Commit inicial, primer cuatrimestre.

6 files changed, 9717 insertions, 0 deletions

diff --git a/algl/n.lyx b/algl/n.lyx
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--- /dev/null
+++ b/algl/n.lyx
@@ -0,0 +1,208 @@
+#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
+\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 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
+Álgebra lineal
+\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{2017}
+\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
+Material clases teóricas, Álgebra Lineal, Universidad de Murcia (anónimo).
+\end_layout
+
+\begin_layout Chapter
+Espacios vectoriales
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n1.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Aplicaciones Lineales
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n2.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Sistemas de ecuaciones lineales
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n3.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Determinantes
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n4.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Diagonalización de endomorfismos
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n5.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/algl/n1.lyx b/algl/n1.lyx
new file mode 100644
index 0000000..47d85c9
--- /dev/null
+++ b/algl/n1.lyx
@@ -0,0 +1,3673 @@
+#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
+\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 swiss
+\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 Section
+Cuerpos
+\end_layout
+
+\begin_layout Standard
+Conjunto 
+\begin_inset Formula $K$
+\end_inset
+
+ con dos operaciones, 
+\series bold
+suma
+\series default
+ (
+\begin_inset Formula $+$
+\end_inset
+
+) y 
+\series bold
+producto
+\series default
+ (
+\begin_inset Formula $\cdot$
+\end_inset
+
+), tales que 
+\begin_inset Formula $\forall a,b,c\in K$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Propiedad asociativa de la suma: 
+\series default
+
+\begin_inset Formula $a+(b+c)=(a+b)+c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Propiedad conmutativa de la suma: 
+\begin_inset Formula $a+b=b+a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Elemento neutro para la suma
+\series default
+ o 
+\series bold
+nulo:
+\series default
+ 
+\begin_inset Formula $\exists!0\in K:\forall a\in K,0+a=a$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Pongamos que existe otro 
+\begin_inset Formula $0$
+\end_inset
+
+ (
+\begin_inset Formula $0'$
+\end_inset
+
+), entonces 
+\begin_inset Formula $0=0+0'=0'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Inverso para la suma
+\series default
+ u 
+\series bold
+opuesto:
+\series default
+ 
+\begin_inset Formula $\forall a\in K,\exists!a':a+a'=0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $-a:=a'$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Pongamos que existe otro opuesto 
+\begin_inset Formula $a''$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a'=0+a'=(a''+a)+a'=a''+(a+a')=a''+0=a''$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Propiedad asociativa del producto:
+\series default
+ 
+\begin_inset Formula $a\cdot(b\cdot c)=(a\cdot b)\cdot c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Propiedad conmutativa del producto:
+\series default
+ 
+\begin_inset Formula $a\cdot b=b\cdot a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Elemento neutro para el producto
+\series default
+ o 
+\series bold
+unidad:
+\series default
+ 
+\begin_inset Formula $\exists!1\in K:\forall a\in K,1\cdot a=a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Inverso para el producto:
+\series default
+ 
+\begin_inset Formula $\forall a\in K\backslash\{0\},\exists!a'':a\cdot a''=1$
+\end_inset
+
+; 
+\begin_inset Formula $a^{-1}:=\frac{1}{a}:=a''$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Propiedad distributiva:
+\series default
+ 
+\begin_inset Formula $a\cdot(b+c)=a\cdot b+a\cdot c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+La congruencia 
+\begin_inset Formula $\mathbb{Z}_{2}=\{0,1\}$
+\end_inset
+
+ con operaciones 
+\begin_inset Formula $0+0=1+1=0$
+\end_inset
+
+, 
+\begin_inset Formula $0+1=1+0=1$
+\end_inset
+
+, 
+\begin_inset Formula $0\cdot0=0\cdot1=1\cdot0=0$
+\end_inset
+
+ y 
+\begin_inset Formula $1\cdot1=1$
+\end_inset
+
+ es un cuerpo.
+ Siempre existe un cuerpo 
+\begin_inset Formula $F_{p^{n}}$
+\end_inset
+
+, formado por 
+\begin_inset Formula $p^{n}$
+\end_inset
+
+ elementos, donde 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ y 
+\begin_inset Formula $p$
+\end_inset
+
+ es primo.
+ Algunas propiedades: 
+\begin_inset Formula $\forall a,b,c\in K$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a+b=a+c\implies b=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $0a=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $(-a)b=a(-b)=-(ab)$
+\end_inset
+
+; 
+\begin_inset Formula $(-1)a=-a$
+\end_inset
+
+; 
+\begin_inset Formula $(-a)(-b)=ab$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $ab=0\implies a=0\lor b=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $ab=ac\implies a=0\lor b=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+El cuerpo de los números complejos
+\end_layout
+
+\begin_layout Standard
+Si consideramos 
+\begin_inset Formula $\mathbb{R}\times\mathbb{R}=\{(a,b)|a,b\in\mathbb{R}\}$
+\end_inset
+
+, con 
+\begin_inset Formula $(a,b)+(c,d)=(a+c,b+d)$
+\end_inset
+
+ y 
+\begin_inset Formula $(a,b)\cdot(c,d)=(ac-bd,ad+bc)$
+\end_inset
+
+, obtenemos el cuerpo de los números complejos (
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+).
+ El conjunto de elementos de 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ con forma 
+\begin_inset Formula $(a,0)$
+\end_inset
+
+ es una copia del cuerpo 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+unidad imaginaria
+\series default
+ a 
+\begin_inset Formula $i=(0,1)$
+\end_inset
+
+, de forma que 
+\begin_inset Formula $i^{2}=i\cdot i=(-1,0)=-1$
+\end_inset
+
+.
+ Dado que 
+\begin_inset Formula $(b,0)i=(0,b)$
+\end_inset
+
+, y como 
+\begin_inset Formula $(b,0)$
+\end_inset
+
+ es el número real 
+\begin_inset Formula $b$
+\end_inset
+
+, tenemos 
+\begin_inset Formula $(a,b)=a+bi$
+\end_inset
+
+, lo que denominamos la 
+\series bold
+forma binomial.
+
+\series default
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ es la 
+\series bold
+componente real,
+\series default
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ la 
+\series bold
+componente imaginaria.
+
+\series default
+ Si 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+, llamamos 
+\series bold
+conjugado
+\series default
+ de 
+\begin_inset Formula $z$
+\end_inset
+
+ a 
+\begin_inset Formula $\overline{z}=a-bi$
+\end_inset
+
+, de forma que 
+\begin_inset Formula $z=\overline{z}\iff z\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Podemos representar un número complejo 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+ como un punto del plano, con coordenadas 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+.
+ La distancia del punto al origen de coordenadas, llamada 
+\series bold
+módulo
+\series default
+, es 
+\begin_inset Formula $r=|z|=\sqrt{a^{2}+b^{2}}=\sqrt{z\overline{z}}$
+\end_inset
+
+.
+ El ángulo con el eje 
+\begin_inset Formula $OX$
+\end_inset
+
+, llamado 
+\series bold
+argumento
+\series default
+, cumple que 
+\begin_inset Formula $a+bi=r(\cos(\alpha)+i\sin(\alpha))$
+\end_inset
+
+.
