q
+\]
+
+\end_inset
+
+y entonces 
+\begin_inset Formula $q=-\frac{\Delta}{\delta}$
+\end_inset
+
+.
+ Así, si 
+\begin_inset Formula $\Delta=0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $q=0$
+\end_inset
+
+ y estamos en un caso degenerado, mientras que si 
+\begin_inset Formula $\Delta\neq0$
+\end_inset
+
+ estamos en el correspondiente caso no degenerado.
+ Si 
+\begin_inset Formula $\delta=0$
+\end_inset
+
+, tras la primera reducción y suponiendo 
+\begin_inset Formula $\lambda_{1}=0$
+\end_inset
+
+ tendríamos
+\begin_inset Formula 
+\[
+\Delta=\left|\begin{array}{ccc}
+0 & 0 & m\\
+0 & \lambda_{2} & n\\
+m & n & d
+\end{array}\right|=-m^{2}\lambda_{2}
+\]
+
+\end_inset
+
+Así, si 
+\begin_inset Formula $\Delta=0$
+\end_inset
+
+ tenemos 
+\begin_inset Formula $m=0$
+\end_inset
+
+ y estamos en un caso degenerado, mientras que si 
+\begin_inset Formula $\Delta\neq0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $m^{2}\neq0$
+\end_inset
+
+ y la ecuación se reduce a 
+\begin_inset Formula $\lambda_{2}y^{2}+2mx=0$
+\end_inset
+
+, es decir, 
+\begin_inset Formula $y^{2}+2\frac{m}{\lambda_{2}}x=0$
+\end_inset
+
+, y la ecuación se debe a que 
+\begin_inset Formula $\frac{m}{\lambda_{2}}=\frac{1}{\lambda_{2}}\sqrt{-\frac{\Delta}{\lambda_{2}}}=\sqrt{-\frac{\Delta}{\lambda_{2}^{3}}}=\sqrt{-\frac{\Delta}{s^{3}}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Elementos geométricos
+\end_layout
+
+\begin_layout Standard
+Una cónica es 
+\series bold
+centrada
+\series default
+ si 
+\begin_inset Formula $\delta\neq0$
+\end_inset
+
+, y llamamos 
+\series bold
+centro de simetría
+\series default
+ de una cónica a todo punto 
+\begin_inset Formula $(x_{0},y_{0})$
+\end_inset
+
+ tal que la traslación dada por 
+\begin_inset Formula $x'=x-x_{0}$
+\end_inset
+
+ e 
+\begin_inset Formula $y'=y-y_{0}$
+\end_inset
+
+ elimina los términos en 
+\begin_inset Formula $x$
+\end_inset
+
+ e 
+\begin_inset Formula $y$
+\end_inset
+
+ de la ecuación.
+ Una cónica centrada tiene un único centro de simetría que es la solución
+ del sistema
+\begin_inset Formula 
+\[
+AX=-B
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Si escribimos la traslación como
+\begin_inset Formula 
+\begin{eqnarray*}
+\left(\begin{array}{c}
+X\\
+\hline 1
+\end{array}\right)=N\left(\begin{array}{c}
+X'\\
+\hline 1
+\end{array}\right) & N=\left(\begin{array}{c|c}
+I & X_{0}\\
+\hline 0 & 1
+\end{array}\right) & X_{0}=\left(\begin{array}{c}
+x_{0}\\
+y_{0}
+\end{array}\right)
+\end{eqnarray*}
+
+\end_inset
+
+la matriz de la cónica tras la traslación es
+\begin_inset Formula 
+\[
+N^{t}\overline{A}N=\left(\begin{array}{c|c}
+* & AX_{0}+B\\
+\hline * & *
+\end{array}\right)
+\]
+
+\end_inset
+
+luego debe ser 
+\begin_inset Formula $AX_{0}+B=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $AX=-B$
+\end_inset
+
+, sistema que tiene solución única porque 
+\begin_inset Formula $|A|\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+ejes
+\series default
+ de una cónica a los del referencial ortonormal en que la cónica tiene ecuación
+ reducida.
+ Las direcciones de los ejes son los subespacios propios de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Las direcciones de los ejes tras la doble reducción son 
+\begin_inset Formula $<(1,0)>$
+\end_inset
+
+ y 
+\begin_inset Formula $<(0,1)>$
+\end_inset
+
+ y multiplicando por la matriz de cambio de base 
+\begin_inset Formula $Q$
+\end_inset
+
+, cuyas columnas son los vectores propios de 
+\begin_inset Formula $A$
+\end_inset
+
+, obtenemos los ejes en el referencial actual.
+\end_layout
+
+\begin_layout Standard
+Dada una elipse real o hipérbola 
+\begin_inset Formula ${\cal G}$
+\end_inset
+
+ de matriz 
+\begin_inset Formula $\overline{A}$
+\end_inset
+
+, si 
+\begin_inset Formula $\lambda_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $\lambda_{2}$
+\end_inset
+
+ son los valores propios de 
+\begin_inset Formula $A$
+\end_inset
+
+, los semiejes principal y secundario de la cónica son 
+\begin_inset Formula $\{a,b\}=\left\{ \sqrt{\left|\frac{\Delta}{\delta\lambda_{1}}\right|},\sqrt{\left|\frac{\Delta}{\delta\lambda_{1}}\right|}\right\} $
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Llevamos 
+\begin_inset Formula ${\cal G}$
+\end_inset
+
+ a un referencial ortonormal donde 
+\begin_inset Formula ${\cal G}\equiv\lambda_{1}x^{2}+\lambda_{2}y^{2}=q$
+\end_inset
+
+ con 
+\begin_inset Formula $q\neq0$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\frac{x^{2}}{\frac{q}{\lambda_{1}}}+\frac{y^{2}}{\frac{q}{\lambda_{2}}}=1$
+\end_inset
+
+, luego 
+\begin_inset Formula $\{a,b\}=\left\{ \sqrt{\left|\frac{q}{\lambda_{1}}\right|},\sqrt{\left|\frac{q}{\lambda_{2}}\right|}\right\} $
+\end_inset
+
+, pero 
+\begin_inset Formula $q=-\frac{\Delta}{\delta}$
+\end_inset
+
+, de donde se deduce la ecuación.
+\end_layout
+
+\begin_layout Standard
+El eje de una parábola tiene por dirección el subespacio de vectores propios
+ correspondiente al valor propio nulo.
+ Para hallar el vértice, si el eje tiene pendiente 
+\begin_inset Formula $k$
+\end_inset
+
+, lo más fácil es derivar implícitamente 
+\begin_inset Formula $y$
+\end_inset
+
+ en función de 
+\begin_inset Formula $x$
+\end_inset
+
+ y buscar un punto de la parábola en el que esta valga 
+\begin_inset Formula $-\frac{1}{k}$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+La validez de este procedimiento se desprende del teorema de la función
+ implícita, estudiado en FVV3.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/aalg/n3.lyx b/aalg/n3.lyx
new file mode 100644
index 0000000..d0e0932
--- /dev/null
+++ b/aalg/n3.lyx
@@ -0,0 +1,1955 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\usepackage{tikz}
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style french
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Un 
+\series bold
+plano afín
+\series default
+ es una terna 
+\begin_inset Formula $\mathbb{A}=({\cal P},{\cal L},\epsilon)$
+\end_inset
+
+ formada por los conjuntos 
+\begin_inset Formula ${\cal P},{\cal L}\neq\emptyset$
+\end_inset
+
+, cuyos elementos se llaman 
+\series bold
+puntos
+\series default
+ y 
+\series bold
+rectas
+\series default
+, respectivamente, y la 
+\series bold
+relación de incidencia
+\series default
+ 
+\begin_inset Formula $\epsilon\subseteq{\cal P}\times{\cal L}$
+\end_inset
+
+, que satisface que 
+\begin_inset Formula $\forall P,Q\in{\cal P},\ell\in{\cal L}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+
+\family sans
+\begin_inset Formula $P\neq Q\implies\exists!\ell\in{\cal L}:P,Q\epsilon\ell$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists P,Q,R\in{\cal P}:\nexists\ell\in{\cal L}:P,Q,R\epsilon\ell$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $P\not\epsilon\ell\implies\exists!m\in{\cal L}:(P\epsilon m\land\nexists Q\in{\cal P}:Q\epsilon\ell,m)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $P\in{\cal P}$
+\end_inset
+
+ es 
+\series bold
+incidente
+\series default
+ con 
+\begin_inset Formula $\ell\in{\cal L}$
+\end_inset
+
+ si 
+\begin_inset Formula $P\epsilon\ell$
+\end_inset
+
+, y 
+\begin_inset Formula $\ell,m\in{\cal L}$
+\end_inset
+
+ son 
+\series bold
+paralelas
+\series default
+ si 
+\begin_inset Formula $\nexists P:P\epsilon\ell,m$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $V$
+\end_inset
+
+ es un 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio vectorial y 
+\begin_inset Formula $\dim_{\mathbb{K}}V\geq2$
+\end_inset
+
+, definimos el plano afín 
+\begin_inset Formula $\mathbb{A}(V):=({\cal P}(V),{\cal L}(V),\in)$
+\end_inset
+
+ con 
+\begin_inset Formula ${\cal P}(V):=V$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal L}(V)=\{\vec{v}+<\vec{w}>\}_{\vec{v},\vec{w}\in V,\vec{w}\neq0}$
+\end_inset
+
+, y escribimos 
+\begin_inset Formula $\mathbb{A}^{n}(\mathbb{K}):=\mathbb{A}(\mathbb{K}^{n})$
+\end_inset
+
+.
+ Llamamos a 
+\begin_inset Formula $\mathbb{A}^{2}(\mathbb{R})$
+\end_inset
+
+ el 
+\series bold
+plano afín usual
+\series default
+.
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+plano proyectivo
+\series default
+ es una terna 
+\begin_inset Formula $\mathbb{P}=({\cal P},{\cal L},\epsilon)$
+\end_inset
+
+ similar a un plano afín pero cambiando el último axioma por que 
+\begin_inset Formula $\forall\ell,m\in{\cal L}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+3.
+\end_layout
+
+\end_inset
+
+
+\family sans
+
+\begin_inset Formula $\exists P\in{\cal P}:P\epsilon\ell,m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+4.
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\exists P,Q,R\in{\cal P}:(P\neq Q\neq R\neq P\land P,Q,R\in\ell)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+principio de dualidad para planos proyectivos
+\series default
+ afirma que si 
+\begin_inset Formula $\pi:=({\cal P},{\cal L},\epsilon)$
+\end_inset
+
+ es un plano proyectivo entonces 
+\begin_inset Formula $\pi^{*}:=({\cal L},{\cal P},\epsilon^{*})$
+\end_inset
+
+ con 
+\begin_inset Formula $\ell\epsilon^{*}P\iff P\epsilon\ell$
+\end_inset
+
+ también lo es.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall\ell,m\in{\cal L},(\ell\neq m\implies\exists!P\in{\cal P}:P\epsilon\ell,m)$
+\end_inset
+
+: El axioma 3 asegura que 
+\begin_inset Formula $P$
+\end_inset
+
+ existe.
+ Ahora bien, si existiera otro 
+\begin_inset Formula $Q\neq P$
+\end_inset
+
+ con 
+\begin_inset Formula $Q\epsilon\ell,m$
+\end_inset
+
+, por el axioma 1 se tendría 
+\begin_inset Formula $\ell=m\#$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists\ell,m,n\in{\cal L}:\nexists P\in{\cal P}:P\epsilon\ell,m,n$
+\end_inset
+
+: El axioma 2 nos dice que hay 3 puntos 
+\begin_inset Formula $Q,R,S\in{\cal P}$
+\end_inset
+
+ para los que 
+\begin_inset Formula $\nexists\ell\in{\cal L}:Q,R,S\epsilon\ell$
+\end_inset
+
+.
+ Si fueran 
+\begin_inset Formula $Q=R\neq S$
+\end_inset
+
+, el axioma 1 nos dice que existe una recta que los contiene, y si fueran
+ 
+\begin_inset Formula $Q=R=S$
+\end_inset
+
+, podríamos tomar uno de los puntos del axioma 4 (para alguna recta) como
+ punto distinto a este para aplicar el axioma 1.
+ Por tanto los 3 puntos son distintos.
+ Sean ahora 
+\begin_inset Formula $\ell:=QR$
+\end_inset
+
+, 
+\begin_inset Formula $m:=RS$
+\end_inset
+
+ y 
+\begin_inset Formula $n:=SQ$
+\end_inset
+
+ (aplicando el axioma 1).
+ Si hubiera un punto 
+\begin_inset Formula $P\epsilon\ell,m,n$
+\end_inset
+
+ (podemos suponer 
+\begin_inset Formula $P\neq Q,R$
+\end_inset
+
+), entonces por el axioma 1 
+\begin_inset Formula $n=PQ=\ell=PR=m$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $Q,R,S\in\ell\#$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall P,Q\in{\cal P},\exists\ell\in{\cal L}:P,Q\epsilon\ell$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $P\neq Q$
+\end_inset
+
+, esto nos lo asegura el axioma 1.
+ Para poder aplicarlo con 
+\begin_inset Formula $P=Q$
+\end_inset
+
+, tomamos un punto de los dados por el axioma 4 que sea distinto a 
+\begin_inset Formula $P$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall P\in{\cal P},\exists\ell,m,n\in{\cal L}:(\ell\neq m\neq n\neq\ell\land P\epsilon\ell,m,n)$
+\end_inset
+
+.
+ Tomamos los puntos 
+\begin_inset Formula $Q,R,S$
+\end_inset
+
+ dados por el axioma 2, que ya hemos visto que deben ser distintos.
+ Podemos suponer 
+\begin_inset Formula $P\neq Q,R$
+\end_inset
+
+, y entonces podemos suponer 
+\begin_inset Formula $PQ\neq PR$
+\end_inset
+
+.
+ En efecto, si fueran iguales sería 
+\begin_inset Formula $P\epsilon QR$
+\end_inset
+
+ y 
+\begin_inset Formula $S\not\epsilon QR=PR$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $P\neq S$
+\end_inset
+
+ y además 
+\begin_inset Formula $PS\neq PR$
+\end_inset
+
+, y podríamos tomar 
+\begin_inset Formula $S$
+\end_inset
+
+ en vez de 
+\begin_inset Formula $R$
+\end_inset
+
+.
