From 1f7f9bcc7660fba0827a62c3068d5c7082f025d7 Mon Sep 17 00:00:00 2001 From: Juan Marín Noguera Date: Thu, 20 Feb 2020 20:21:46 +0100 Subject: Otras dos asignaturas --- aalg/n.lyx | 199 +++ aalg/n1.lyx | 3205 +++++++++++++++++++++++++++++++++++++++++++ aalg/n2.lyx | 1570 ++++++++++++++++++++++ aalg/n3.lyx | 1955 +++++++++++++++++++++++++++ aalg/n4.lyx | 4335 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 5 files changed, 11264 insertions(+) create mode 100644 aalg/n.lyx create mode 100644 aalg/n1.lyx create mode 100644 aalg/n2.lyx create mode 100644 aalg/n3.lyx create mode 100644 aalg/n4.lyx (limited to 'aalg') 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 $rr$ +\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 $\Vert\overrightarrow{OP'}\Vert1$ +\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 + + + + + + + +\begin_inset Text + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $\Delta\neq0$ +\end_inset + +: No degenerada +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $\Delta=0$ +\end_inset + +: Degenerada +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $\delta>0$ +\end_inset + +: Ecuación elíptica +\end_layout + +\end_inset + + +\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 + + +\begin_inset Text + +\begin_layout Plain Layout +Punto +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $\delta<0$ +\end_inset + +: Ecuación hiperbólica +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +Hipérbola +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +Dos rectas secantes +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $\delta=0$ +\end_inset + +: Ecuación parabólica +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +Parábola +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +Recta doble, o dos rectas paralelas reales o imaginarias +\end_layout + +\end_inset + + + + +\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 $\sigma([])=<(u,0)>\forall u\in W$ +\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 $[]$ +\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 $[],[]$ +\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 $\sigma([])$ +\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 $\sigma([])=<(u_{i},0)>\in W\times\{0\}$ +\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 $P=$ +\end_inset + +, +\begin_inset Formula $Q=$ +\end_inset + +, +\begin_inset Formula $R=$ +\end_inset + + y +\begin_inset Formula $U=$ +\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 $,,$ +\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 $P=:$ +\end_inset + +, +\begin_inset Formula $Q=:$ +\end_inset + + y +\begin_inset Formula $R=:$ +\end_inset + +, podemos escribir +\begin_inset Formula $U=:$ +\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 $=P=$ +\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 $=<\lambda_{1}v'_{1}+\lambda_{2}v'_{2}+\lambda_{3}v'_{3}>=U=$ +\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 $E:=$ +\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 $E:=$ +\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_{+}:=, & V_{-}:=, & V_{0}:= +\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 $E:==^{\bot}=^{\bot}=$ +\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 $w'\in^{\bot}$ +\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 $$ +\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 + + + + + + + + +\begin_inset Text + +\begin_layout Plain Layout +Rango +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +Signatura +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +Ecuación reducida +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +Tipo de cónica +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +3 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $(3,0)/(0,3)$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $x^{2}+y^{2}+z^{2}=0$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +No degenerada imaginaria +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $(2,1)/(1,2)$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $x^{2}+y^{2}-z^{2}=0$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +No degenerada real +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +2 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $(2,0)/(0,2)$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $x^{2}+y^{2}=0$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +Punto +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $(1,1)$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $x^{2}-y^{2}=0$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +Par de rectas distintas +\end_layout + +\end_inset + + + + +\begin_inset Text + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $(1,0)/(0,1)$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +\begin_inset Formula $x^{2}=0$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Text + +\begin_layout Plain Layout +Recta doble +\end_layout + +\end_inset + + + + +\end_inset + + +\end_layout + +\end_body +\end_document -- cgit v1.2.3