+ Esta es la 
+\series bold
+forma polar
+\series default
+ o 
+\series bold
+módulo argumental
+\series default
+ del complejo.
+ La multiplicación en forma polar es: 
+\begin_inset Formula $[r(\cos(\alpha)+i\sin(\alpha))][r'(\cos(\alpha')+i\sin(\alpha'))]=rr'(\cos(\alpha+\alpha')+i\sin(\alpha+\alpha'))$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+Teorema Fundamental del Álgebra
+\series default
+ nos dice que todo polinomio 
+\begin_inset Formula $P(x)=a_{0}+a_{1}X+a_{2}X^{2}+\dots+a_{n}X^{n}$
+\end_inset
+
+, con 
+\begin_inset Formula $n\geq1$
+\end_inset
+
+, 
+\begin_inset Formula $a_{i}\in\mathbb{C}$
+\end_inset
+
+ y 
+\begin_inset Formula $a_{n}\neq0$
+\end_inset
+
+, tiene raíz compleja.
+\end_layout
+
+\begin_layout Subsection
+Característica de un cuerpo.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\forall a\in K,n\in\mathbb{N},na=\underset{n\text{ veces}}{a+a+\cdots+a}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+En particular, 
+\begin_inset Formula 
+\[
+n1=\underset{n\text{ veces}}{1+1+\cdots+1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Por tanto, 
+\begin_inset Formula $na=(n1)a$
+\end_inset
+
+, 
+\begin_inset Formula $n1+m1=(n+m)1$
+\end_inset
+
+ y 
+\begin_inset Formula $(n1)(m1)=(nm)1$
+\end_inset
+
+ para cualquier cuerpo 
+\begin_inset Formula $K$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Un cuerpo tiene 
+\series bold
+característica cero
+\series default
+ si 
+\begin_inset Formula $\forall n>0,n1\neq0$
+\end_inset
+
+.
+ De lo contrario, se dice que tiene 
+\series bold
+característica 
+\begin_inset Formula $n$
+\end_inset
+
+
+\series default
+, siendo 
+\begin_inset Formula $n$
+\end_inset
+
+ el menor natural tal que 
+\begin_inset Formula $n1=0$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $na=(n1)a\implies na=0$
+\end_inset
+
+.
+ Dado que 
+\begin_inset Formula $ab=0\iff a=0\lor b=0$
+\end_inset
+
+, tenemos que 
+\begin_inset Formula $\exists p,q<n:n=pq\implies0=n1=(p1)(q1)\implies p1=0\lor q1=0$
+\end_inset
+
+, por lo que la característica de un cuerpo es cero o un nº primo.
+\end_layout
+
+\begin_layout Section
+Espacios vectoriales
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+espacio vectorial
+\series default
+ sobre 
+\begin_inset Formula $K$
+\end_inset
+
+, o 
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial, es una terna 
+\begin_inset Formula $(V,+,\cdot)$
+\end_inset
+
+ donde 
+\begin_inset Formula $V\neq\emptyset$
+\end_inset
+
+, 
+\begin_inset Formula $+$
+\end_inset
+
+ es una operación 
+\begin_inset Formula $V\times V\longrightarrow V$
+\end_inset
+
+ y 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ es una operación 
+\begin_inset Formula $K\times V\longrightarrow V$
+\end_inset
+
+, tales que 
+\begin_inset Formula $\forall u,v,w\in V,\alpha,\beta\in K$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Suma asociativa:
+\series default
+ 
+\begin_inset Formula $u+(v+w)=(u+v)+w$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Suma conmutativa:
+\series default
+ 
+\begin_inset Formula $u+v=v+u$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Vector cero
+\series default
+ o 
+\series bold
+nulo:
+\series default
+ 
+\begin_inset Formula $\exists0_{V}:\forall u\in V,0_{V}+u=u$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Opuesto para la suma:
+\series default
+ 
+\begin_inset Formula $\forall u\in V,\exists u':u+u'=0_{V}$
+\end_inset
+
+; 
+\begin_inset Formula $u':=-u$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\alpha\cdot(u+v)=\alpha\cdot u+\alpha\cdot v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $(\alpha+\beta)\cdot u=\alpha\cdot u+\beta\cdot u$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $(\alpha\cdot\beta)\cdot u=\alpha\cdot(\beta\cdot u)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $1_{K}\cdot u=u$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+vectores
+\series default
+ a los elementos de 
+\begin_inset Formula $V$
+\end_inset
+
+ y 
+\series bold
+escalares
+\series default
+ a los de 
+\begin_inset Formula $K$
+\end_inset
+
+.
+ Todo cuerpo es espacio vectorial sobre sí mismo.
+ 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ es espacio vectorial sobre 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+, y 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ sobre 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El plano real 
+\begin_inset Formula $\mathbb{R}^{2}=\{(x,y)|x,y\in\mathbb{R}\}$
+\end_inset
+
+, con las operaciones 
+\begin_inset Formula $(x,y)+(x',y')=(x+x',y+y')$
+\end_inset
+
+ y 
+\begin_inset Formula $\alpha(x,y)=(\alpha x,\alpha y)$
+\end_inset
+
+, es un 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+-espacio vectorial.
+ El conjunto 
+\begin_inset Formula $K^{n}$
+\end_inset
+
+ de las 
+\begin_inset Formula $n$
+\end_inset
+
+-uplas de elementos de 
+\begin_inset Formula $K$
+\end_inset
+
+, 
+\begin_inset Formula $K^{n}=\{(x_{1},x_{2},\dots,x_{n}|x_{1},x_{2},\dots,x_{n}\in K\}$
+\end_inset
+
+ es un 
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial con las operaciones 
+\begin_inset Formula $(x_{1},\dots,x_{n})+(y_{1},\dots,y_{n})=(x_{1}+y_{1},\dots,x_{n}+y_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $\alpha(x_{1},\dots x_{n})=(\alpha x_{1},\dots,\alpha x_{n})$
+\end_inset
+
+.
+ También, si 
+\begin_inset Formula $V_{1},\dots,V_{n}$
+\end_inset
+
+ son 
+\begin_inset Formula $K$
+\end_inset
+
+-espacios vectoriales, el conjunto 
+\begin_inset Formula $V_{1}\times\dots\times V_{n}=\{(v_{1},v_{2},\dots,v_{n})|v_{1}\in V_{1},v_{2}\in V_{2},\dots,v_{n}\in V_{n}\}$
+\end_inset
+
+, con operaciones similares, es un 
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial llamado 
+\series bold
+espacio vectorial producto
+\series default
+ de los 
+\begin_inset Formula $V_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una 
+\series bold
+matriz
+\series default
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ de tamaño 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ (con 
+\begin_inset Formula $m,n\in\mathbb{N}$
+\end_inset
+
+) sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ es una disposición en 
+\begin_inset Formula $m$
+\end_inset
+
+ filas y 
+\begin_inset Formula $n$
+\end_inset
+
+ columnas de 
+\begin_inset Formula $m\cdot n$
+\end_inset
+
+ elementos de 
+\begin_inset Formula $K$
+\end_inset
+
+.