+ Ahora tomamos 
+\begin_inset Formula $QR$
+\end_inset
+
+ que, por el axioma 4, contiene un tercer punto 
+\begin_inset Formula $T\neq Q,R$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $P\neq T$
+\end_inset
+
+ (si fuera 
+\begin_inset Formula $P=T$
+\end_inset
+
+ se tendría 
+\begin_inset Formula $PQ=PR\#$
+\end_inset
+
+) y 
+\begin_inset Formula $PT\neq PQ,PR$
+\end_inset
+
+ (si fuera, por ejemplo, 
+\begin_inset Formula $PT=PQ$
+\end_inset
+
+, se tendría 
+\begin_inset Formula $PQ=TQ=TR=QR\#$
+\end_inset
+
+).
+ Por tanto, 
+\begin_inset Formula $\ell:=PQ$
+\end_inset
+
+, 
+\begin_inset Formula $m:=PT$
+\end_inset
+
+ y 
+\begin_inset Formula $n:=PR$
+\end_inset
+
+ cumplen las condiciones.
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $\pi=({\cal P},{\cal L},\epsilon)$
+\end_inset
+
+ y 
+\begin_inset Formula $\pi'=({\cal P}',{\cal L}',\epsilon')$
+\end_inset
+
+ dos planos proyectivos, un 
+\series bold
+isomorfismo
+\series default
+ de 
+\begin_inset Formula $\pi$
+\end_inset
+
+ a 
+\begin_inset Formula $\pi'$
+\end_inset
+
+ es un par 
+\begin_inset Formula $(f:{\cal P}\rightarrow{\cal P}',f':{\cal L}\rightarrow{\cal L}')$
+\end_inset
+
+ de biyecciones tal que 
+\begin_inset Formula $\forall P\in{\cal P},\ell\in{\cal L},(P\epsilon\ell\implies f(P)\epsilon'f'(\ell))$
+\end_inset
+
+.
+ Si existe, decimos que 
+\begin_inset Formula $\pi$
+\end_inset
+
+ y 
+\begin_inset Formula $\pi'$
+\end_inset
+
+ son 
+\series bold
+isomorfos
+\series default
+, si y sólo si existe una biyección 
+\begin_inset Formula $f:{\cal P}\rightarrow{\cal P}'$
+\end_inset
+
+ que lleva ternas de puntos alineados a ternas de puntos alineados.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Obvio.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Dada una recta 
+\begin_inset Formula $\ell$
+\end_inset
+
+, por el axioma 4 existen tres puntos 
+\begin_inset Formula $P,Q,R$
+\end_inset
+
+ distintos sobre la recta.
+ Definimos 
+\begin_inset Formula $f':{\cal L}\rightarrow{\cal L}'$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f'(\ell):=\overline{f(P)f(Q)}$
+\end_inset
+
+.
+ Para ver que está bien definida, sean 
+\begin_inset Formula $P',Q'\epsilon\ell$
+\end_inset
+
+, 
+\begin_inset Formula $P'\neq Q'$
+\end_inset
+
+ con 
+\begin_inset Formula $\{P,Q\}\neq\{P',Q'\}$
+\end_inset
+
+ (podemos suponer 
+\begin_inset Formula $P\neq P',Q'$
+\end_inset
+
+ y 
+\begin_inset Formula $P'\neq P,Q$
+\end_inset
+
+).
+ Entonces 
+\begin_inset Formula $f'(\ell)=\overline{f(P')f(Q')}$
+\end_inset
+
+, pero como 
+\begin_inset Formula $P,Q,P',Q'$
+\end_inset
+
+ están alineados, 
+\begin_inset Formula $f(P),f(Q),f(P'),f(Q')$
+\end_inset
+
+ también lo están, y 
+\begin_inset Formula $\overline{f(P')f(Q')}=\overline{f(P)f(Q)}$
+\end_inset
+
+.
+ Sean 
+\begin_inset Formula $\ell:=\overline{PQ}$
+\end_inset
+
+ y 
+\begin_inset Formula $R\epsilon\ell$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f'(\ell)=\overline{f(P)f(Q)}$
+\end_inset
+
+, pero como 
+\begin_inset Formula $P,Q,R$
+\end_inset
+
+ están alineados, 
+\begin_inset Formula $f(R)\epsilon'\overline{f(P)f(Q)}=f'(\ell)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Construcción de 
+\begin_inset Formula $\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si en el espacio afín 
+\begin_inset Formula $\mathbb{A}:=\mathbb{A}(W)$
+\end_inset
+
+ para cierto espacio vectorial 
+\begin_inset Formula $W$
+\end_inset
+
+ definimos la relación de equivalencia 
+\begin_inset Formula $\ell\sim\ell':\iff\ell\parallel\ell'$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\overline{\mathbb{A}}:=({\cal P}',{\cal L}',\in)$
+\end_inset
+
+ con 
+\begin_inset Formula ${\cal P}':={\cal P}\cup({\cal L}/\sim)$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal L}':=\{\ell\cup\{[\ell]\}\}_{\ell\in{\cal L}}\cup\{{\cal L}/\sim\}$
+\end_inset
+
+ es un plano proyectivo al que llamamos 
+\series bold
+extensión proyectiva
+\series default
+ de 
+\begin_inset Formula $\mathbb{A}$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+puntos afines
+\series default
+ a los de 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+ y 
+\series bold
+puntos del infinito
+\series default
+ a los de 
+\begin_inset Formula ${\cal L}/\sim$
+\end_inset
+
+.
+ De igual modo, llamamos 
+\series bold
+rectas extendidas
+\series default
+ a las 
+\begin_inset Formula $\overline{\ell}:=\ell\cup\{[\ell]\}$
+\end_inset
+
+ y 
+\series bold
+recta del infinito
+\series default
+ a 
+\begin_inset Formula $\ell_{\infty}:={\cal L}/\sim$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dado el 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio vectorial 
+\begin_inset Formula $W\equiv\mathbb{K}^{3}$
+\end_inset
+
+, si 
+\begin_inset Formula ${\cal P}(W):=\{\text{rectas vectoriales de }W\}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal L}(W):=\{\text{planos vectoriales de }W\}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $({\cal P}(W),{\cal L}(W),\subseteq)$
+\end_inset
+
+ es un plano proyectivo.
+ Llamamos 
+\series bold
+plano proyectivo en 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+
+\series default
+ a 
+\begin_inset Formula $\mathbb{P}^{2}(\mathbb{K}):=({\cal P}(\mathbb{K}^{3}),{\cal L}(\mathbb{K}^{3}),\subseteq)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $<\vec{v}>\neq<\vec{w}>\implies\exists!\pi\in{\cal L}(W):<\vec{v}>,<\vec{w}>\subseteq\pi$
+\end_inset
+
+: 
+\begin_inset Formula $\vec{v}$
+\end_inset
+
+ y 
+\begin_inset Formula $\vec{w}$
+\end_inset
+
+ son LI, luego necesariamente 
+\begin_inset Formula $\pi=<\vec{v},\vec{w}>$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists\vec{u},\vec{v},\vec{w}\in W:\nexists\pi\in{\cal L}(W):<\vec{u},\vec{v},\vec{w}>\subseteq\pi$
+\end_inset
+
+: Basta tomar una base de 
+\begin_inset Formula $W$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists\vec{u}\in W:<\vec{u}>\subseteq\pi$
+\end_inset
+
+: Si 
+\begin_inset Formula $\pi=<\vec{v},\vec{w}>$
+\end_inset
+
+, basta tomar 
+\begin_inset Formula $\vec{v}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists\vec{u},\vec{v},\vec{w}\in W:(<\vec{u}>\neq<\vec{v}>\neq<\vec{w}>\neq<\vec{u}>\land<\vec{u},\vec{v},\vec{w}>\in\pi)$
+\end_inset
+
+: Si 
+\begin_inset Formula $\pi=<\vec{v},\vec{w}>$
+\end_inset
+
+, basta tomar 
+\begin_inset Formula $\vec{v}$
+\end_inset
+
+, 
+\begin_inset Formula $\vec{w}$
+\end_inset
+
+ y 
+\begin_inset Formula $\vec{v}+\vec{w}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{sloppypar}
+\end_layout
+
+\end_inset
+
+Dado un 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio vectorial 
+\begin_inset Formula $W$
+\end_inset
+
+ de dimensión 2, 
+\begin_inset Formula $\overline{\mathbb{A}(W)}$
+\end_inset
+
+ es isomorfo a 
+\begin_inset Formula $\mathbb{P}(W\times\mathbb{K})$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula ${\cal P}:=W\cup\{[\ell]\}_{\ell\text{ recta afín de }W}$
+\end_inset
+
+ el conjunto de puntos de 
+\begin_inset Formula $\overline{\mathbb{A}(W)}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal P}':=\{\text{rectas vectoriales de }W\times\mathbb{K}\}$
+\end_inset
+
+ el conjunto de puntos de 
+\begin_inset Formula $\mathbb{P}(W\times\mathbb{K})$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $\sigma:{\cal P}\rightarrow{\cal P}'$
+\end_inset
+
+ dada por 
+\begin_inset Formula $\sigma(u)=<(u,1)>\forall u\in W$
+\end_inset
+
+ y 
+\begin_inset Formula $\sigma([<u>])=<(u,0)>\forall u\in W$
+\end_inset
+
+, una biyección cuya inversa viene dada por 
+\begin_inset Formula $\sigma^{-1}(<(u,0)>)=[<(u,0)>]\forall u\in W$
+\end_inset
+
+ y 
+\begin_inset Formula $\sigma^{-1}(<(u,\lambda)>)=\frac{u}{\lambda}\forall u\in W,\lambda\neq0$
+\end_inset
+
+.
+ Veamos que lleva ternas de puntos alineados a ternas de puntos alineados:
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{sloppypar}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Si los tres puntos son afines y suponemos 
+\begin_inset Formula $u_{1}\neq0,u_{2}$
+\end_inset
+
+, que estén alineados significa que 
+\begin_inset Formula $u_{2}=\lambda u_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $u_{3}=\mu u_{1}$
+\end_inset
+
+ para 
+\begin_inset Formula $\lambda\neq1$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\sigma(u_{1})=<(u_{1},1)>$
+\end_inset
+
+, 
+\begin_inset Formula $\sigma(u_{2})=<(\lambda u_{1},1)>$
+\end_inset
+
+ y 
+\begin_inset Formula $\sigma(u_{3})=<(\mu u_{1},1)>$
+\end_inset
+
+, pero 
+\begin_inset Formula $\frac{\lambda-\mu}{\lambda-1}(u_{1},1)+\frac{\mu-1}{\lambda-1}(\lambda u_{1},1)=(\mu u_{1},1)$
+\end_inset
+
+, luego las tres rectas se encuentran en un plano.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $u_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $u_{2}$
+\end_inset
+
+ son afines y 
+\begin_inset Formula $[<u_{3}>]$
+\end_inset
+
+ es del infinito, que estén alineados significa que 
+\begin_inset Formula $u_{2}=u_{1}+\lambda u_{3}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\sigma(u_{1})=<(u_{1},1)>$
+\end_inset
+
+, 
+\begin_inset Formula $\sigma(u_{2})=<(u_{1}+\lambda u_{3},1)>$
+\end_inset
+
+ y 
+\begin_inset Formula $\sigma(u_{3})=<(u_{3},0)>$
+\end_inset
+
+.
+ Pero 
+\begin_inset Formula $(u_{1},1)+\lambda(u_{3},0)=(u_{1}+\lambda_{3},1)$
+\end_inset
+
+, luego las tres rectas están en el mismo plano.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $u_{1}$
+\end_inset
+
+ es afín y 
+\begin_inset Formula $[<u_{2}>],[<u_{3}>]$
+\end_inset
+
+ son del infinito, que estén alineados significa que 
+\begin_inset Formula $u_{2}=u_{3}$
+\end_inset
+
+, y entonces es claro que hay una recta que une 
+\begin_inset Formula $\sigma(u_{1})$
+\end_inset
+
+ con 
+\begin_inset Formula $\sigma([<u_{2}=u_{3}>])$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si los tres puntos son del infinito, siempre están alineados, pero entonces
+ para 
+\begin_inset Formula $i\in\{1,2,3\}$
+\end_inset
+
+, 
+\begin_inset Formula $\sigma([<u_{i}>])=<(u_{i},0)>\in W\times\{0\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Referencias proyectivas
+\end_layout
+
+\begin_layout Standard
+Tres puntos 
+\begin_inset Formula $P,Q,R\in\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+ son (
+\series bold
+proyectivamente
+\series default
+)
+\series bold
+ independientes
+\series default
+ si los vectores que los representan forman una base de 
+\begin_inset Formula $\mathbb{K}^{3}$
+\end_inset
+
+.
+ Una 
+\series bold
+referencia proyectiva
+\series default
+ o 
+\series bold
+referencial proyectivo
+\series default
+ en 
+\begin_inset Formula $\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+ es una cuaterna 
+\begin_inset Formula ${\cal R}:=(P,Q,R,U)$
+\end_inset
+
+ de puntos tales que tres puntos cualesquiera de ellos son independientes.
+\end_layout
+
+\begin_layout Standard
+Todo referencial proyectivo de 
+\begin_inset Formula $\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+ admite una base 
+\begin_inset Formula ${\cal B}:=(v_{1},v_{2},v_{3})$
+\end_inset
+
+ de 
+\begin_inset Formula $\mathbb{K}^{3}$
+\end_inset
+
+ tal que
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula $P=<v_{1}>$
+\end_inset
+
+, 
+\begin_inset Formula $Q=<v_{2}>$
+\end_inset
+
+, 
+\begin_inset Formula $R=<v_{3}>$
+\end_inset
+
+ y 
+\begin_inset Formula $U=<v_{1}+v_{2}+v_{3}>$
+\end_inset
+
+, única salvo multiplicación simultánea de los 3 vectores por un escalar
+ no nulo.
+ A esta base la llamamos 
+\series bold
+base asociada
+\series default
+ al referencial 
+\begin_inset Formula ${\cal R}$
+\end_inset
+
+, y el punto 
+\begin_inset Formula $U$
+\end_inset
+
+ es el 
+\series bold
+punto unidad
+\series default
+ del referencial.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $<v_{1}>,<v_{2}>,<v_{3}>$
+\end_inset
+
+ son no alineados en 
+\begin_inset Formula $\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $(v_{1},v_{2},v_{3})$
+\end_inset
+
+ es una base.
+ Entonces, si 
+\begin_inset Formula $P=:<u_{1}>$
+\end_inset
+
+, 
+\begin_inset Formula $Q=:<u_{2}>$
+\end_inset
+
+ y 
+\begin_inset Formula $R=:<u_{3}>$
+\end_inset
+
+, podemos escribir 
+\begin_inset Formula $U=:<u>$
+\end_inset
+
+ con 
+\begin_inset Formula $u:=\alpha_{1}u_{1}+\alpha_{2}u_{2}+\alpha_{3}u_{3}$
+\end_inset
+
+ con 
+\begin_inset Formula $(\alpha_{1},\alpha_{2},\alpha_{3})\neq(0,0,0)$
+\end_inset
+
+.
+ Entonces hacemos 
+\begin_inset Formula $v_{i}:=\alpha_{i}u_{i}$
+\end_inset
+
+ para 
+\begin_inset Formula $i\in\{1,2,3\}$
+\end_inset
+
+, y sabemos que 
+\begin_inset Formula $\alpha_{1},\alpha_{2},\alpha_{3}\neq0$
+\end_inset
+
+, pues si fuera algún 
+\begin_inset Formula $\alpha_{i}=0$
+\end_inset
+
+, 
+\begin_inset Formula $u$
+\end_inset
+
+ sería linealmente dependiente con 
+\begin_inset Formula $u_{j}$
+\end_inset
+
+ y 
+\begin_inset Formula $u_{k}$
+\end_inset
+
+ para 
+\begin_inset Formula $j,k\neq i$
+\end_inset
+
+ y serían alineados, luego 
+\begin_inset Formula $(v_{1},v_{2},v_{3})$
+\end_inset
+
+ es una base que satisface las condiciones.
+ Ahora bien, si existe 
+\begin_inset Formula ${\cal B}'=(v'_{1},v'_{2},v'_{3})$
+\end_inset
+
+ que también satisface las condiciones, necesariamente 
+\begin_inset Formula $<v'_{1}>=P=<v_{1}>$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $v'_{1}=\lambda_{1}v_{1}$
+\end_inset
+
+ para algún 
+\begin_inset Formula $\lambda_{1}\neq0$
+\end_inset
+
+, y lo mismo sucede con 
+\begin_inset Formula $v'_{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $v'_{3}$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $<v'_{1}+v'_{2}+v'_{3}>=<\lambda_{1}v'_{1}+\lambda_{2}v'_{2}+\lambda_{3}v'_{3}>=U=<v_{1}+v_{2}+v_{3}>$
+\end_inset
+
+, y es claro que 
+\begin_inset Formula $\lambda_{1}=\lambda_{2}=\lambda_{3}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dado 
+\begin_inset Formula $P\in\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+, decimos que sus 
+\series bold
+coordenadas homogéneas
+\series default
+ respecto a la base 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ a
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+so
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+cia
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+da al referencial 
+\begin_inset Formula ${\cal R}$
+\end_inset
+
+ son 
+\begin_inset Formula $x,y,z$
+\end_inset
+
+ (
+\begin_inset Formula $P:=[x,y,z]$
+\end_inset
+
+) si 
+\begin_inset Formula $P=:<\vec{u}>$
+\end_inset
+
+ con 
+\begin_inset Formula $[\vec{u}]_{{\cal B}}=(x,y,z)$
+\end_inset
+
+.
+ Estas son únicas salvo multiplicación de las tres por un escalar no nulo.
+ Tres puntos de coordenadas homogéneas 
+\begin_inset Formula $[a,b,c]$
+\end_inset
+
+, 
+\begin_inset Formula $[d,e,f]$
+\end_inset
+
+ y 
+\begin_inset Formula $[g,h,i]$
+\end_inset
+
+ son proyectivamente independientes si y sólo si
+\begin_inset Formula 
+\[
+\left|\begin{array}{ccc}
+a & b & c\\
+d & e & f\\
+g & h & i
+\end{array}\right|\neq0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\begin_inset Formula $[a,b,c]^{*}$
+\end_inset
+
+ a la recta en 
+\begin_inset Formula $\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+ dada por 
+\begin_inset Formula $ax+by+cz=0$
+\end_inset
+
+.
+ Las rectas 
+\begin_inset Formula $\ell:=[a_{1},b_{1},c_{1}]^{*}$
+\end_inset
+
+, 
+\begin_inset Formula $m:=[a_{2},b_{2},c_{2}]^{*}$
+\end_inset
+
+ y 
+\begin_inset Formula $n:=[a_{3},b_{3},c_{3}]^{*}$
+\end_inset
+
+ son 
+\series bold
+congruentes
+\series default
+ (se cortan) si y sólo si
+\begin_inset Formula 
+\[
+\left|\begin{array}{ccc}
+a_{1} & b_{1} & c_{1}\\
+a_{2} & b_{2} & c_{2}\\
+a_{3} & b_{3} & c_{3}
+\end{array}\right|=0
+\]
+
+\end_inset
+
+
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula 
+\begin{multline*}
+\exists P\in\ell,m,n\iff\exists(x_{0},y_{0},x_{0})\neq0:\forall i\in\{1,2,3\},a_{i}x_{0}+b_{i}y_{0}+c_{i}z_{0}=0\iff\\
+\iff\dim\left\{ \left(\begin{array}{ccc}
+a_{1} & b_{1} & c_{1}\\
+a_{2} & b_{2} & c_{2}\\
+a_{3} & b_{3} & c_{3}
+\end{array}\right)\left(\begin{array}{c}
+x\\
+y\\
+z
+\end{array}\right)=\left(\begin{array}{c}
+0\\
+0\\
+0
+\end{array}\right)\right\} >0
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Teoremas de Desargues y Pappus
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+teorema de Desargues
+\series default
+ afirma que, dados dos triángulos 
+\begin_inset Formula $ABC$
+\end_inset
+
+ y 
+\begin_inset Formula $A'B'C'$
+\end_inset
+
+ sin vértices ni lados comunes, si las rectas que unen vértices correspondientes
+ (
+\begin_inset Formula $AA'$
+\end_inset
+
+, 
+\begin_inset Formula $BB'$
+\end_inset
+
+ y 
+\begin_inset Formula $CC'$
+\end_inset
+
+) se cortan en un punto, los puntos de corte de lados correspondientes están
+ alineados.
+ Un plano proyectivo es 
+\series bold
+desarguesiano
+\series default
+ si satisface este teorema.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+teorema de Pappus
+\series default
+ afirma que, dados tres puntos distintos 
+\begin_inset Formula $A,B,C$
+\end_inset
+
+ en una recta y 
+\begin_inset Formula $A',B',C'$
+\end_inset
+
+ en otra, los puntos 
+\begin_inset Formula $L\in AB'\cap A'B$
+\end_inset
+
+, 
+\begin_inset Formula $M\in AC'\cap A'C$
+\end_inset
+
+ y 
+\begin_inset Formula $N\in BC'\cap B'C$
+\end_inset
+
+ están alineados.
+ Un plano proyectivo es 
+\series bold
+papiano
+\series default
+ si satisface este teorema.
+\end_layout
+
+\begin_layout Standard
+Un plano proyectivo 
+\begin_inset Formula $\pi$
+\end_inset
+
+ es papiano y desarguesiano si y sólo si es isomorfo a 
+\begin_inset Formula $\mathbb{P}(V)$
+\end_inset
+
+ para algún 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio vectorial tridimensional, si y sólo si es isomorfo a 
+\begin_inset Formula $\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+.
+ En tal caso, el cuerpo 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+ de las dos últimas condiciones es el mismo y está unívocamente determinado
+ por 
+\begin_inset Formula $\pi$
+\end_inset
+
+ salvo isomorfismo.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $2\iff3]$
+\end_inset
+
+ Sea 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ una base del 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio tridimensional 
+\begin_inset Formula $V$
+\end_inset
+
+, 
+\begin_inset Formula $[\cdot]_{{\cal B}}:V\longrightarrow\mathbb{K}^{3}$
+\end_inset
+
+ define un isomorfismo entre los puntos de 
+\begin_inset Formula $\mathbb{P}(V)$
+\end_inset
+
+ y los de 
+\begin_inset Formula $\mathbb{P}(\mathbb{K}^{3})=\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $3\implies1]$
+\end_inset
+
+ Probemos el teorema de Desargues.
+ Sean 
+\begin_inset Formula $O:=[\vec{o}]$
+\end_inset
+
+ el punto de corte entre las tres rectas, 
+\begin_inset Formula $A:=[\vec{a}]$
+\end_inset
+
+, 
+\begin_inset Formula $B:=[\vec{b}]$
+\end_inset
+
+ y 
+\begin_inset Formula $C:=[\vec{c}]$
+\end_inset
+
+ con 
+\begin_inset Formula $\vec{a},\vec{b},\vec{c},\vec{o}\neq0$
+\end_inset
+
+, como 
+\begin_inset Formula $O,A,A'$
+\end_inset
+
+ están alineados, debe ser 
+\begin_inset Formula $A'=[\lambda\vec{o}+\mu\vec{a}]$
+\end_inset
+
+ con 
+\begin_inset Formula $\lambda\neq0$
+\end_inset
+
+ (si fuera 
+\begin_inset Formula $\lambda=0$
+\end_inset
+
+ sería 
+\begin_inset Formula $A=A'$
+\end_inset
+
+ y 
+\begin_inset Formula $AA'$
+\end_inset
+
+ no tendría sentido) y, dividiendo por 
+\begin_inset Formula $\lambda$
+\end_inset
+
+, 
+\begin_inset Formula $A'=:[\vec{o}+\alpha\vec{a}]$
+\end_inset
+
+.
+ Análogamente 
+\begin_inset Formula $B'=:[\vec{o}+\beta\vec{b}]$
+\end_inset
+
+ y 
+\begin_inset Formula $C'=:[\vec{o}+\gamma\vec{c}]$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $\alpha\vec{a}-\beta\vec{b}=(\vec{o}+\alpha\vec{a})-(\vec{o}+\beta\vec{b})$
+\end_inset
+
+, tenemos que 
+\begin_inset Formula $AB\cap A'B'=\{[\alpha\vec{a}-\beta\vec{b}]\}$
+\end_inset
+
+, y del mismo modo 
+\begin_inset Formula $AC\cap A'C'=\{[\alpha\vec{a}-\gamma\vec{c}]\}$
+\end_inset
+
+ y 
+\begin_inset Formula $BC\cap B'C'=\{[\beta\vec{b}-\gamma\vec{c}]\}$
+\end_inset
+
+.
+ Estos tres puntos están alineados, pues 
+\begin_inset Formula $(\alpha\vec{a}-\beta\vec{b})-(\alpha\vec{a}-\gamma\vec{c})+(\beta\vec{b}-\gamma\vec{c})=0$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Para el teorema de Pappus, consideremos la referencia proyectiva 
+\begin_inset Formula ${\cal R}:=(A',A,B,B')$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $A'=[1,0,0]$
+\end_inset
+
+, 
+\begin_inset Formula $A=[0,1,0]$
+\end_inset
+
+, 
+\begin_inset Formula $B=[0,0,1]$
+\end_inset
+
+ y 
+\begin_inset Formula $B'=[1,1,1]$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $C\epsilon AB$
+\end_inset
+
+, debe ser 
+\begin_inset Formula $C=[0,\alpha,\beta]$
+\end_inset
+
+ con 
+\begin_inset Formula $\alpha,\beta\neq0$
+\end_inset
+
+, luego 
+\begin_inset Formula $C=[0,1,c]$
+\end_inset
+
+ para algún 
+\begin_inset Formula $c\neq0$
+\end_inset
+
+.
+ De forma parecida, 
+\begin_inset Formula $C'=[c',1,1]$
+\end_inset
+
+.
+ Entonces
+\begin_inset Formula 
+\begin{eqnarray*}
+AB':x=z & AC':x=c'z & BC':x=c'y\\
+A'B:y=0 & A'C:z=cy & B'C:(c-1)x-cy+z=0
+\end{eqnarray*}
+
+\end_inset
+
+de donde 
+\begin_inset Formula $AB'\cap A'B=\{[1,0,1]\}$
+\end_inset
+
+, 
+\begin_inset Formula $AC'\cap A'C=\{[cc',1,c]\}$
+\end_inset
+
+ y 
+\begin_inset Formula $BC'\cap B'C=\{[c',1,c+c'-cc']\}$
+\end_inset
+
+, y los tres puntos están alineados porque
+\begin_inset Formula 
+\[
+\left|\begin{array}{ccc}
+1 & 0 & 1\\
+cc' & 1 & c\\
+c' & 1 & c+c'-cc'
+\end{array}\right|=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Ampliación proyectiva
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\begin_inset Formula $\mathbb{K}[x_{1},\dots,x_{n}]$
+\end_inset
+
+ al conjunto de polinomios de 
+\begin_inset Formula $n$
+\end_inset
+
+ variables sobre 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+, y decimos que 
+\begin_inset Formula $F\in\mathbb{K}[x_{1},\dots,x_{n}]$
+\end_inset
+
+ es 
+\series bold
+homogéneo
+\series default
+ si todos sus monomios tienen el mismo grado.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{sloppypar}
+\end_layout
+
+\end_inset
+
+Dado 
+\begin_inset Formula $f\in\mathbb{K}[x_{1},\dots,x_{n}]$
+\end_inset
+
+, su 
+\series bold
+homogeneización
+\series default
+ es el polinomio homogéneo 
+\begin_inset Formula $f^{*}\in\mathbb{K}[x_{1},\dots,x_{n+1}]$
+\end_inset
+
+ dado por 
+\begin_inset Formula $f^{*}(x_{1},\dots,x_{n+1})=x_{n+1}^{d}f(\frac{x_{1}}{x_{n+1}},\dots,\frac{x_{n}}{x_{n+1}})$
+\end_inset
+
+, siendo 
+\begin_inset Formula $d$
+\end_inset
+
+ el grado de 
+\begin_inset Formula $f$
+\end_inset
+
+, es decir, el máximo de los grados de sus monomios.