+ La matriz es 
+\series bold
+cuadrada
+\series default
+ si 
+\begin_inset Formula $m=n$
+\end_inset
+
+, 
+\series bold
+fila
+\series default
+ si 
+\begin_inset Formula $m=1$
+\end_inset
+
+ y 
+\series bold
+columna
+\series default
+ si 
+\begin_inset Formula $n=1$
+\end_inset
+
+.
+ Llamamos 
+\begin_inset Formula $M_{m,n}(K)$
+\end_inset
+
+ al conjunto de las matrices 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ sobre 
+\begin_inset Formula $K$
+\end_inset
+
+, y 
+\begin_inset Formula $M_{n,n}(K)=M_{n}(K)$
+\end_inset
+
+.
+ Se representan de la siguiente forma:
+\begin_inset Formula 
+\[
+A=\left(\begin{array}{cccc}
+a_{11} & a_{12} & \cdots & a_{1n}\\
+a_{21} & a_{22} & \cdots & a_{2n}\\
+\vdots & \vdots & \ddots & \vdots\\
+a_{m1} & a_{m2} & \cdots & a_{mn}
+\end{array}\right),\,a_{ij}\in K\forall1\leq i\leq m,1\leq j\leq n
+\]
+
+\end_inset
+
+
+\begin_inset Formula $\forall A=(a_{ij}),B=(b_{ij})\in M_{m,n}(K),A+B=(c_{ij})\in M_{m,n}(K)$
+\end_inset
+
+, donde 
+\begin_inset Formula $c_{ij}=a_{ij}+b_{ij}$
+\end_inset
+
+ para cada elemento de la matriz.
+ También, 
+\begin_inset Formula $\forall\alpha\in K,A\in M_{m,n}(K),\alpha A=(c_{ij})\in M_{m,n}(K)$
+\end_inset
+
+, donde 
+\begin_inset Formula $c_{ij}=\alpha a_{ij}$
+\end_inset
+
+.
+ Con estas operaciones, 
+\begin_inset Formula $M_{m,n}(K)$
+\end_inset
+
+ es un 
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial.
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+polinomio
+\series default
+ en una 
+\series bold
+indeterminada
+\series default
+ 
+\begin_inset Formula $X$
+\end_inset
+
+ con 
+\series bold
+coeficientes
+\series default
+ en 
+\begin_inset Formula $K$
+\end_inset
+
+ es una expresión de la forma
+\begin_inset Formula 
+\[
+a_{0}+a_{1}X+a_{2}X^{2}+\dots+a_{n}X^{n}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Donde 
+\begin_inset Formula $n$
+\end_inset
+
+ es un entero no negativo, 
+\begin_inset Formula $a_{i}\in K\forall i=0,1,\dots n$
+\end_inset
+
+.
+ El conjunto de polinomios sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ se llama 
+\begin_inset Formula $K(X)$
+\end_inset
+
+ y es un espacio vectorial con las operaciones: 
+\begin_inset Formula 
+\begin{eqnarray*}
+(a_{0}+\dots+a_{n}X^{n})+(b_{0}+\dots+b_{n}X^{n}) & = & (a_{0}+b_{0})+\dots+(a_{n}+b_{n})X^{n}\\
+\alpha(a_{0}+a_{1}X+\dots+a_{n}X^{n}) & = & \alpha a_{0}+\alpha a_{1}X+\dots+\alpha a_{n}X^{n}
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $\mathcal{S}\neq\emptyset$
+\end_inset
+
+, el conjunto 
+\begin_inset Formula $\mathcal{F}(\mathcal{S},K)=\{f:\mathcal{S}\rightarrow K\}$
+\end_inset
+
+, formado por todas las aplicaciones de 
+\begin_inset Formula $\mathcal{S}$
+\end_inset
+
+ en 
+\begin_inset Formula $K$
+\end_inset
+
+, con operaciones 
+\begin_inset Formula $(f+g)(s)=f(s)+g(s)$
+\end_inset
+
+ y 
+\begin_inset Formula $(\alpha f)(s)=\alpha f(s)$
+\end_inset
+
+, es un 
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial.
+ Con estas mismas operaciones, el conjunto 
+\begin_inset Formula $\mathcal{C}([a,b],\mathbb{R})=\{f:[a,b]\rightarrow\mathbb{R}|f\text{ continua}\}$
+\end_inset
+
+ es un 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+-espacio vectorial.
+\end_layout
+
+\begin_layout Standard
+Propiedades de los espacios vectoriales: 
+\begin_inset Formula $\forall u,v,w\in V,\alpha,\beta\in K$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $u+v=u+w\implies v=w$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+u+v=u+w\implies(-u)+(u+v)=(-u)+(u+w)=((-u)+u)+v=((-u)+u)+w=v=w
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $0u=0_{V}$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0u+0u=(0+0)u=0u=0u+0_{V}\implies0u=0_{V}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\alpha0_{V}=0_{V}$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\alpha0_{V}+0_{V}=\alpha0_{V}=\alpha(0_{V}+0_{V})=\alpha0_{V}+\alpha0_{V}\implies\alpha0_{V}=0_{V}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $u\in V,\alpha u=0_{V}\implies\alpha=0\lor u=0_{V}$
+\end_inset
+
+; 
+\begin_inset Formula $\alpha u=\alpha v\implies\alpha=0\lor u=v$
+\end_inset
+
+; 
+\begin_inset Formula $\alpha u=\beta u\implies\alpha=\beta\lor u=0_{V}$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\begin{array}{c}
+\alpha u=0_{V}\\
+\alpha\neq0
+\end{array}\implies u=(\alpha^{-1}\cdot\alpha)u=\alpha^{-1}(\alpha u)=\alpha^{-1}\cdot0_{V}=0_{V}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\begin{array}{c}
+\alpha u=\alpha v\\
+\alpha\neq0
+\end{array}\implies\alpha^{-1}(\alpha u)=\alpha^{-1}(\alpha v)=1\cdot u=1\cdot v=u=v
+\]
+
+\end_inset
+
+Para la última demostración, usamos (5):
+\begin_inset Formula 
+\[
+\begin{array}{c}
+\begin{array}{c}
+\alpha u=\beta u\\
+u\neq0_{V}
+\end{array}\implies0_{V}=\alpha u+(-(\beta u))\overset{(5)}{=}\alpha u+(-\beta)u=(\alpha+(-\beta))u\implies\\
+\implies\alpha+(-\beta)=0\implies\alpha=\beta
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $u\in V,(-\alpha)u=\alpha(-u)=-(\alpha u)$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+(-\alpha)u+\alpha u=(-\alpha+\alpha)u=0u=0_{V}\implies(-\alpha)u=-(\alpha u)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\alpha(-u)+\alpha u=\alpha(-u+u)=\alpha\cdot0_{V}=0_{V}\implies\alpha(-u)=-(\alpha u)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+También podemos llamar 
+\begin_inset Formula $0$
+\end_inset
+
+ a 
+\begin_inset Formula $0_{V}$
+\end_inset
+
+ si esto no conlleva ambigüedad.