+ La 
+\series bold
+deshomogeneización
+\series default
+ de 
+\begin_inset Formula $F\in\mathbb{K}[x_{1},\dots,x_{n+1}]$
+\end_inset
+
+ es 
+\begin_inset Formula $F_{*}\in\mathbb{K}[x_{1},\dots,x_{n}]$
+\end_inset
+
+ dado por 
+\begin_inset Formula $F_{*}(x_{1},\dots,x_{n})=F(x_{1},\dots,x_{n},1)$
+\end_inset
+
+.
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{sloppypar}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Dado 
+\begin_inset Formula $f\in\mathbb{K}[x_{1},\dots,x_{n}]$
+\end_inset
+
+, 
+\begin_inset Formula $(f^{*})_{*}=f$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Si 
+\begin_inset Formula $f(x_{1},\dots,x_{n}):=\sum_{i=1}^{k}\prod_{j=1}^{d_{i}}x_{a_{ij}}$
+\end_inset
+
+, entonces 
+\begin_inset Formula 
+\[
+f^{*}(x_{1},\dots,x_{n+1})=\sum_{i=1}^{k}x_{n+1}^{\max\{d_{i}\}}\prod_{j=1}^{d_{i}}\frac{x_{a_{ij}}}{x_{n+1}}=\sum_{i=1}^{k}x_{n+1}^{\max\{d_{i}\}-d_{i}}\prod_{j=1}^{d_{i}}x_{a_{ij}}
+\]
+
+\end_inset
+
+ y 
+\begin_inset Formula $(f^{*})_{*}(x_{1},\dots,x_{n})=\sum_{i=1}^{k}\prod_{j=1}^{d_{i}}x_{a_{ij}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Dado 
+\begin_inset Formula $F\in\mathbb{K}[x_{1},\dots,x_{n+1}]$
+\end_inset
+
+ homogéneo, 
+\begin_inset Formula $F=x_{n+1}^{k}(F_{*})^{*}$
+\end_inset
+
+, siendo 
+\begin_inset Formula $k$
+\end_inset
+
+ la mayor potencia de 
+\begin_inset Formula $x_{n+1}$
+\end_inset
+
+ que divide a todos los monomios de 
+\begin_inset Formula $F$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Si 
+\begin_inset Formula $F(x_{1},\dots,x_{n+1}):=\sum_{i=1}^{k}x_{n+1}^{b_{i}}\prod_{j=1}^{d-b_{i}}x_{a_{ij}}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $F_{*}(x_{1},\dots,x_{n})=\sum_{i=1}^{k}\prod_{j=1}^{d-b_{i}}x_{a_{ij}}$
+\end_inset
+
+ y 
+\begin_inset Formula 
+\begin{eqnarray*}
+(F_{*})^{*}(x_{1},\dots,x_{n+1}) & = & \sum_{i=1}^{k}x_{n+1}^{\max\{d-b_{i}\}}\prod_{j=1}^{d-b_{i}}\frac{x_{a_{ij}}}{x_{n+1}}=\sum_{i=1}^{k}x_{n+1}^{d-\min\{b_{i}\}-d+b_{i}}\prod_{j=1}^{d-b_{i}}x_{a_{ij}}\\
+ & = & \frac{1}{x_{n+1}^{\min\{b_{i}\}}}\sum_{i=1}^{k}x_{n+1}^{b_{i}}\prod_{j=1}^{d-b_{i}}x_{a_{ij}}=\frac{F}{x_{n+1}^{\min\{b_{i}\}}}
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $f\in\mathbb{K}[x,y]$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal L}:=\{(x,y)\in\mathbb{A}^{2}(\mathbb{K}):f(x,y)=0\}$
+\end_inset
+
+, llamamos 
+\series bold
+ampliación proyectiva
+\series default
+ o 
+\series bold
+completación proyectiva
+\series default
+ de 
+\begin_inset Formula ${\cal L}$
+\end_inset
+
+ a 
+\begin_inset Formula $\overline{{\cal L}}:=\{<(x,y,z)>\in\mathbb{P}^{2}(\mathbb{K}):f^{*}(x,y,z)=0\}$
+\end_inset
+
+, y para 
+\begin_inset Formula $\hat{{\cal L}}\subseteq\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+, la 
+\series bold
+parte afín
+\series default
+ de 
+\begin_inset Formula $\hat{{\cal L}}$
+\end_inset
+
+ es 
+\begin_inset Formula $\hat{{\cal L}}^{\text{afín}}:=\{(x,y)\in\mathbb{A}^{2}(\mathbb{K}):<(x,y,1)>\in\hat{{\cal L}}\}$
+\end_inset
+
+.
+ Vemos que para 
+\begin_inset Formula $F\in\mathbb{K}[x,y,z]$
+\end_inset
+
+ homogéneo y 
+\begin_inset Formula $\hat{{\cal L}}:=\{F(x,y,z)=0\}$
+\end_inset
+
+, 
+\begin_inset Formula $\hat{{\cal L}}^{\text{afín}}=\{(x,y):F(x,y,1)=0\}=\{(x,y):F_{*}(x,y)=0\}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\overline{\hat{{\cal L}}^{\text{afín}}}=\{<(a,b,c)>:(F_{*})^{*}(a,b,c)=0\}=\hat{{\cal L}}\cup\{<(x,y,0)>:F(x,y,0)=0\}$
+\end_inset
+
+, y si 
+\begin_inset Formula $F$
+\end_inset
+
+ no es divisible por 
+\begin_inset Formula $z$
+\end_inset
+
+ es 
+\begin_inset Formula $\overline{\hat{{\cal L}}^{\text{afín}}}=\hat{{\cal L}}$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/aalg/n4.lyx b/aalg/n4.lyx
new file mode 100644
index 0000000..5bd177f
--- /dev/null
+++ b/aalg/n4.lyx
@@ -0,0 +1,4335 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\usepackage{tikz}
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style french
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Una 
+\series bold
+forma bilineal
+\series default
+ en un 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio vectorial 
+\begin_inset Formula $V$
+\end_inset
+
+ es una aplicación 
+\begin_inset Formula $\langle\cdot\rangle:V\times V\rightarrow\mathbb{K}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall u,u_{1},u_{2},v,v_{1},v_{2}\in V,\lambda\in\mathbb{K}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\langle u_{1}+u_{2},v\rangle=\langle u_{1},v\rangle+\langle u_{2},v\rangle$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\langle u,v_{1}+v_{2}\rangle=\langle u,v_{1}\rangle+\langle u,v_{2}\rangle$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\langle\lambda u,v\rangle=\langle u,\lambda v\rangle=\lambda\langle u,v\rangle$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una forma bilineal es 
+\series bold
+simétrica
+\series default
+ si 
+\begin_inset Formula $\forall u,v\in V,\langle u,v\rangle=\langle v,u\rangle$
+\end_inset
+
+, y es 
+\series bold
+alternada
+\series default
+ si 
+\begin_inset Formula $\forall u\in V,\langle u,u\rangle=0$
+\end_inset
+
+.
+ En 
+\begin_inset Formula $\mathbb{K}=\mathbb{R}$
+\end_inset
+
+, una forma bilineal simétrica tal que 
+\begin_inset Formula $\forall u\neq0,\langle u,u\rangle>0$
+\end_inset
+
+ es un 
+\series bold
+producto escalar
+\series default
+.
+ Llamamos 
+\series bold
+espacio bilineal
+\series default
+ o 
+\series bold
+cuadrático
+\series default
+ a un par 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ formado por un espacio vectorial 
+\begin_inset Formula $V$
+\end_inset
+
+ y una forma bilineal simétrica 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en él.
+ Llamamos 
+\begin_inset Formula ${\cal B}(V)$
+\end_inset
+
+ al conjunto de formas bilineales simétricas en 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ una forma bilineal sobre el espacio vectorial 
+\begin_inset Formula $V$
+\end_inset
+
+ con base 
+\begin_inset Formula $(e_{1},\dots,e_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $A:=(a_{ij}:=\langle e_{i},e_{j}\rangle)\in{\cal M}_{n}(\mathbb{K})$
+\end_inset
+
+, entonces si 
+\begin_inset Formula $x=\sum x_{i}e_{i}$
+\end_inset
+
+ e 
+\begin_inset Formula $y=\sum y_{i}e_{i}$
+\end_inset
+
+, se tiene 
+\begin_inset Formula 
+\[
+\langle x,y\rangle=\langle\sum_{i}x_{i}e_{i},\sum_{j}y_{j}e_{j}\rangle=\sum_{i,j}\langle x_{i}e_{i},y_{j}e_{j}\rangle=\sum_{i,j}x_{i}y_{j}a_{ij}
+\]
+
+\end_inset
+
+y por tanto 
+\begin_inset Formula $\langle X,Y\rangle=X^{t}AY$
+\end_inset
+
+.
+ La matriz 
+\begin_inset Formula $A$
+\end_inset
+
+ es simétrica si 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ lo es, y se llama 
+\series bold
+matriz de la forma bilineal
+\series default
+ en la base dada.
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+forma cuadrática
+\series default
+ en un 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio vectorial 
+\begin_inset Formula $V$
+\end_inset
+
+ es una aplicación 
+\begin_inset Formula $q:V\rightarrow\mathbb{K}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall u\in V,\lambda\in\mathbb{K}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $q(\lambda u)=\lambda^{2}q(u)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\langle\cdot\rangle:V\times V\rightarrow K$
+\end_inset
+
+ dada por 
+\begin_inset Formula $\langle u,v\rangle:=\frac{1}{2}(q(u+v)-q(u)-q(v))$
+\end_inset
+
+ es una forma bilineal simétrica en 
+\begin_inset Formula $V$
+\end_inset
+
+, la 
+\series bold
+forma bilineal asociada
+\series default
+ o 
+\series bold
+forma polar
+\series default
+ de 
+\begin_inset Formula $q$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\begin_inset Formula ${\cal Q}(V)$
+\end_inset
+
+ al conjunto de formas cuadráticas en 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ La aplicación 
+\begin_inset Formula ${\cal Q}(V)\rightarrow{\cal B}(V)$
+\end_inset
+
+ que asocia a cada forma cuadrática su forma polar es biyectiva y su inversa
+ asocia a cada forma bilineal simétrica la forma cuadrática dada por 
+\begin_inset Formula $q(u):=\langle u,u\rangle$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $\langle\cdot\rangle\in{\cal B}(V)$
+\end_inset
+
+ y 
+\begin_inset Formula $q(u):=\langle u,u\rangle$
+\end_inset
+
+, es claro que 
+\begin_inset Formula $q(\lambda u)=\lambda^{2}q(u)$
+\end_inset
+
+.
+ Por otra parte,
+\begin_inset Formula 
+\begin{multline*}
+\frac{1}{2}(q(u+v)-q(u)-q(v))=\frac{1}{2}(\langle u+v,u+v\rangle-\langle u,u\rangle-\langle v,v\rangle)=\frac{1}{2}\cdot2\langle u,v\rangle=\langle u,v\rangle
+\end{multline*}
+
+\end_inset
+
+Sean ahora 
+\begin_inset Formula $q$
+\end_inset
+
+ una forma cuadrática, 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ su forma bilineal asociada y 
+\begin_inset Formula $q'\in{\cal Q}(V)$
+\end_inset
+
+ dada por 
+\begin_inset Formula $q'(u)=\langle u,u\rangle$
+\end_inset
+
+, 
+\begin_inset Formula $q'(u)=\langle u,u\rangle=\frac{1}{2}(q(2u)-q(u)-q(u))=\frac{1}{2}(4q(u)-2q(u))=q(u)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Esta correspondencia permite asociar una matriz 
+\begin_inset Formula $A:=(a_{ij})\in{\cal M}_{n}(\mathbb{K})$
+\end_inset
+
+ a una forma cuadrática 
+\begin_inset Formula $q$
+\end_inset
+
+ en un 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio vectorial de dimensión 
+\begin_inset Formula $n<+\infty$
+\end_inset
+
+, pues si 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es la forma polar de 
+\begin_inset Formula $q$
+\end_inset
+
+, 
+\begin_inset Formula $q(u)=\langle u,u\rangle=u^{t}Au$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Cambios de base
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ un espacio bilineal, 
+\begin_inset Formula ${\cal C}:=(u_{1},\dots,u_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}:=(v_{1},\dots,v_{n})$
+\end_inset
+
+ bases de 
+\begin_inset Formula $V$
+\end_inset
+
+ donde 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ tiene matrices respectivas 
+\begin_inset Formula $A:=(a_{ij})$
+\end_inset
+
+ y 
+\begin_inset Formula $B:=(b_{ij})$
+\end_inset
+
+, 
+\begin_inset Formula $X$
+\end_inset
+
+ e 
+\begin_inset Formula $Y$
+\end_inset
+
+ las matrices columna de las coordenadas de dos vectores en la base 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+, 
+\begin_inset Formula $X'$
+\end_inset
+
+ e 
+\begin_inset Formula $Y'$
+\end_inset
+
+ las de los mismos vectores en la base 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ y 
+\begin_inset Formula $P$
+\end_inset
+
+ la matriz de cambio de base de 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ a 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $X=PX'$
+\end_inset
+
+ e 
+\begin_inset Formula $Y=PY'$
+\end_inset
+
+, luego 
+\begin_inset Formula $X^{t}AY=(PX')^{t}A(PY')=(X')^{t}(P^{t}AP)Y'$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $B=P^{t}AP$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Standard
+Dos matrices 
+\begin_inset Formula $A,B\in{\cal M}_{n}(\mathbb{K})$
+\end_inset
+
+ son 
+\series bold
+congruentes
+\series default
+ si existe una matriz invertible 
+\begin_inset Formula $P$
+\end_inset
+
+ tal que 
+\begin_inset Formula $B=P^{t}AP$
+\end_inset
+
+, y escribimos 
+\begin_inset Formula $A\sim B$
+\end_inset
+
+.
+ Esta es una relación de equivalencia.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+sremember{AlgL}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $A,B\in M_{n}(K)$
+\end_inset
+
+ son 
+\series bold
+semejantes
+\series default
+ si 
+\begin_inset Formula $\exists P\in M_{n}(K):B=P^{-1}AP$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+eremember
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Dos formas bilineales 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en 
+\begin_inset Formula $V$
+\end_inset
+
+ y 
+\begin_inset Formula $\langle\cdot\rangle'$
+\end_inset
+
+ en 
+\begin_inset Formula $V'$
+\end_inset
+
+ son 
+\series bold
+equivalentes
+\series default
+, escrito 
+\begin_inset Formula $\langle\cdot\rangle\sim\langle\cdot\rangle'$
+\end_inset
+
+, si existen bases respectivas de 
+\begin_inset Formula $V$
+\end_inset
+
+ y 
+\begin_inset Formula $V'$
+\end_inset
+
+ respecto de las cuales las formas bilineales tiene la misma matriz asociada.