+\end_layout
+
+\begin_layout Section
+Subespacios vectoriales
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $U\subseteq V$
+\end_inset
+
+ (
+\begin_inset Formula $U\neq\emptyset$
+\end_inset
+
+) es un 
+\series bold
+subespacio vectorial
+\series default
+ de 
+\begin_inset Formula $V$
+\end_inset
+
+ (
+\begin_inset Formula $U\leq V$
+\end_inset
+
+) si 
+\begin_inset Formula $\forall u,v\in U,u+v\in U$
+\end_inset
+
+ y 
+\begin_inset Formula $\forall u\in U,\alpha\in K,\alpha u\in U$
+\end_inset
+
+.
+ De esta forma 
+\begin_inset Formula $U$
+\end_inset
+
+ es también un 
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial.
+\end_layout
+
+\begin_layout Standard
+Los subconjuntos 
+\begin_inset Formula $\{0\}$
+\end_inset
+
+ y 
+\begin_inset Formula $V$
+\end_inset
+
+ son subespacios vectoriales de 
+\begin_inset Formula $V$
+\end_inset
+
+, y 
+\begin_inset Formula $\{(x_{1},\dots,x_{n})\in K^{n}|x_{1}+\dots+x_{n}=0\}$
+\end_inset
+
+ es un subespacio vectorial de 
+\begin_inset Formula $K^{n}$
+\end_inset
+
+.
+ El conjunto 
+\begin_inset Formula $\mathcal{P}_{n}$
+\end_inset
+
+ de polinomios con coeficientes reales con grado menor o igual a 
+\begin_inset Formula $n$
+\end_inset
+
+, junto con el 
+\begin_inset Formula $0$
+\end_inset
+
+, es un subespacio vectorial de 
+\begin_inset Formula $\mathbb{R}[X]$
+\end_inset
+
+.
+ También lo es 
+\begin_inset Formula $U_{a,b}=\{f\in\mathcal{C}([a,b],\mathbb{R}):f(a)=f(b)\}$
+\end_inset
+
+ respecto de 
+\begin_inset Formula $\mathcal{C}([a,b],\mathbb{R})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Combinaciones lineales.
+ Sistemas de generadores
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $u\in V$
+\end_inset
+
+ es 
+\series bold
+combinación lineal
+\series default
+ de 
+\begin_inset Formula $v_{1},\dots,v_{n}\in V$
+\end_inset
+
+ si 
+\begin_inset Formula $\exists\alpha_{1},\dots,\alpha_{n}\in K:u=\alpha_{1}v_{1}+\dots+\alpha_{n}v_{n}$
+\end_inset
+
+.
+ Se dice que es combinación lineal de vectores de 
+\begin_inset Formula $\mathcal{S}$
+\end_inset
+
+ (con 
+\begin_inset Formula $\mathcal{S}\subseteq V$
+\end_inset
+
+) si 
+\begin_inset Formula $\exists n\in\mathbb{N},v_{1},\dots,v_{n}\in\mathcal{S}:u=\alpha_{1}v_{1}+\dots+\alpha_{n}v_{n}$
+\end_inset
+
+.
+ Los escalares 
+\begin_inset Formula $\alpha_{i}$
+\end_inset
+
+ se llaman 
+\series bold
+coeficientes
+\series default
+ de la combinación.
+ Así, un subconjunto no vacío 
+\begin_inset Formula $U\subseteq V$
+\end_inset
+
+ es un subespacio vectorial de 
+\begin_inset Formula $V$
+\end_inset
+
+ si cualquier combinación lineal de vectores de 
+\begin_inset Formula $U$
+\end_inset
+
+ también está en 
+\begin_inset Formula $U$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $\mathcal{S}$
+\end_inset
+
+ es un subconjunto no vacío de 
+\begin_inset Formula $V$
+\end_inset
+
+, el conjunto 
+\begin_inset Formula $<\mathcal{S}>$
+\end_inset
+
+ de todas las combinaciones lineales de vectores en 
+\begin_inset Formula $\mathcal{S}$
+\end_inset
+
+ es el menor subespacio vectorial tal que 
+\begin_inset Formula $\mathcal{S}\subseteq<{\cal S}>$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $u\in\mathcal{S}\implies1\cdot u\in<\mathcal{S}>$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $u,v\in<\mathcal{S}>$
+\end_inset
+
+, entonces existirán 
+\begin_inset Formula $u_{1},\dots,u_{k},v_{1},\dots,v_{l}\in\mathcal{S}$
+\end_inset
+
+ y 
+\begin_inset Formula $\alpha_{1},\dots\alpha_{k},\beta_{1},\dots,\beta_{l}\in K$
+\end_inset
+
+ tales que 
+\begin_inset Formula $u=\alpha_{1}u_{1}+\dots+\alpha_{k}u_{k}$
+\end_inset
+
+ y 
+\begin_inset Formula $v=\beta_{1}v_{1}+\dots+\beta_{l}+v_{l}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $u+v$
+\end_inset
+
+ también es combinación lineal de vectores en 
+\begin_inset Formula $\mathcal{S}$
+\end_inset
+
+ y por tanto está en 
+\begin_inset Formula $<S>$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $\alpha\in K$
+\end_inset
+
+, tenemos que 
+\begin_inset Formula $u=\alpha\alpha_{1}u_{1}+\dots+\alpha\alpha_{k}u_{k}\in<\mathcal{S}>$
+\end_inset
+
+.
+ Finalmente, si 
+\begin_inset Formula $U$
+\end_inset
+
+ es un subespacio vectorial de 
+\begin_inset Formula $V$
+\end_inset
+
+ que contiene a 
+\begin_inset Formula $\mathcal{S}$
+\end_inset
+
+, como toda combinación de vectores de 
+\begin_inset Formula $U$
+\end_inset
+
+ está en 
+\begin_inset Formula $U$
+\end_inset
+
+, entonces 
+\begin_inset Formula $<\mathcal{S}>\subseteq U$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Un subconjunto 
+\begin_inset Formula $\mathcal{S}\subseteq V$
+\end_inset
+
+ es un 
+\series bold
+sistema de generadores
+\series default
+ de 
+\begin_inset Formula $V$
+\end_inset
+
+ si 
+\begin_inset Formula $<\mathcal{S}>=V$
+\end_inset
+
+.
+ 
+\begin_inset Formula $V$
+\end_inset
+
+ es 
+\series bold
+de dimensión finita
+\series default
+ o 
+\series bold
+finitamente generado
+\series default
+ si tiene un sistema de generadores finito.
+ Estas definiciones también son válidas para subespacios vectoriales.
+ 
+\begin_inset Formula $U$
+\end_inset
+
+ es el subespacio 
+\series bold
+generado
+\series default
+ por 
+\begin_inset Formula $\mathcal{S}$
+\end_inset
+
+ si 
+\begin_inset Formula $U=<\mathcal{S}>$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Dependencia e independencia lineal
+\end_layout
+
+\begin_layout Standard
+Un conjunto 
+\begin_inset Formula ${\cal S}\subseteq V$
+\end_inset
+
+ es 
+\series bold
+linealmente independiente
+\series default
+ si la única forma de obtener el vector nulo como combinación lineal de
+ vectores de 
+\begin_inset Formula ${\cal S}$
+\end_inset
+
+ es tomando todos los coeficientes nulos.
+ De lo contrario es 
+\series bold
+linealmente dependiente
+\series default
+.
+ Así, 
+\begin_inset Formula $\{v\}$
+\end_inset
+
+ es linealmente independiente si y sólo si 
+\begin_inset Formula $v\neq0$
+\end_inset
+
+, con lo que cualquier conjunto 
+\begin_inset Formula $\{0\}\subseteq{\cal S}$
+\end_inset
+
+ es linealmente dependiente.
+ En 
+\begin_inset Formula $\mathbb{C_{R}}$
+\end_inset
+
+, 
+\begin_inset Formula $\{1,i\}$
+\end_inset
+
+ (
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ con escalares reales) es linealmente independiente, mientras que en 
+\begin_inset Formula $\mathbb{C_{C}}$
+\end_inset
+
+ es linealmente dependiente porque 
+\begin_inset Formula $1+(i)i=0$
+\end_inset
+
+.
+ Un subconjunto de un conjunto linealmente independiente también es linealmente
+ dependiente.
+\end_layout
+
+\begin_layout Standard
+Un conjunto 
+\begin_inset Formula $\{v_{1},\dots,v_{m}\}$
+\end_inset
+
+ con al menos dos vectores es linealmente dependiente si y sólo si alguno
+ de ellos es combinación lineal del resto.
+\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
+
+Se tiene que existen 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{m}$
+\end_inset
+
+ no todos nulos con 
+\begin_inset Formula $\sum_{i=1}^{m}\alpha_{i}v_{i}=\alpha_{1}v_{1}+\dots+\alpha_{m}v_{m}=0$
+\end_inset
+
+.
+ Suponemos 
+\begin_inset Formula $\alpha_{j}\neq0$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\alpha_{j}v_{j}=-\sum_{i=1,i\neq j}^{m}\alpha_{i}v_{i}=-\alpha_{1}v_{1}-\dots-\alpha_{j-1}v_{j-1}-\alpha_{j+1}v_{j+1}-\dots-\alpha_{m}v_{m}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $v_{j}=-\sum_{i=1,i\neq j}^{m}(\alpha_{j}^{-1}\alpha_{i}v_{i})$
+\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 
+\begin_inset Formula $v_{j}$
+\end_inset
+
+ es combinación lineal de 
+\begin_inset Formula $\{v_{i}\}_{1\leq i\leq m}$
+\end_inset
+
+, existen escalares 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{j-1},\alpha_{j+1},\dots,\alpha_{m}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $v_{j}=\alpha_{1}v_{1}\dots,\alpha_{j-1}v_{j-1}+\alpha_{j+1}v_{j+1}+\dots+\alpha_{m}v_{m}=\sum_{i=1,i\neq j}^{m}\alpha_{i}v_{i}$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $0=\alpha_{1}v_{1}+\dots+\alpha_{j-1}v_{j-1}+(-1)v_{j}+\alpha_{j+1}v_{j+1}+\dots+\alpha_{m}v_{m}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Bases.
+ Dimensión
+\end_layout
+
+\begin_layout Standard
+Una 
+\series bold
+base
+\series default
+ de un espacio vectorial es un sistema de generadores linealmente independiente.
+ Así, 
+\begin_inset Formula $\{1\}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $K_{K}$
+\end_inset
+
+, 
+\begin_inset Formula $\{(1,0,\dots,0),(0,1,0,\dots,0),\dots,(0,0,\dots,0,1)\}$
+\end_inset
+
+ es 
+\series bold
+base canónica
+\series default
+ de 
+\begin_inset Formula $K^{n}$
+\end_inset
+
+ y 
+\begin_inset Formula $\{1,X,\dots,X^{n},\dots\}$
+\end_inset
+
+ lo es de 
+\begin_inset Formula $\mathbb{R}[X]$
+\end_inset
+
+.
+ Si llamamos 
+\begin_inset Formula $A_{ij}\in M_{m,n}(K)$
+\end_inset
+
+ a la matriz con un 1 en el lugar 
+\begin_inset Formula $ij$
+\end_inset
+
+ y 0 en el resto, entonces 
+\begin_inset Formula $\{A_{ij}:1\leq i\leq m,1\leq j\leq n\}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $M_{m,n}(K)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $V$
+\end_inset
+
+ si y sólo si todo 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ se expresa de modo único como combinación lineal de 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+.
+\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
+
+Como 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ es base, es sistema de generadores, por lo que todo vector de 
+\begin_inset Formula $V$
+\end_inset
+
+ es combinación lineal de vectores de 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+.
+ Ahora, sea 
+\begin_inset Formula $v=\sum_{i=1}^{n}\alpha_{i}v_{i}=\sum_{i=1}^{n}\beta_{i}v_{i}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $0=(\alpha_{1}v_{1}+\dots+\alpha_{n}v_{n})-(\beta_{1}v_{1}+\dots+\beta_{n}v_{n})$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $0=(\alpha_{1}-\beta_{1})v_{1}+\dots+(\alpha_{n}-\beta_{n})v_{n}$
+\end_inset
+
+, y como 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ es linealmente independiente, se tiene que 
+\begin_inset Formula $\alpha_{1}-\beta_{1}=\dots=\alpha_{n}-\beta_{n}=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $v$
+\end_inset
+
+ se expresa de modo único.
+\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 ${\cal B}$
+\end_inset
+
+ es entonces sistema de generadores y queda demostrar que es linealmente
+ dependiente.
+ Sean 
+\begin_inset Formula $v_{1},\dots,v_{m}\in{\cal B}$
+\end_inset
+
+, como el 0 se representa de modo único, se tiene que si 
+\begin_inset Formula $\alpha_{1}v_{1}+\dots+\alpha_{m}v_{m}=0=0v_{1}+\dots+0v_{m}$
+\end_inset
+
+ entonces 
+\begin_inset Formula $\alpha_{1}=\dots=\alpha_{m}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $S=\{v_{1},v_{2},\dots,v_{m}\}$
+\end_inset
+
+ es un conjunto linealmente independiente y 
+\begin_inset Formula $u\notin<S>$
+\end_inset
+
+, entonces 
+\begin_inset Formula $S\cup\{u\}$
+\end_inset
+
+ es linealmente independiente.
+ 
+\series bold
+Demostración:
+\series default
+ Supongamos 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{m},\beta$
+\end_inset
+
+ tales que 
+\begin_inset Formula $\alpha_{1}v_{1}+\dots+\alpha_{m}v_{m}+\beta u=0$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $\beta\neq0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $\beta u=-(\alpha v_{1}+\dots+\alpha_{m}v_{m})$
+\end_inset
+
+, de donde 
+\begin_inset Formula $u=-\beta^{-1}\alpha_{1}v_{1}-\dots-\beta^{-1}\alpha_{m}v_{m}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $u\in<S>$
+\end_inset
+
+
+\begin_inset Formula $\#$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $\beta=0$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $\alpha_{1}v_{1}+\dots+\alpha_{m}v_{m}=0$
+\end_inset
+
+, de donde 
+\begin_inset Formula $\alpha_{1}=\dots=\alpha_{m}=0$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $S\cup\{u\}$
+\end_inset
+
+ es linealmente independiente.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema:
+\series default
+ Todo espacio vectorial 
+\begin_inset Formula $V\neq\{0\}$
+\end_inset
+
+ tiene una base.