+\end_layout
+
+\begin_layout Standard
+Dos espacios bilineales 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ y 
+\begin_inset Formula $(V',\langle\cdot\rangle')$
+\end_inset
+
+ son 
+\series bold
+isométricos
+\series default
+ si existe un isomorfismo 
+\begin_inset Formula $f:V\rightarrow V'$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall u,v\in V,\langle u,v\rangle=\langle f(u),f(v)\rangle'$
+\end_inset
+
+, y decimos que 
+\begin_inset Formula $f$
+\end_inset
+
+ es una 
+\series bold
+isometría
+\series default
+.
+\end_layout
+
+\begin_layout Standard
+Dados dos espacios bilineales 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ y 
+\begin_inset Formula $(V',\langle\cdot\rangle')$
+\end_inset
+
+, si 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ son bases respectivas de 
+\begin_inset Formula $V$
+\end_inset
+
+ y 
+\begin_inset Formula $V'$
+\end_inset
+
+ y 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $A'$
+\end_inset
+
+ son las matrices respectivas de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ y 
+\begin_inset Formula $\langle\cdot\rangle'$
+\end_inset
+
+ respecto de 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+A\sim A'\iff\langle\cdot\rangle\sim\langle\cdot\rangle'\iff(V,\langle\cdot\rangle),(V',\langle\cdot\rangle')\text{ isométricos}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Existe 
+\begin_inset Formula $P$
+\end_inset
+
+ invertible tal que 
+\begin_inset Formula $A'=P^{t}AP$
+\end_inset
+
+, luego en la base 
+\begin_inset Formula ${\cal B}''$
+\end_inset
+
+ en la que 
+\begin_inset Formula $P$
+\end_inset
+
+ es matriz de cambio de 
+\begin_inset Formula ${\cal B}''$
+\end_inset
+
+ a 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+, 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ tiene matriz 
+\begin_inset Formula $A'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $2\implies3]$
+\end_inset
+
+ Si 
+\begin_inset Formula ${\cal C}=:(v_{1},\dots,v_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal C}'=:(v'_{1},\dots,v'_{n})$
+\end_inset
+
+ son bases en las que 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ y 
+\begin_inset Formula $\langle\cdot\rangle'$
+\end_inset
+
+ tienen la misma matriz asociada 
+\begin_inset Formula $C:=(c_{ij})$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\langle v_{i},v_{j}\rangle=c_{ij}=\langle v'_{i},v'_{j}\rangle$
+\end_inset
+
+, luego el isomorfismo 
+\begin_inset Formula $v_{i}\mapsto v'_{i}$
+\end_inset
+
+ es una isometría.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $3\implies1]$
+\end_inset
+
+ Sea 
+\begin_inset Formula $f:V\rightarrow V'$
+\end_inset
+
+ una isometría y 
+\begin_inset Formula ${\cal B}:=(v_{1},\dots,v_{n})$
+\end_inset
+
+, entonces 
+\begin_inset Formula ${\cal B}':=(f(v_{1}),\dots,f(v_{n}))$
+\end_inset
+
+ es una base de 
+\begin_inset Formula $V'$
+\end_inset
+
+ y, como 
+\begin_inset Formula $\langle v_{i},v_{j}\rangle=\langle f(v_{i}),f(v_{j})\rangle'=:c_{ij}$
+\end_inset
+
+, ambas formas bilineales tienen la misma matriz 
+\begin_inset Formula $C:=(c_{ij})$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $A\sim C\sim A'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Ortogonalidad
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ un espacio bilineal y 
+\begin_inset Formula $E$
+\end_inset
+
+ un subespacio de 
+\begin_inset Formula $V$
+\end_inset
+
+, llamamos 
+\series bold
+subespacio ortogonal
+\series default
+ a 
+\begin_inset Formula $E$
+\end_inset
+
+ al subespacio 
+\begin_inset Formula $E^{\bot}:=\{v\in V:\forall e\in E,\langle v,e\rangle=0\}$
+\end_inset
+
+.
+ Dos vectores 
+\begin_inset Formula $u,v\in V$
+\end_inset
+
+ son 
+\series bold
+ortogonales
+\series default
+, 
+\series bold
+perpendiculares
+\series default
+ o 
+\series bold
+conjugados
+\series default
+ si 
+\begin_inset Formula $\langle u,v\rangle=0$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+radical
+\series default
+ de 
+\begin_inset Formula $V$
+\end_inset
+
+ a 
+\begin_inset Formula $Rad(V):=V^{\bot}$
+\end_inset
+
+.
+ Una forma bilineal simétrica 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en 
+\begin_inset Formula $V$
+\end_inset
+
+ es 
+\series bold
+no degenerada
+\series default
+ si 
+\begin_inset Formula $Rad(V)=0$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es la matriz asociada a 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en la base 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ de 
+\begin_inset Formula $V$
+\end_inset
+
+, 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es no degenerada si y sólo si 
+\begin_inset Formula $|A|\neq0$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula ${\cal B}=:(u_{1},\dots,u_{n})$
+\end_inset
+
+, un vector 
+\begin_inset Formula 
+\begin{multline*}
+u:=\sum\alpha_{i}u_{i}\in Rad(V)\iff\langle u,v\rangle=0\forall v\in V\iff\langle u,u_{i}\rangle=0\forall i\iff\\
+\iff\forall i,\left(\begin{array}{ccccc}
+0 & \cdots & \overset{\underset{\downarrow}{i}}{1} & \cdots & 0\end{array}\right)A\left(\begin{array}{c}
+x_{1}\\
+\vdots\\
+x_{n}
+\end{array}\right)=0
+\end{multline*}
+
+\end_inset
+
+Por tanto el radical está formado por los vectores cuyas coordenadas constituyen
+ el núcleo de 
+\begin_inset Formula $A$
+\end_inset
+
+, que se reduce al vector 0 si y sólo si 
+\begin_inset Formula $|A|\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Un vector es 
+\series bold
+isótropo
+\series default
+ si 
+\begin_inset Formula $\langle u,u\rangle=0$
+\end_inset
+
+, y un subespacio 
+\begin_inset Formula $U\leq V$
+\end_inset
+
+ es (
+\series bold
+totalmente
+\series default
+) 
+\series bold
+isótropo
+\series default
+ si todo vector de 
+\begin_inset Formula $U$
+\end_inset
+
+ es isótropo, y es 
+\series bold
+anisótropo
+\series default
+ si no contiene vectores isótropos no nulos.
+ Si todos los vectores son isótropos, entonces 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es idénticamente nula, pues en tal caso 
+\begin_inset Formula $0=\langle u+v,u+v\rangle=\langle u,u\rangle+\langle v,v\rangle+2\langle u,v\rangle=2\langle u,v\rangle$
+\end_inset
+
+ para cualesquiera 
+\begin_inset Formula $u,v\in V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Diagonalización
+\end_layout
+
+\begin_layout Standard
+Dado un espacio bilineal 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ y 
+\begin_inset Formula $E\leq V$
+\end_inset
+
+, si 
+\begin_inset Formula $\langle\cdot\rangle|_{E}$
+\end_inset
+
+ es no degenerada entonces 
+\begin_inset Formula $V=E\oplus E^{\bot}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ La suma es directa porque 
+\begin_inset Formula $E\cap E^{\bot}=Rad(E)=0$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula ${\cal B}:=(e_{1},\dots,e_{m})$
+\end_inset
+
+ una base de 
+\begin_inset Formula $E$
+\end_inset
+
+ y 
+\begin_inset Formula $A\in{\cal M}_{m}(\mathbb{R})$
+\end_inset
+
+ la matriz de 
+\begin_inset Formula $\langle\cdot\rangle|_{E}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $|A|\neq0$
+\end_inset
+
+ y, dado 
+\begin_inset Formula $u\in V$
+\end_inset
+
+, el sistema
+\begin_inset Formula 
+\[
+A\left(\begin{array}{c}
+x_{1}\\
+\vdots\\
+x_{m}
+\end{array}\right)=\left(\begin{array}{c}
+\langle u,e_{1}\rangle\\
+\vdots\\
+\langle u,e_{m}\rangle
+\end{array}\right)
+\]
+
+\end_inset
+
+tiene solución única y 
+\begin_inset Formula $x:=\sum x_{i}e_{i}\in E$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $v:=u-x$
+\end_inset
+
+, 
+\begin_inset Formula $v\in E^{\bot}\iff\forall i,\langle e_{i},v\rangle=0$
+\end_inset
+
+, pero 
+\begin_inset Formula $\langle e_{i},v\rangle=\langle e_{i},u\rangle-\sum_{j}x_{j}\langle e_{i},e_{j}\rangle=\langle e_{i},u\rangle-\sum_{j}a_{ij}x_{j}=0$
+\end_inset
+
+, luego todo vector 
+\begin_inset Formula $u\in V$
+\end_inset
+
+ se puede descomponer en un vector 
+\begin_inset Formula $x\in E$
+\end_inset
+
+ y otro 
+\begin_inset Formula $v\in E^{\bot}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, para todo espacio bilineal 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ existe una base ortogonal, y por tanto la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es siempre de la forma 
+\begin_inset Formula $\text{diag}(d_{1},\dots,d_{m},0,\dots,0)$
+\end_inset
+
+ (matriz diagonal) con 
+\begin_inset Formula $d_{i}\neq0\forall i\in\{1,\dots,m\}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $V$
+\end_inset
+
+ tiene dimensión 1 toda base es ortogonal.
+ Supongamos que la dimensión de 
+\begin_inset Formula $V$
+\end_inset
+
+ es 
+\begin_inset Formula $n>1$
+\end_inset
+
+ y el teorema se cumple para dimensión 
+\begin_inset Formula $n-1$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es nula, toda base es ortogonal.
+ De lo contrario existe un vector 
+\begin_inset Formula $e_{1}$
+\end_inset
+
+ no isótropo y, si 
+\begin_inset Formula $E:=<e_{1}>$
+\end_inset
+
+, 
+\begin_inset Formula $\langle\cdot\rangle|_{E}$
+\end_inset
+
+ es no degenerada, por lo que tenemos 
+\begin_inset Formula $V=E\oplus E^{\bot}$
+\end_inset
+
+ y, por la hipótesis de inducción, 
+\begin_inset Formula $E^{\bot}$
+\end_inset
+
+ tiene una base 
+\begin_inset Formula $(e_{2},\dots,e_{n})$
+\end_inset
+
+ ortogonal y la base 
+\begin_inset Formula $(e_{1},\dots,e_{n})$
+\end_inset
+
+ de 
+\begin_inset Formula $V$
+\end_inset
+
+ también lo es.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $A,B\in{\cal M}_{n}(\mathbb{K})$
+\end_inset
+
+ son congruentes si y sólo si una se puede obtener de la otra por operaciones
+ elementales, las mismas por filas que por columnas.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Existe 
+\begin_inset Formula $P$
+\end_inset
+
+ invertible tal que 
+\begin_inset Formula $P^{t}AP=B$
+\end_inset
+
+.
+ Al ser invertible debe ser producto de matrices elementales, 
+\begin_inset Formula $P^{t}=:E_{1}\cdots E_{k}$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $B=E_{k}\cdots E_{1}AE_{1}^{t}\cdots E_{k}^{t}$
+\end_inset
+
+, pero la traspuesta de una matriz elemental que representa una operación
+ por filas representa la misma operación por columnas.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $B=E_{k}\cdots E_{1}AE_{1}^{t}\cdots E_{k}^{t}$
+\end_inset
+
+, basta tomar 
+\begin_inset Formula $P:=E_{1}^{t}\cdots E_{k}^{t}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así, para obtener a partir de una matriz simétrica 
+\begin_inset Formula $A$
+\end_inset
+
+ una matriz diagonal congruente:
+\end_layout
+
+\begin_layout Standard
+
+\family sans
+\begin_inset Box Frameless
+position "t"
+hor_pos "c"
+has_inner_box 1
+inner_pos "t"
+use_parbox 0
+use_makebox 0
+width "100col%"
+special "none"
+height "1in"
+height_special "totalheight"
+thickness "0.4pt"
+separation "3pt"
+shadowsize "4pt"
+framecolor "black"
+backgroundcolor "none"
+status open
+
+\begin_layout Plain Layout
+
+\family sans
+\series bold
+operación
+\series default
+ diagonalizar(var 
+\begin_inset Formula $A$
+\end_inset
+
+: 
+\begin_inset Formula ${\cal M}_{n}(\mathbb{K})$
+\end_inset
+
+)
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+si
+\series default
+ 
+\begin_inset Formula $n>1$
+\end_inset
+
+ 
+\series bold
+y
+\series default
+ 
+\begin_inset Formula $A\neq0$
+\end_inset
+
+ 
+\series bold
+entonces
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+si
+\series default
+ la primera columna es no nula 
+\series bold
+entonces
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+si
+\series default
+ no hay ningún 
+\begin_inset Formula $i$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{ii}\neq0$
+\end_inset
+
+ 
+\series bold
+entonces
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Sumar a la fila 
+\begin_inset Formula $1$
+\end_inset
+
+ la fila 
+\begin_inset Formula $i$
+\end_inset
+
+, para algún 
+\begin_inset Formula $i$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{i1}\neq0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Sumar a la columna 
+\begin_inset Formula $1$
+\end_inset
+
+ la columna 
+\begin_inset Formula $i$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+finsi
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Tomar 
+\begin_inset Formula $i$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{ii}\neq0$
+\end_inset
+
+; intercambiar filas 1 e 
+\begin_inset Formula $i$
+\end_inset
+
+ y columnas 1 e 
+\begin_inset Formula $i$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Hacer ceros en la primera columna con operaciones fila
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Hacer las mismas operaciones columna 
+\begin_inset Formula $//$
+\end_inset
+
+ 
+\emph on
+Lo que hace ceros en la primera fila
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+finsi
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+diagonalizar(A[2..n,2..n])
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+finsi
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Para recordar los cambios, escribimos una matriz identidad al lado de 
+\begin_inset Formula $A$
+\end_inset
+
+ y registramos en ella las operaciones elementales de filas, o bien las
+ de columnas.