+ 
+\series bold
+Demostración
+\series default
+ para espacios finitamente generados.
+ Sea 
+\begin_inset Formula $V=<u_{1},\dots,u_{m}>$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+ el conjunto de subconjuntos 
+\begin_inset Formula ${\cal L}\subseteq\{u_{1},\dots,u_{m}\}$
+\end_inset
+
+ linealmente independientes.
+ Sabemos que 
+\begin_inset Formula ${\cal A}\neq\emptyset$
+\end_inset
+
+ porque como 
+\begin_inset Formula $V\neq\{0\}$
+\end_inset
+
+ existe un 
+\begin_inset Formula $u_{i_{0}}\neq0$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $\{u_{i_{0}}\}$
+\end_inset
+
+ es linealmente independiente.
+ Sea 
+\begin_inset Formula ${\cal B}=\{v_{1},\dots,v_{n}\}\in{\cal A}$
+\end_inset
+
+ un elemento de 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+ con un máximo de vectores, entonces 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $V$
+\end_inset
+
+ si es sistema de generadores de 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $u_{i}$
+\end_inset
+
+ un elemento del conjunto de generadores de 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $u_{i}\in{\cal B}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $u_{i}\in<{\cal B}>$
+\end_inset
+
+.
+ Si no, entonces 
+\begin_inset Formula ${\cal B}\cup\{u_{i}\}\subseteq\{u_{1},\dots,u_{m}\}$
+\end_inset
+
+ tiene un elemento más que 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+, que es un elemento de 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+ con el número máximo de vectores, por lo que 
+\begin_inset Formula ${\cal B}\cup\{u_{i}\}\notin{\cal A}$
+\end_inset
+
+ y será linealmente de
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+pen
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+dien
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+te, pero entonces 
+\begin_inset Formula $u_{i}\in<{\cal B}>$
+\end_inset
+
+
+\begin_inset Formula $\#$
+\end_inset
+
+.
+ Acabamos de probar que 
+\begin_inset Formula $\{u_{1},\dots,u_{m}\}\subseteq<{\cal B}>$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $V=<u_{1},\dots,u_{m}>\subseteq<{\cal B}>\subseteq V$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $<{\cal B}>=V$
+\end_inset
+
+ y ya hemos demostrado que 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de Steinitz:
+\series default
+ Si 
+\begin_inset Formula $\{u_{1},\dots,u_{n}\}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $V$
+\end_inset
+
+ y 
+\begin_inset Formula $\{v_{1},\dots,v_{m}\}$
+\end_inset
+
+ es un conjunto li
+\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
+
+al
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+men
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+te independiente, entonces se pueden sustituir 
+\begin_inset Formula $m$
+\end_inset
+
+ vectores de 
+\begin_inset Formula $\{u_{1},\dots,u_{n}\}$
+\end_inset
+
+ por los vectores 
+\begin_inset Formula $\{v_{1},\dots,v_{m}\}$
+\end_inset
+
+ y obtener una nueva base de 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $m\le n$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Vemos que, como 
+\begin_inset Formula $\{v_{1},\dots,v_{m}\}$
+\end_inset
+
+ es linealmente independiente, entonces 
+\begin_inset Formula $v_{1}\neq0$
+\end_inset
+
+, y tenemos que 
+\begin_inset Formula $v_{1}=\alpha_{1}u_{1}+\dots+a_{n}u_{n}$
+\end_inset
+
+ con algún 
+\begin_inset Formula $\alpha_{i}\neq0$
+\end_inset
+
+.
+ Podemos suponer que 
+\begin_inset Formula $\alpha_{1}\neq0$
+\end_inset
+
+, y queremos probar que 
+\begin_inset Formula $\{v_{1},u_{2},\dots,u_{n}\}$
+\end_inset
+
+ es base.
+ Primero probamos que es sistema de generadores: 
+\begin_inset Formula $\alpha_{1}u_{1}=v_{1}-\alpha_{2}u_{2}-\dots-\alpha_{n}u_{n}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $u_{1}=\alpha_{1}^{-1}v_{1}-\alpha_{1}^{-1}\alpha_{2}u_{2}-\dots-\alpha_{1}^{-1}\alpha_{n}u_{n}$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $u_{1}$
+\end_inset
+
+ es combinación lineal de 
+\begin_inset Formula $\{v_{1},u_{2},\dots,u_{n}\}$
+\end_inset
+
+ y 
+\begin_inset Formula $<v_{1},u_{2},\dots,u_{n}>$
+\end_inset
+
+ contiene un sistema de generadores, por lo que 
+\begin_inset Formula $\{v_{1},u_{2},\dots,u_{n}\}$
+\end_inset
+
+ también lo es.
+ Ahora bien, sean 
+\begin_inset Formula $\beta_{1},\dots,\beta_{n}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $\beta_{1}v_{1}+\dots+\beta_{n}u_{n}=0$
+\end_inset
+
+.
+ Entonces
+\begin_inset Formula 
+\begin{eqnarray*}
+0 & = & \beta_{1}v_{1}+\dots+\beta_{n}u_{n}\\
+ & = & \beta_{1}(\alpha_{1}u_{1}+\dots+\alpha_{n}u_{n})+\beta_{2}u_{2}+\dots+\beta_{n}u_{n}\\
+ & = & \beta_{1}\alpha_{1}u_{1}+(\beta_{1}\alpha_{1}+\beta_{2})u_{2}+\dots+(\beta_{1}\alpha_{n}+\beta_{n})u_{n}
+\end{eqnarray*}
+
+\end_inset
+
+Por tanto, como 
+\begin_inset Formula $\alpha_{1}\neq0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\beta_{1}=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\beta_{2}=\dots=\beta_{n}=0$
+\end_inset
+
+ y el nuevo conjunto es también linealmente independiente.
+ De aquí además podemos concluir que todo conjunto de vectores linealmente
+ independiente de un espacio vectorial puede completarse a una base (añadiendo
+ vectores fuera del subespacio generado por este conjunto).
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema:
+\series default
+ Todas las bases de un espacio vectorial no nulo tienen el mismo número
+ de elementos.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $\{u_{1},\dots,u_{n}\}$
+\end_inset
+
+ y 
+\begin_inset Formula $\{v_{1},\dots,v_{m}\}$
+\end_inset
+
+ bases de 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $\{v_{1},\dots,v_{m}\}$
+\end_inset
+
+ es linealmente independiente, por el teorema de Steinitz tenemos que 
+\begin_inset Formula $m\leq n$
+\end_inset
+
+.
+ Análogamente, como 
+\begin_inset Formula $\{u_{1},\dots,u_{n}\}$
+\end_inset
+
+ también lo es, entonces 
+\begin_inset Formula $n\leq m$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $m=n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Definimos la 
+\series bold
+dimensión
+\series default
+ de 
+\begin_inset Formula $V$
+\end_inset
+
+ (
+\begin_inset Formula $\dim_{K}(V)$
+\end_inset
+
+ o 
+\begin_inset Formula $\dim(V)$
+\end_inset
+
+) como el número de elementos de cualquier base de 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $V=\{0\}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\dim(V)=0$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $V$
+\end_inset
+
+ no es finitamente generado, entonces 
+\begin_inset Formula $\dim_{K}(V)=\infty$
+\end_inset
+
+.