+ La 
+\series bold
+diagonalización por completación de cuadrados
+\series default
+ es igual pero trabajando con la forma cuadrática:
+\end_layout
+
+\begin_layout Standard
+
+\family sans
+\begin_inset Box Frameless
+position "t"
+hor_pos "c"
+has_inner_box 1
+inner_pos "t"
+use_parbox 0
+use_makebox 0
+width "100col%"
+special "none"
+height "1in"
+height_special "totalheight"
+thickness "0.4pt"
+separation "3pt"
+shadowsize "4pt"
+framecolor "black"
+backgroundcolor "none"
+status open
+
+\begin_layout Plain Layout
+
+\family sans
+\series bold
+operación
+\series default
+ diagonalizar(var 
+\begin_inset Formula $q$
+\end_inset
+
+: 
+\begin_inset Formula ${\cal Q}(\mathbb{K}^{n})$
+\end_inset
+
+) 
+\begin_inset Formula $//$
+\end_inset
+
+ 
+\emph on
+Trabajamos con coordenadas
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+si
+\series default
+ 
+\begin_inset Formula $n>1$
+\end_inset
+
+ 
+\series bold
+y
+\series default
+ 
+\begin_inset Formula $q\neq0$
+\end_inset
+
+ 
+\series bold
+entonces
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+si
+\series default
+ el valor de 
+\begin_inset Formula $q$
+\end_inset
+
+ depende de 
+\begin_inset Formula $x_{1}$
+\end_inset
+
+ 
+\series bold
+entonces
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+si
+\series default
+ no hay ningún elemento 
+\begin_inset Formula $a_{ii}x_{i}^{2}$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{ii}\neq0$
+\end_inset
+
+ 
+\series bold
+entonces
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Tomar un término 
+\begin_inset Formula $a_{ij}x_{i}x_{j}$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{ij}\neq0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Hacer el cambio 
+\begin_inset Formula $x_{i}=:x'_{i}+x'_{j}$
+\end_inset
+
+, 
+\begin_inset Formula $x_{j}=:x'_{i}-x'_{j}$
+\end_inset
+
+ y 
+\begin_inset Formula $x_{k}=:x'_{k},k\neq i,j$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+finsi
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Tomar 
+\begin_inset Formula $i$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{ii}\neq0$
+\end_inset
+
+; intercambiar 
+\begin_inset Formula $x_{i}$
+\end_inset
+
+ y 
+\begin_inset Formula $x_{1}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Tomar 
+\begin_inset Formula $p$
+\end_inset
+
+ y 
+\begin_inset Formula $r$
+\end_inset
+
+ de 
+\begin_inset Formula $q(x_{1},\dots,x_{n})=:a_{11}x_{1}^{2}+x_{1}p(x_{2},\dots,x_{n})+r(x_{2},\dots,x_{n})$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Reescribir 
+\begin_inset Formula $q$
+\end_inset
+
+ como 
+\begin_inset Formula $a_{11}(x_{1}+\frac{p(x_{2},\dots,x_{n})}{2a_{11}})^{2}-\frac{p(x_{2},\dots,x_{n})}{4a_{11}}+r(x_{2},\dots,x_{n})$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+Hacer el cambio 
+\begin_inset Formula $x'_{1}:=x_{1}+\frac{p(x_{2},\dots,x_{n})}{2a_{11}}$
+\end_inset
+
+ y 
+\begin_inset Formula $x'_{j}:=x_{j},j\neq1$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+finsi
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+diagonalizar(
+\begin_inset Formula $q(0,x_{2},\dots,x_{n})$
+\end_inset
+
+)
+\end_layout
+
+\begin_layout Plain Layout
+
+\family sans
+\begin_inset space \hspace{}
+\length 5ex
+\end_inset
+
+
+\series bold
+finsi
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, todo endomorfismo simétrico 
+\begin_inset Formula $f:V\rightarrow V$
+\end_inset
+
+ diagonaliza con una base ortonormal de vectores propios.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{m}$
+\end_inset
+
+ los valores propios de 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $U:=V_{(\alpha_{1})}\oplus\dots\oplus V_{(\alpha_{m})}$
+\end_inset
+
+, siendo 
+\begin_inset Formula $V_{(\alpha_{i})}$
+\end_inset
+
+ el subespacio propio correspondiente al valor propio 
+\begin_inset Formula $\alpha_{i}$
+\end_inset
+
+.
+ Para ver que 
+\begin_inset Formula $U=V$
+\end_inset
+
+, primero observamos que 
+\begin_inset Formula $f(U)\subseteq U$
+\end_inset
+
+, pues si 
+\begin_inset Formula $v_{i}\in V_{(\alpha_{i})}$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f(\sum\lambda_{i}v_{i})=\sum\lambda_{i}\alpha_{i}v_{i}\in U$
+\end_inset
+
+.
+ Por otro lado, si 
+\begin_inset Formula $u\in U$
+\end_inset
+
+ y 
+\begin_inset Formula $w\in U^{\bot}$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f(u)\in U$
+\end_inset
+
+ y 
+\begin_inset Formula $\langle f(w),u\rangle=0=\langle w,f(u)\rangle$
+\end_inset
+
+.
+ Consideremos el endomorfismo simétrico 
+\begin_inset Formula $f|_{U^{\bot}}$
+\end_inset
+
+.
+ Como todos los vectores propios de 
+\begin_inset Formula $f$
+\end_inset
+
+ están en 
+\begin_inset Formula $U$
+\end_inset
+
+, el endomorfismo 
+\begin_inset Formula $f|_{U^{\bot}}$
+\end_inset
+
+ no tiene vectores propios y por tanto 
+\begin_inset Formula $U^{\bot}=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $U=V$
+\end_inset
+
+.
+ Si tomamos una base ortonormal de cada 
+\begin_inset Formula $V_{(\alpha_{i})}$
+\end_inset
+
+, al juntarlas obtenemos una base de 
+\begin_inset Formula $V$
+\end_inset
+
+ ortonormal.
+\end_layout
+
+\begin_layout Standard
+De aquí que toda matriz simétrica real 
+\begin_inset Formula $A\in{\cal M}_{m\times n}(\mathbb{R})$
+\end_inset
+
+ admite una matriz ortogonal 
+\begin_inset Formula $P$
+\end_inset
+
+ tal que 
+\begin_inset Formula $P^{-1}AP=P^{t}AP$
+\end_inset
+
+ es diagonal.
+\end_layout
+
+\begin_layout Section
+Rango
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ un espacio bilineal y 
+\begin_inset Formula $A$
+\end_inset
+
+ la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en cierta base, llamamos 
+\series bold
+rango
+\series default
+ de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ a 
+\begin_inset Formula $\text{rg}(\langle\cdot\rangle):=\text{rg}(A)=\dim(V)-\dim Rad(\langle\cdot\rangle)$
+\end_inset
+
+.
+ Dadas las formas bilineales 
+\begin_inset Formula $\langle\cdot\rangle\sim\langle\cdot\rangle'$
+\end_inset
+
+ en un 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio vectorial 
+\begin_inset Formula $V$
+\end_inset
+
+, 
+\begin_inset Formula $\text{rg}(\langle\cdot\rangle)=\text{rg}(\langle\cdot\rangle')$
+\end_inset
+
+ y, si 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ son las matrices respectivas de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ y 
+\begin_inset Formula $\langle\cdot\rangle'$
+\end_inset
+
+, existe 
+\begin_inset Formula $\lambda\in\mathbb{K}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $|B|=\lambda^{2}|A|$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $A\sim B$
+\end_inset
+
+, existe 
+\begin_inset Formula $P$
+\end_inset
+
+ invertible tal que 
+\begin_inset Formula $B=P^{t}AP$
+\end_inset
+
+, luego 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ tienen igual rango y 
+\begin_inset Formula $|B|=\lambda^{2}|A|$
+\end_inset
+
+ con 
+\begin_inset Formula $\lambda:=|P|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+%
+\backslash
+begin{sloppypar}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Un cuerpo es 
+\series bold
+algebraicamente cerrado
+\series default
+ si cualquier polinomio con coeficientes en 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+ tiene todas sus raíces en 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+.
+ Como 
+\series bold
+teorema
+\series default
+, dos formas bilineales simétricas 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ y 
+\begin_inset Formula $\langle\cdot\rangle'$
+\end_inset
+
+ con igual rango en un 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio vectorial 
+\begin_inset Formula $V$
+\end_inset
+
+, siendo 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+ algebraicamente cerrado, son equivalentes.
+ 
+\series bold
+Demostración:
+\series default
+ Sabemos que en cierta base, la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es 
+\begin_inset Formula $D:=\text{diag}(d_{1},\dots,d_{m},0,\dots,0)$
+\end_inset
+
+, siendo 
+\begin_inset Formula $m:=\text{rg}(\langle\cdot\rangle)$
+\end_inset
+
+, con 
+\begin_inset Formula $d_{1},\dots,d_{m}\neq0$
+\end_inset
+
+.
+ Tomando la matriz in
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+ver
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+ti
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+ble 
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula 
+\[
+P:=\text{diag}(\frac{1}{\sqrt{d_{1}}},\dots,\frac{1}{\sqrt{d_{m}}},1,\dots,1)
+\]
+
+\end_inset
+
+ tenemos que 
+\begin_inset Formula 
+\[
+P^{t}DP=\text{diag}(\overset{m}{\overbrace{1,\dots,1}},0,\dots,0)
+\]
+
+\end_inset
+
+Haciendo lo mismo con 
+\begin_inset Formula $\langle\cdot\rangle'$
+\end_inset
+
+ obtenemos que su matriz en cierta base también es congruente con esta misma
+ matriz, luego ambas son congruentes.
+\end_layout
+
+\begin_layout Standard
+Por tanto, dadas dos matrices simétricas 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ sobre un cuerpo algebraicamente cerrado, 
+\begin_inset Formula $A\sim B\iff\text{rg}(A)=\text{rg}(B)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Cuerpos ordenados y signatura
+\end_layout
+
+\begin_layout Standard
+Un cuerpo 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+ es 
+\series bold
+ordenado
+\series default
+ si existe un 
+\begin_inset Formula $P\subseteq\mathbb{K}$
+\end_inset
+
+, cuyos elementos se llaman 
+\series bold
+positivos
+\series default
+, tal que:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\mathbb{K}=P\dot{\cup}\{0\}\dot{\cup}-P$
+\end_inset
+
+.
+ A los elementos de 
+\begin_inset Formula $-P:=\{-x\}_{x\in P}$
+\end_inset
+
+ los llamamos 
+\series bold
+negativos
+\series default
+.
+\end_layout
+
+\begin_layout Enumerate
+Para 
+\begin_inset Formula $x,y\in P$
+\end_inset
+
+, 
+\begin_inset Formula $x+y,xy\in P$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Por ejemplo, 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+ son ordenados, mientras que 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ no lo es.
+ Escribimos 
+\begin_inset Formula $x\geq0$
+\end_inset
+
+ si 
+\begin_inset Formula $x$
+\end_inset
+
+ es positivo o 
+\begin_inset Formula $x=0$
+\end_inset
+
+, y definimos la relación de orden total 
+\begin_inset Formula $x\leq y:\iff y-x\geq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una forma bilineal 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en un 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+-espacio vectorial 
+\begin_inset Formula $V$
+\end_inset
+
+, siendo 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+ ordenado, es:
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Semidefinida positiva
+\series default
+ si 
+\begin_inset Formula $\forall u\in V,\langle u,u\rangle\geq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Semidefinida negativa
+\series default
+ si 
+\begin_inset Formula $\forall u\in V,\langle u,u\rangle\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Definida positiva
+\series default
+ si 
+\begin_inset Formula $\forall u\neq0,\langle u,u\rangle>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Definida negativa
+\series default
+ si 
+\begin_inset Formula $\forall u\neq0,\langle u,u\rangle<0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Las mismas definiciones se aplican a una forma cuadrática.
+ Sean 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ un espacio bilineal sobre un cuerpo 
+\begin_inset Formula $\mathbb{\mathbb{K}}$
+\end_inset
+
+, 
+\begin_inset Formula $A:=(a_{ij})$
+\end_inset
+
+ la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en cierta base 
+\begin_inset Formula ${\cal C}:=(e_{1},\dots,e_{n})$
+\end_inset
+
+ y definimos
+\begin_inset Formula 
+\[
+d_{1}=a_{11},d_{2}=\left|\begin{array}{cc}
+a_{11} & a_{12}\\
+a_{21} & a_{22}
+\end{array}\right|,\dots,d_{n}=|A|
+\]
+
+\end_inset
+
+ Si los 
+\begin_inset Formula $d_{1},\dots,d_{n}$
+\end_inset
+
+ son todos no nulos, hay una base en que la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es 
+\begin_inset Formula $\text{diag}(d_{1},\frac{d_{2}}{d_{1}},\dots,\frac{d_{n}}{d_{n-1}})$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $E:=<e_{1},\dots,e_{n-1}>$
+\end_inset
+
+, la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en 
+\begin_inset Formula $E$
+\end_inset
+
+ es la matriz 
+\begin_inset Formula $A$
+\end_inset
+
+ sin la última fila y columna, cuyo determinante es 
+\begin_inset Formula $d_{n-1}\neq0$
+\end_inset
+
+, luego es no degenerada y 
+\begin_inset Formula $V=E\oplus E^{\bot}$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $v\in E^{\bot}\backslash\{0\}$
+\end_inset
+
+, 
+\begin_inset Formula $(e_{1},\dots,e_{n-1},v)$
+\end_inset
+
+ es una base de 
+\begin_inset Formula $V$
+\end_inset
+
+, y la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en esta base es
+\begin_inset Formula 
+\[
+B:=\left(\begin{array}{cccc}
+a_{11} & \cdots & a_{1,n-1} & 0\\
+\vdots & \ddots & \vdots & \vdots\\
+a_{n-1,1} & \cdots & a_{n-1,n-1} & 0\\
+0 & \cdots & 0 & b
+\end{array}\right)
+\]
+
+\end_inset
+
+Tenemos 
+\begin_inset Formula $A\sim B$
+\end_inset
+
+, luego existe 
+\begin_inset Formula $P$
+\end_inset
+
+ invertible con 
+\begin_inset Formula $A=P^{t}BP$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $\lambda:=|P|$
+\end_inset
+
+, 
+\begin_inset Formula $|A|=\lambda^{2}|B|$
+\end_inset
+
+ y 
+\begin_inset Formula $d_{n}=\lambda^{2}d_{n-1}b$
+\end_inset
+
+, y entonces la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en la base 
+\begin_inset Formula $(e_{1},\dots,e_{n-1},w:=\lambda v)$
+\end_inset
+
+ es como 
+\begin_inset Formula $B$
+\end_inset
+
+ pero cambiando 
+\begin_inset Formula $b$
+\end_inset
+
+ por 
+\begin_inset Formula $\frac{d_{n}}{d_{n-1}}$
+\end_inset
+
+.
+ El resultado sigue por inducción.
+\end_layout
+
+\begin_layout Standard
+De aquí que, si además 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+ es ordenado, la forma bilineal es definida positiva si y sólo si 
+\begin_inset Formula $d_{1},\dots,d_{n}>0$
+\end_inset
+
+, y es definida negativa si y sólo si 
+\begin_inset Formula $d_{1}<0$
+\end_inset
+
+, 
+\begin_inset Formula $d_{2}>0$
+\end_inset
+
+, 
+\begin_inset Formula $d_{3}<0$
+\end_inset
+
+, etc.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+teorema de Sylvester
+\series default
+ afirma que si 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ es un espacio bilineal sobre un cuerpo 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+ ordenado, 
+\begin_inset Formula $V$
+\end_inset
+
+ se descompone en suma directa ortogonal como 
+\begin_inset Formula $V=V_{+}\oplus V_{-}\oplus V_{0}$
+\end_inset
+
+, donde 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ restringida a 
+\begin_inset Formula $V_{+}$
+\end_inset
+
+, a 
+\begin_inset Formula $V_{-}$
+\end_inset
+
+ y a 
+\begin_inset Formula $V_{0}$
+\end_inset
+
+ es definida positiva, definida negativa y nula, respectivamente.
+ Además, 
+\begin_inset Formula $p:=\dim(V_{+})$
+\end_inset
+
+ y 
+\begin_inset Formula $m:=\dim(V_{-})$
+\end_inset
+
+ son únicos, y al par 
+\begin_inset Formula $(p,m)$
+\end_inset
+
+ lo llamamos la 
+\series bold
+signatura
+\series default
+ de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula ${\cal C}:=(e_{1},\dots,e_{n})$
+\end_inset
+
+ una base de 
+\begin_inset Formula $V$
+\end_inset
+
+ donde la matriz de la forma bilineal es 
+\begin_inset Formula 
+\[
+\text{diag}(d_{1},\dots,d_{p},d_{p+1},\dots,d_{p+m},0,\dots,0)
+\]
+
+\end_inset
+
+con 
+\begin_inset Formula $d_{i}>0$
+\end_inset
+
+ para 
+\begin_inset Formula $i\in\{1,\dots,p\}$
+\end_inset
+
+ y 
+\begin_inset Formula $d_{i}<0$
+\end_inset
+
+ para 
+\begin_inset Formula $i\in\{p+1,\dots,p+m\}$
+\end_inset
+
+.
+ Es claro que la descomposición dada por 
+\begin_inset Formula 
+\begin{eqnarray*}
+V_{+}:=<e_{1},\dots,e_{p}>, & V_{-}:=<e_{p+1},\dots,e_{p+m}>, & V_{0}:=<e_{p+m+1},\dots,e_{n}>
+\end{eqnarray*}
+
+\end_inset
+
+ cumple las condiciones.
+ Para la unicidad, supongamos 
+\begin_inset Formula $V=V_{+}\oplus V_{-}\oplus V_{0}=W_{+}\oplus W_{-}\oplus W_{0}$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $\pi_{+}:V\rightarrow V_{+}$
+\end_inset
+
+ la proyección canónica de 
+\begin_inset Formula $V$
+\end_inset
+
+ sobre 
+\begin_inset Formula $V_{+}$
+\end_inset
+
+, 
+\begin_inset Formula $\ker(\pi|_{W_{+}})=\ker(\pi)\cap W_{+}=(V_{-}\oplus V_{0})\cap W_{+}$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $u\in(V_{-}\oplus V_{0})\cap W_{+}$
+\end_inset
+
+, tenemos que 
+\begin_inset Formula $u=u_{-}+u_{0}$
+\end_inset
+
+ con 
+\begin_inset Formula $u_{-}\in V_{-}$
+\end_inset
+
+ y 
+\begin_inset Formula $u_{0}\in V_{0}$
+\end_inset
+
+ y como 
+\begin_inset Formula $u\in W_{+}$
+\end_inset
+
+, 
+\begin_inset Formula $\langle u,u\rangle\geq0$
+\end_inset
+
+, pero 
+\begin_inset Formula 
+\[
+0\leq\langle u,u\rangle=\langle u_{-},u_{-}\rangle+2\langle u_{-},u_{0}\rangle+\langle u_{0},u_{0}\rangle=\langle u_{-},u_{-}\rangle\leq0
+\]
+
+\end_inset
+
+de donde 
+\begin_inset Formula $\langle u,u\rangle=0$
+\end_inset
+
+ y, por ser 
+\begin_inset Formula $u\in W_{+}$
+\end_inset
+
+, 
+\begin_inset Formula $u=0$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $\pi|_{W_{+}}$
+\end_inset
+
+ es inyectiva y 
+\begin_inset Formula $\dim W_{+}\leq\dim V_{+}$
+\end_inset
+
+.
+ De forma parecida podemos probar que 
+\begin_inset Formula $\dim W_{-}\leq\dim V_{-}$
+\end_inset
+
+ y 
+\begin_inset Formula $\dim W_{0}\leq\dim W_{0}$
+\end_inset
+
+, probando el teorema.
+\end_layout
+
+\begin_layout Standard
+De aquí que, si 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ y 
+\begin_inset Formula $(V,\langle\cdot\rangle')$
+\end_inset
+
+ son espacios bilineales isométricos sobre un cuerpo 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+ ordenado, entonces 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ y 
+\begin_inset Formula $\langle\cdot\rangle'$
+\end_inset
+
+ tienen la misma signatura.
+ La 
+\series bold
+ley de inercia de Sylvester
+\series default
+ afirma que, si 
+\begin_inset Formula $\mathbb{K}=\mathbb{R}$
+\end_inset
+
+, el recíproco de esto también se cumple.
+ En efecto, si 
+\begin_inset Formula $(p,m)$
+\end_inset
+
+ es la signatura de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ y 
+\begin_inset Formula $\langle\cdot\rangle'$
+\end_inset
+
+, existe una base de 
+\begin_inset Formula $V$
+\end_inset
+
+ en que la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es 
+\begin_inset Formula $\text{diag}(d_{1},\dots,d_{p},d_{p+1},\dots,d_{p+m},0,\dots,0)$
+\end_inset
+
+, siendo 
+\begin_inset Formula $d_{1},\dots,d_{p}>0$
+\end_inset
+
+ y 
+\begin_inset Formula $d_{p+1},\dots,d_{p+m}<0$
+\end_inset
+
+, pero los positivos difieren de 1 en un cuadrado y los negativos de 
+\begin_inset Formula $-1$
+\end_inset
+
+ en un cuadrado, por lo que hay una base en que la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es 
+\begin_inset Formula 
+\[
+D:=\text{diag}(\overset{p}{\overbrace{1,\dots,1}},\overset{m}{\overbrace{-1,\dots,-1}},0,\dots,0)
+\]
+
+\end_inset
+
+y, análogamente, hay una base en que la matriz de 
+\begin_inset Formula $\langle\cdot\rangle'$
+\end_inset
+
+ es 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Descomposición de Witt
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ un espacio bilineal, llamamos 
+\series bold
+simetría respecto al vector
+\series default
+ 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ no isótropo a la isometría 
+\begin_inset Formula $s_{v}:V\rightarrow V$
+\end_inset
+
+ dada por
+\begin_inset Formula 
+\[
+s_{v}(u)=-u+2\frac{\langle u,v\rangle}{\langle v,v\rangle}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $u,v\in V$
+\end_inset
+
+ no isótropos con 
+\begin_inset Formula $\langle u,u\rangle=\langle v,v\rangle$
+\end_inset
+
+, existe una isometría 
+\begin_inset Formula $f:V\rightarrow V$
+\end_inset
+
+ con 
+\begin_inset Formula $f(u)=v$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $u+v$
+\end_inset
+
+ es no isótropo,
+\begin_inset Formula 
+\[
+s_{u+v}(u)=-u+\frac{2\langle u,u\rangle+2\langle u,v\rangle}{\langle u,u\rangle+\langle v,v\rangle+2\langle u,v\rangle}(u+v)=-u+\frac{2\langle u,u\rangle+2\langle u,v\rangle}{2\langle u,u\rangle+2\langle u,v\rangle}(u+v)=v
+\]
+
+\end_inset
+
+Si 
+\begin_inset Formula $u+v$
+\end_inset
+
+ es isótropo, 
+\begin_inset Formula $u-v$
+\end_inset
+
+ no lo es, pues 
+\begin_inset Formula $\langle u+v,u+v\rangle+\langle u-v,u-v\rangle=4\langle u,u\rangle\neq0$
+\end_inset
+
+, y entonces definimos 
+\begin_inset Formula $t(w):=-w$
+\end_inset
+
+ y vemos que 
+\begin_inset Formula $(t\circ s_{u-v})(u)=v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, si 
+\begin_inset Formula $D_{1}:=\text{diag}(a_{1},\dots,a_{r},b_{r+1},\dots,b_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $D_{2}:=\text{diag}(a_{1},\dots,a_{r},c_{r+1},\dots,c_{n})$
+\end_inset
+
+ son matrices con 
+\begin_inset Formula $a_{1},\dots,a_{r}\neq0$
+\end_inset
+
+, si 
+\begin_inset Formula $D_{1}$
+\end_inset
+
+ es congruente con 
+\begin_inset Formula $D_{2}$
+\end_inset
+
+ entonces 
+\begin_inset Formula $\text{diag}(b_{r+1},\dots,b_{n})$
+\end_inset
+
+ lo es con 
+\begin_inset Formula $\text{diag}(c_{r+1},\dots,c_{n})$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Basta ver que esto se cumple con 
+\begin_inset Formula $r=1$
+\end_inset
+
+.
+ Sean 
+\begin_inset Formula $D_{1}=\text{diag}(a,b_{2},\dots,b_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $D_{2}=\text{diag}(a,c_{2},\dots,c_{n})$
+\end_inset
+
+ matrices de una forma bilineal 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en las bases 
+\begin_inset Formula $(u_{1},\dots,u_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $(v_{1},\dots,v_{n})$
+\end_inset
+
+, respectivamente.
+ Entonces 
+\begin_inset Formula $\langle u_{1},u_{1}\rangle=a=\langle v_{1},v_{1}\rangle\neq0$
+\end_inset
+
+ y existe una isometría 
+\begin_inset Formula $s$
+\end_inset
+
+ con 
+\begin_inset Formula $s(u_{1})=v_{1}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $\{s(u_{1}),\dots,s(u_{n})\}$
+\end_inset
+
+ es base ortogonal de 
+\begin_inset Formula $V$
+\end_inset
+
+ y 
+\begin_inset Formula $E:=<s(u_{2}),\dots,s(u_{n})>=<s(u_{1})>^{\bot}=<v_{1}>^{\bot}=<v_{2},\dots,v_{n}>$
+\end_inset
+
+.
+ La matriz de 
+\begin_inset Formula $\langle\cdot\rangle|_{E}$
+\end_inset
+
+ es 
+\begin_inset Formula $\text{diag}(b_{2},\dots,b_{n})$
+\end_inset
+
+ en 
+\begin_inset Formula $(s(u_{2}),\dots,s(u_{n}))$
+\end_inset
+
+ y es 
+\begin_inset Formula $\text{diag}(c_{2},\dots,c_{n})$
+\end_inset
+
+ en 
+\begin_inset Formula $(v_{2},\dots,v_{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+corolario de cancelación de Witt
+\series default
+ afirma que si 
+\begin_inset Formula $U_{1},U_{2}\leq V$
+\end_inset
+
+ son tales que 
+\begin_inset Formula $\langle\cdot\rangle|_{U_{1}}$
+\end_inset
+
+ y 
+\begin_inset Formula $\langle\cdot\rangle|_{U_{2}}$
+\end_inset
+
+ son no degeneradas y 
+\begin_inset Formula $U_{1}$
+\end_inset
+
+ es isométrico a 
+\begin_inset Formula $U_{2}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $U_{1}^{\bot}$
+\end_inset
+
+ es isométrico a 
+\begin_inset Formula $U_{2}^{\bot}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Tenemos 
+\begin_inset Formula $V=U_{1}\oplus U_{1}^{\bot}=U_{2}\oplus U_{2}^{\bot}$
+\end_inset
+
+, existen bases respectivas 
+\begin_inset Formula $(u_{1},\dots,u_{r})$
+\end_inset
+
+ y 
+\begin_inset Formula $(v_{1},\dots,v_{r})$
+\end_inset
+
+ de 
+\begin_inset Formula $U_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $U_{2}$
+\end_inset
+
+ respecto de las cuales la matriz de 
+\begin_inset Formula $\langle\cdot\rangle|_{U_{1}}$
+\end_inset
+
+ y de 
+\begin_inset Formula $\langle\cdot\rangle|_{U_{2}}$
+\end_inset
+
+ es 
+\begin_inset Formula $\text{diag}(a_{1},\dots,a_{r})$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{1},\dots,a_{r}\neq0$
+\end_inset
+
+.
+ Sean 
+\begin_inset Formula $(u_{r+1},\dots,u_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $(v_{r+1},\dots,v_{n})$
+\end_inset
+
+ bases respectivas de 
+\begin_inset Formula $U_{1}^{\bot}$
+\end_inset
+
+ y 
+\begin_inset Formula $U_{2}^{\bot}$
+\end_inset
+
+, si 
+\begin_inset Formula $D_{1}:=\text{diag}(a_{1},\dots,a_{r},b_{r+1},\dots,b_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $D_{2}:=\text{diag}(a_{1},\dots,a_{r},c_{r+1},\dots,c_{n})$
+\end_inset
+
+ son las matrices de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ res
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+pec
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+to de 
+\begin_inset Formula $(u_{1},\dots,u_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $(v_{1},\dots,v_{n})$
+\end_inset
+
+, respectivamente, entonces 
+\begin_inset Formula $D_{1}\sim D_{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $\text{diag}(b_{r+1},\dots,b_{n})\sim\text{diag}(c_{r+1},\dots,c_{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+plano hiperbólico
+\series default
+ es un espacio bilineal 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ de dimensión 2 donde 
+\begin_inset Formula $V$
+\end_inset
+
+ contiene vectores isótropos no nulos y 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es no degenerada.