+ Por ejemplo, 
+\begin_inset Formula $\dim_{K}(K)=1$
+\end_inset
+
+, 
+\begin_inset Formula $\dim_{K}(K^{n})=n$
+\end_inset
+
+, 
+\begin_inset Formula $\dim(M_{m,n}(K))=mn$
+\end_inset
+
+ y 
+\begin_inset Formula $\dim(\mathbb{R}[X])=\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema:
+\series default
+ Si 
+\begin_inset Formula $\dim(V)=n$
+\end_inset
+
+ entonces:
+\end_layout
+
+\begin_layout Enumerate
+Todo conjunto linealmente independiente de 
+\begin_inset Formula $n$
+\end_inset
+
+ vectores es una base.
+\begin_inset Newline newline
+\end_inset
+
+Consecuencia del teorema de Steinitz.
+\end_layout
+
+\begin_layout Enumerate
+Todo conjunto de generadores de 
+\begin_inset Formula $n$
+\end_inset
+
+ vectores es una base.
+\begin_inset Newline newline
+\end_inset
+
+Siempre va a haber una base contenida en él y que va a tener 
+\begin_inset Formula $n$
+\end_inset
+
+ vectores.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $U\neq\emptyset$
+\end_inset
+
+ es un subespacio vectorial de 
+\begin_inset Formula $V$
+\end_inset
+
+ entonces 
+\begin_inset Formula $\dim(U)\leq n$
+\end_inset
+
+ y además 
+\begin_inset Formula $\dim(U)=n\iff U=V$
+\end_inset
+
+.
+\series bold
+
+\begin_inset Newline newline
+\end_inset
+
+
+\series default
+Si 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ es base de 
+\begin_inset Formula $U$
+\end_inset
+
+ entonces es un conjunto de vectores de 
+\begin_inset Formula $V$
+\end_inset
+
+ linealmente independiente y tiene como máximo 
+\begin_inset Formula $n$
+\end_inset
+
+ elementos, por lo que 
+\begin_inset Formula $\dim(U)\leq\dim(V)$
+\end_inset
+
+.
+ Además, si 
+\begin_inset Formula $\dim(U)=\dim(V)$
+\end_inset
+
+ entonces 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ tiene 
+\begin_inset Formula $n$
+\end_inset
+
+ elementos y, por la primera propiedad, también es base de 
+\begin_inset Formula $V$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $U=<{\cal B}'>=V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Definimos el 
+\series bold
+rango
+\series default
+ de un conjunto de vectores como
+\begin_inset Formula 
+\[
+\text{rang}(\{u_{1},\dots,u_{m}\})=\dim(<u_{1},\dots,u_{m}>)
+\]
+
+\end_inset
+
+Así, si 
+\begin_inset Formula $\{v_{1},\dots,v_{m}\}\subseteq V$
+\end_inset
+
+, es fácil comprobar que:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{rang}(\{v_{1},\dots,v_{i},\dots,v_{j},\dots,v_{m}\})=\text{rang}(\{v_{1},\dots,v_{j},\dots,v_{i},\dots,v_{m}\})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\alpha\neq0\implies\text{rang}(\{v_{1},\dots v_{i},\dots,v_{m}\})=\text{rang}(\{v_{1},\dots,\alpha v_{i},\dots,v_{m}\})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{rang}(\{v_{1},\dots,v_{i},\dots,v_{j},\dots,v_{m}\})=\text{rang}(\{v_{1},\dots,v_{i}+\alpha v_{j},\dots,v_{j},\dots,v_{m}\})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una forma de determinar el rango de un conjunto de vectores es ir haciendo
+ estas operaciones y eliminando posibles vectores nulos hasta encontrar
+ un conjunto linealmente independiente.
+\end_layout
+
+\begin_layout Subsection
+Operaciones elementales.
+ Matrices escalonadas.
+ Método Gauss-Jordan
+\end_layout
+
+\begin_layout Standard
+En una matriz 
+\begin_inset Formula $A\in M_{m,n}(K)$
+\end_inset
+
+, a intercambiar dos columnas, multiplicar una fila por un 
+\begin_inset Formula $0\neq\alpha\in K$
+\end_inset
+
+ o añadir una fila a otra multiplicada por un 
+\begin_inset Formula $\alpha\in K$
+\end_inset
+
+ se les llama 
+\series bold
+operaciones elementales por filas
+\series default
+.
+ Las 
+\series bold
+operaciones elementales por columnas
+\series default
+ se definen de forma análoga.
+ Si 
+\begin_inset Formula $B$
+\end_inset
+
+ es la matriz resultante de aplicar una serie de operaciones elementales
+ por filas a 
+\begin_inset Formula $A$
+\end_inset
+
+, entonces el subespacio de 
+\begin_inset Formula $K^{n}$
+\end_inset
+
+ generado por las filas de 
+\begin_inset Formula $A$
+\end_inset
+
+ es el mismo que el generado por las filas de 
+\begin_inset Formula $B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una matriz 
+\begin_inset Formula $A=(a_{ij})\in M_{m,n}(K)$
+\end_inset
+
+ está en forma 
+\series bold
+escalonada por filas
+\series default
+ si las filas nulas, de haberlas, son las últimas, el primer elemento no
+ nulo de cada fila no nula es un 1 (llamado 
+\series bold
+pivote
+\series default
+) y el pivote de cada fila no nula está en una columna posterior a la de
+ cada uno de los pivotes anteriores.
+ En las matrices escalonadas por filas, las filas no nulas son linealmente
+ independientes.
+\end_layout
+
+\begin_layout Standard
+Si en cada columna que contenga un pivote el resto de elementos son nulos,
+ la matriz está en forma 
+\series bold
+escalonada reducida por filas
+\series default
+\SpecialChar endofsentence
+ Cambiando filas por columnas tendríamos una matriz en forma 
+\series bold
+escalonada por columnas
+\series default
+ o 
+\series bold
+escalonada reducida por columnas
+\series default
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Método de eliminación Gauss-Jordan:
+\series default
+ Toda matriz se puede llevar a forma escalonada (también escalonada reducida)
+ mediante operaciones elementales por filas.
+ Algoritmo:
+\end_layout
+
+\begin_layout Enumerate
+Encontrar el primer elemento 
+\begin_inset Formula $a$
+\end_inset
+
+ no nulo de la primera columna no nula e intercambiar su fila con la primera.
+\end_layout
+
+\begin_layout Enumerate
+Multiplicarla por 
+\begin_inset Formula $a^{-1}$
+\end_inset
+
+ para obtener un pivote.
+\end_layout
+
+\begin_layout Enumerate
+Hacer operaciones 
+\begin_inset Formula $Fila_{i}-cFila_{1}$
+\end_inset
+
+, donde 
+\begin_inset Formula $c$
+\end_inset
+
+ es el elemento de cada fila debajo del pivote.
+\end_layout
+
+\begin_layout Enumerate
+Repetir el proceso con la matriz resultado de eliminar la primera fila hasta
+ terminar la matriz.