+ Un espacio bilineal 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ de dimensión 2 es un plano hiperbólico si y sólo si existe una base de
+ 
+\begin_inset Formula $V$
+\end_inset
+
+ respecto la cual la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es 
+\begin_inset Formula $\text{diag}(1,-1)$
+\end_inset
+
+.
+ Por tanto todos los planos hiperbólicos sobre un mismo cuerpo son isométricos.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sean 
+\begin_inset Formula $u\neq0$
+\end_inset
+
+ isótropo, 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ tal que 
+\begin_inset Formula $(u,v)$
+\end_inset
+
+ es una base y 
+\begin_inset Formula $v':=\frac{v}{\langle u,v\rangle}$
+\end_inset
+
+, la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en 
+\begin_inset Formula $(u,v')$
+\end_inset
+
+ es
+\begin_inset Formula 
+\[
+A:=\left(\begin{array}{cc}
+0 & 1\\
+1 & a
+\end{array}\right)
+\]
+
+\end_inset
+
+con 
+\begin_inset Formula $a:=\langle v',v'\rangle$
+\end_inset
+
+.
+ Sea ahora 
+\begin_inset Formula $w:=xu+v'$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\langle w,w\rangle=1$
+\end_inset
+
+, entonces 
+\begin_inset Formula $1=\langle xu+v',xu+v'\rangle=x^{2}\langle u,u\rangle+\langle v',v'\rangle+2x\langle u,v'\rangle=a+2x$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $w=\frac{1-a}{2}u+v'$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $w'\in<w>^{\bot}$
+\end_inset
+
+, la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en la base 
+\begin_inset Formula $(w,w')$
+\end_inset
+
+ es
+\begin_inset Formula 
+\[
+B:=\left(\begin{array}{cc}
+1 & 0\\
+0 & b
+\end{array}\right)
+\]
+
+\end_inset
+
+con 
+\begin_inset Formula $b:=\langle w',w'\rangle$
+\end_inset
+
+.
+ Las matrices 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ son congruentes, luego sus determinantes difieren en un cuadrado y 
+\begin_inset Formula $b=-\lambda^{2}$
+\end_inset
+
+ para cierto 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $w''=\frac{w'}{\lambda}$
+\end_inset
+
+, 
+\begin_inset Formula $\langle w'',w''\rangle=-1$
+\end_inset
+
+ y la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ en 
+\begin_inset Formula $(w,w'')$
+\end_inset
+
+ es
+\begin_inset Formula 
+\[
+\left(\begin{array}{cc}
+1 & 0\\
+0 & -1
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si identificamos los vectores con sus coordenadas respecto a la base en
+ la que la matriz de 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es 
+\begin_inset Formula $\text{diag}(1,-1)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $(1,-1)$
+\end_inset
+
+ es isótropo no nulo y, si hubiera un 
+\begin_inset Formula $v:=(v_{1},v_{2})$
+\end_inset
+
+ con 
+\begin_inset Formula $\langle u,v\rangle=0\forall u$
+\end_inset
+
+, en particular 
+\begin_inset Formula $\langle(1,0),v\rangle=v_{1}=0$
+\end_inset
+
+ y 
+\begin_inset Formula $\langle(0,1),v\rangle=-v_{2}=0$
+\end_inset
+
+ y sería 
+\begin_inset Formula $v=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es no degenerada.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $\dim(V)\geq2$
+\end_inset
+
+, 
+\begin_inset Formula $V$
+\end_inset
+
+ contiene vectores isótropos no nulos y 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ es no degenerada, entonces 
+\begin_inset Formula $V$
+\end_inset
+
+ contiene un plano hiperbólico.
+ En efecto, sea 
+\begin_inset Formula $u\neq0$
+\end_inset
+
+ isótropo, existe 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ con 
+\begin_inset Formula $\langle u,v\rangle=0$
+\end_inset
+
+, pues de lo contrario 
+\begin_inset Formula $u\in\text{Rad}(V)=0$
+\end_inset
+
+, y podemos ver que 
+\begin_inset Formula $<u,v>$
+\end_inset
+
+ es un plano hiperbólico.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+teorema de descomposición de Witt
+\series default
+ afirma que, sea 
+\begin_inset Formula $(V,\langle\cdot\rangle)$
+\end_inset
+
+ un espacio bilineal con 
+\begin_inset Formula $\langle\cdot\rangle$
+\end_inset
+
+ no degenerada, entonces 
+\begin_inset Formula 
+\[
+V=:P_{1}\oplus\dots\oplus P_{s}\oplus W
+\]
+
+\end_inset
+
+siendo 
+\begin_inset Formula $P_{1},\dots,P_{k}$
+\end_inset
+
+ planos hiperbólicos y 
+\begin_inset Formula $W$
+\end_inset
+
+ anisótropo, y si 
+\begin_inset Formula $V=Q_{1}\oplus\dots\oplus Q_{t}\oplus W'$
+\end_inset
+
+ es otra descomposición ortogonal de 
+\begin_inset Formula $V$
+\end_inset
+
+ con 
+\begin_inset Formula $Q_{1},\dots,Q_{t}$
+\end_inset
+
+ planos hiperbólicos y 
+\begin_inset Formula $W'$
+\end_inset
+
+ anisótropo, entonces 
+\begin_inset Formula $s=t$
+\end_inset
+
+ y 
+\begin_inset Formula $W$
+\end_inset
+
+ es isométrico a 
+\begin_inset Formula $W'$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+descomposición de Witt
+\series default
+ a cualquiera de este tipo, y llamamos a 
+\begin_inset Formula $s$
+\end_inset
+
+ el 
+\series bold
+índice de Witt
+\series default
+.
+ 
+\series bold
+Demostración:
+\series default
+ Para 
+\begin_inset Formula $\dim(V)=1$
+\end_inset
+
+, si hubiera un vector 
+\begin_inset Formula $u\neq0$
+\end_inset
+
+ isótropo, sería 
+\begin_inset Formula $\langle\lambda u,u\rangle=0$
+\end_inset
+
+ para todo 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\text{Rad}(V)\neq0\#$
+\end_inset
+
+, luego 
+\begin_inset Formula $V$
+\end_inset
+
+ es anisótropo.
+ Si 
+\begin_inset Formula $n:=\dim(V)\geq2$
+\end_inset
+
+ y 
+\begin_inset Formula $V$
+\end_inset
+
+ no es anisótropo, debe contener un plano hiperbólico 
+\begin_inset Formula $P$
+\end_inset
+
+ y 
+\begin_inset Formula $V=P\oplus P^{\bot}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\langle\cdot\rangle|_{P^{\bot}}$
+\end_inset
+
+ es no degenerada y por tanto el resultado sigue por inducción.
+ Para la unicidad, sea 
+\begin_inset Formula $V=P_{1}\oplus\dots\oplus P_{s}\oplus W=Q_{1}\oplus\dots\oplus Q_{t}\oplus W'$
+\end_inset
+
+ y supongamos 
+\begin_inset Formula $t\geq s$
+\end_inset
+
+.
+ Como todos los planos hiperbólicos sobre un mismo cuerpo son isométricos,
+ 
+\begin_inset Formula $P_{1}\oplus\dots\oplus P_{s}$
+\end_inset
+
+ es isométrico a 
+\begin_inset Formula $Q_{1}\oplus\dots\oplus Q_{s}$
+\end_inset
+
+ y, por el teorema de cancelación de Witt, 
+\begin_inset Formula $W$
+\end_inset
+
+ es isométrico a 
+\begin_inset Formula $Q_{s+1}\oplus\dots\oplus Q_{t}\oplus W'$
+\end_inset
+
+.
+ Entonces debe ser 
+\begin_inset Formula $t=s$
+\end_inset
+
+ porque de lo contrario tendríamos un subespacio anisótropo isométrico a
+ uno que no lo es, y por tanto 
+\begin_inset Formula $W$
+\end_inset
+
+ debe ser isométrico a 
+\begin_inset Formula $W'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Cónicas proyectivas y formas cuadráticas
+\end_layout
+
+\begin_layout Standard
+Una 
+\series bold
+cónica proyectiva
+\series default
+ en 
+\begin_inset Formula $\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+ es una clase de equivalencia en el conjunto de polinomios homogéneos de
+ grado 2 en 
+\begin_inset Formula $\mathbb{K}[x,y,z]$
+\end_inset
+
+, o de formas cuadráticas no nulas de dimensión 3, bajo la relación 
+\begin_inset Formula $q\sim q':\iff\exists\lambda\in\mathbb{K}\backslash\{0\}:q'=\lambda q$
+\end_inset
+
+.
+ Escribimos 
+\begin_inset Formula ${\cal C}_{q}:=[q]$
+\end_inset
+
+, y la identificamos con el conjunto de puntos 
+\begin_inset Formula $[a,b,c]$
+\end_inset
+
+ en los que 
+\begin_inset Formula $q(a,b,c)=0$
+\end_inset
+
+.
+ En 
+\begin_inset Formula $\mathbb{P}^{2}(\mathbb{R})$
+\end_inset
+
+, una cónica proyectiva es la ampliación proyectiva de una cónica afín de
+ matriz proyectiva igual a la matriz de la forma cuadrática:
+\end_layout
+
+\begin_layout Itemize
+Dada una elipse de ecuación 
+\begin_inset Formula $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
+\end_inset
+
+, su homogeneización es 
+\begin_inset Formula $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=z^{2}$
+\end_inset
+
+.
+ Los puntos del infinito son aquellos en que 
+\begin_inset Formula $z=0$
+\end_inset
+
+, siendo la única solución cuando 
+\begin_inset Formula $x=y=0$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $[0,0,0]$
+\end_inset
+
+ no existe, la elipse no tiene puntos en el infinito.
+\end_layout
+
+\begin_layout Itemize
+Dada una hipérbola de ecuación 
+\begin_inset Formula $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
+\end_inset
+
+, su homogeneización es 
+\begin_inset Formula $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=z^{2}$
+\end_inset
+
+ y vemos que sus puntos del infinito son aquellos en que 
+\begin_inset Formula $z=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $x=\pm\frac{a}{b}y$
+\end_inset
+
+, con lo que la hipérbola tiene dos puntos del infinito correspondientes
+ a sus asíntotas.
+\end_layout
+
+\begin_layout Itemize
+Dada una parábola de ecuación 
+\begin_inset Formula $y^{2}=2px$
+\end_inset
+
+, su homogeneización es 
+\begin_inset Formula $y^{2}=2pxz$
+\end_inset
+
+, siendo los puntos en el infinito aquellos en que 
+\begin_inset Formula $y=z=0$
+\end_inset
+
+, con lo que la parábola tiene un punto en el infinito correspondiente a
+ su eje.
+\end_layout
+
+\begin_layout Standard
+Dos puntos 
+\begin_inset Formula $P,Q\in\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+ son 
+\series bold
+conjugados
+\series default
+ respecto de una cónica proyectiva de matriz proyectiva 
+\begin_inset Formula $\overline{A}$
+\end_inset
+
+ si 
+\begin_inset Formula $[P]^{t}\overline{A}[Q]=0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $P\in\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+ es un punto 
+\series bold
+singular
+\series default
+ respecto de una cónica proyectiva si es conjugado de cualquier 
+\begin_inset Formula $Q\in\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+, y es 
+\series bold
+regular
+\series default
+ en caso contrario.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula ${\cal Q}$
+\end_inset
+
+ una cónica no degenerada de matriz 
+\begin_inset Formula $\overline{A}$
+\end_inset
+
+, llamamos 
+\series bold
+recta polar
+\series default
+ de 
+\begin_inset Formula $P\in\mathbb{P}^{2}(\mathbb{K})$
+\end_inset
+
+ respecto de 
+\begin_inset Formula ${\cal Q}$
+\end_inset
+
+ a 
+\begin_inset Formula $r_{P}:=\{X\in\mathbb{P}^{2}(\mathbb{K}):[P]^{t}\overline{A}[X]=0\}$
+\end_inset
+
+, y decimos que 
+\begin_inset Formula $P$
+\end_inset
+
+ es el 
+\series bold
+polo
+\series default
+ de la recta 
+\begin_inset Formula $r_{P}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una recta es 
+\series bold
+tangente
+\series default
+ a una cónica si la corta en un único punto.
+ Por el principio de dua
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+li
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+dad del plano proyectivo, podemos describir una cónica mediante los puntos
+ que le pertenecen o como el conjunto de todas sus tangentes.
+\end_layout
+
+\begin_layout Standard
+Una cónica es 
+\series bold
+no degenerada
+\series default
+ si 
+\begin_inset Formula $\Delta:=|\overline{A}|\neq0$
+\end_inset
+
+.
+ Dos cónicas 
+\begin_inset Formula ${\cal C}_{q}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal C}_{q'}$
+\end_inset
+
+ son 
+\series bold
+proyectivamente equivalentes
+\series default
+ si podemos transformar una en la otra mediante un cambio de coordenadas
+ proyectivas, si y sólo si la signatura de la forma bilineal asociada a
+ una es igual u opuesta a la de la otra.
+ Esto resulta en los siguientes tipos de cónicas:
+\end_layout
+
+\begin_layout Standard
+\align center
+\begin_inset Tabular
+<lyxtabular version="3" rows="6" columns="4">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Rango
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Signatura
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Ecuación reducida
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Tipo de cónica
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell multirow="3" alignment="center" valignment="middle" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+3
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $(3,0)/(0,3)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $x^{2}+y^{2}+z^{2}=0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+No degenerada imaginaria
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell multirow="4" alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $(2,1)/(1,2)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $x^{2}+y^{2}-z^{2}=0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+No degenerada real
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell multirow="3" alignment="center" valignment="middle" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+2
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $(2,0)/(0,2)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $x^{2}+y^{2}=0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Punto
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell multirow="4" alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $(1,1)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $x^{2}-y^{2}=0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Par de rectas distintas
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+1
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $(1,0)/(0,1)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $x^{2}=0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Recta doble
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document