+\end_layout
+
+\begin_layout Enumerate
+Para obtener la escalonada reducida, hacer operaciones 
+\begin_inset Formula $Fila_{i}-cFila_{k}$
+\end_inset
+
+, donde 
+\begin_inset Formula $i<k$
+\end_inset
+
+ y 
+\begin_inset Formula $c$
+\end_inset
+
+ es el elemento encima del pivote de la fila 
+\begin_inset Formula $i$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Para obtener la base de un subespacio 
+\begin_inset Formula $K^{n}$
+\end_inset
+
+ generado por 
+\begin_inset Formula $m$
+\end_inset
+
+ vectores, escalonamos la matriz 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ cuyas filas son los vectores generadores del subespacio, y los vectores
+ correspondientes a filas no nulas forman una base.
+ Los vectores fila de cada nueva matriz son combinaciones lineales de los
+ iniciales.
+ Para obtener los coeficientes de estas combinaciones, anexamos la matriz
+ identidad a la derecha de la original, separada por una línea, y le aplicamos
+ también las operaciones, sin considerar estos coeficientes como parte de
+ la matriz a escalonar.
+\end_layout
+
+\begin_layout Section
+Coordenadas.
+ Cambio de base
+\end_layout
+
+\begin_layout Standard
+Las 
+\series bold
+coordenadas
+\series default
+ de un vector 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ respecto a la base 
+\begin_inset Formula ${\cal B}=\{u_{1},\dots,u_{n}\}$
+\end_inset
+
+ son la única 
+\begin_inset Formula $n$
+\end_inset
+
+-upla 
+\begin_inset Formula $[v]_{{\cal B}}=(x_{1},\dots,x_{n})\in K^{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $v=x_{1}u_{1}+\dots+x_{n}u_{n}$
+\end_inset
+
+, de forma que dos vectores de 
+\begin_inset Formula $V$
+\end_inset
+
+ son iguales si y solo si tienen las mismas coordenadas respecto a una base
+ 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+, y operar con vectores en 
+\begin_inset Formula $V$
+\end_inset
+
+ es equivalente a operar en 
+\begin_inset Formula $K^{n}$
+\end_inset
+
+ con las 
+\begin_inset Formula $n$
+\end_inset
+
+-uplas de sus coordenadas, pues 
+\begin_inset Formula $[v+v']_{{\cal B}}=(x_{1}+x'_{1},\dots,x_{n}+x'_{n})=[v]_{{\cal B}}+[v']_{{\cal B}}$
+\end_inset
+
+ y 
+\begin_inset Formula $[rv]_{{\cal B}}=(rx_{1},\dots,rx_{n})=r[v]_{{\cal B}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Cambio de coordenadas
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula ${\cal B}=\{u_{1},\dots,u_{n}\}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'=\{u'_{1},\dots,u'_{n}\}$
+\end_inset
+
+ bases de 
+\begin_inset Formula $V$
+\end_inset
+
+, y llamamos 
+\begin_inset Formula $[u'_{j}]_{{\cal B}}=(p_{1j},\dots,p_{nj})$
+\end_inset
+
+, de forma que 
+\begin_inset Formula $u'_{j}=\sum_{i=1}^{n}p_{ij}u_{i}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $[v]_{{\cal B}'}=(x'_{1},\dots,x'_{n})$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+v=x'_{1}u'_{1}+\dots+x'_{n}u'_{n}=\sum_{j=1}^{n}x'_{j}u'_{j}=\sum_{j=1}^{n}x'_{j}\left(\sum_{i=1}^{n}p_{ij}u_{i}\right)=\sum_{j=1}^{n}\sum_{i=1}^{n}x'_{j}p_{ij}u_{i}=\sum_{i=1}^{n}\left(\sum_{j=1}^{n}x'_{j}p_{ij}\right)u_{i}
+\]
+
+\end_inset
+
+De forma que si 
+\begin_inset Formula $[v]_{{\cal B}}=(x_{1},\dots,x_{n})$
+\end_inset
+
+, entonces 
+\begin_inset Formula $x_{i}=\sum_{j=1}^{n}x'_{j}p_{ij}$
+\end_inset
+
+.
+ A las ecuaciones
+\begin_inset Formula 
+\[
+\left.\begin{array}{ccccccc}
+x_{1} & = & p_{11}x'_{1} & + & \dots & + & p_{1n}x'_{n}\\
+ & \vdots\\
+x_{n} & = & p_{n1}x'_{1} & + & \dots & + & p_{nn}x'_{n}
+\end{array}\right\} 
+\]
+
+\end_inset
+
+las llamamos 
+\series bold
+ecuaciones del cambio de base de 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ a 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Producto de matrices
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $A=(a_{ij})\in M_{m,n}(K)$
+\end_inset
+
+ y 
+\begin_inset Formula $B=(b_{ij})\in M_{n,p}(K)$
+\end_inset
+
+, definimos 
+\begin_inset Formula $AB=(c_{ij})\in M_{m,p}(K)$
+\end_inset
+
+ tal que 
+\begin_inset Formula $c_{ij}=\sum_{k=1}^{n}a_{ik}b_{kj}$
+\end_inset
+
+.
+ En general no es conmutativo, aun si ambos productos se pueden efectuar
+ y fuesen matrices del mismo tamaño.
+ Propiedades:
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Asociativa: 
+\series default
+
+\begin_inset Formula $(AB)C=A(BC)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\begin{array}{c}
+((AB)C)_{ij}=\sum_{k=1}^{p}(AB)_{ik}C_{kj}=\sum_{k=1}^{p}\sum_{l=1}^{n}A_{il}B_{lk}C_{kj}=\sum_{l=1}^{n}\sum_{k=1}^{p}A_{il}B_{lk}C_{kj}=\\
+=\sum_{l=1}^{n}A_{il}\left(\sum_{k=1}^{p}B_{lk}C_{kj}\right)=\sum_{l=1}^{n}A_{il}(BC)_{lj}=(A(BC))_{ij}
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Distributiva respecto de la suma:
+\series default
+ 
+\begin_inset Formula $A(B+C)=AB+AC$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+La 
+\series bold
+matriz identidad
+\series default
+, 
+\begin_inset Formula $I_{n}=(\delta_{ij})$
+\end_inset
+
+ con 
+\begin_inset Formula $\delta_{ii}=1$
+\end_inset
+
+ y 
+\begin_inset Formula $\delta_{ij}=0$
+\end_inset
+
+ si 
+\begin_inset Formula $i\neq j$
+\end_inset
+
+, satisface 
+\begin_inset Formula $AI_{n}=A$
+\end_inset
+
+ y 
+\begin_inset Formula $I_{n}B=B$
+\end_inset
+
+ para cada 
+\begin_inset Formula $A\in M_{m,n}(K)$
+\end_inset
+
+ y 
+\begin_inset Formula $B\in M_{n,m}(K)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+C_{ij}=\sum_{k=1}^{n}A_{ik}\delta_{kj}\underset{\text{(el resto son \ensuremath{=0})}}{=}A_{ij}\delta_{jj}=A_{ij}
+\]
+
+\end_inset
+
+La demostración de que 
+\begin_inset Formula $I_{n}B=B$
+\end_inset
+
+ es análoga.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\alpha(AB)=(\alpha A)B=A(\alpha B)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $A\in M_{n}(K)$
+\end_inset
+
+ es 
+\series bold
+invertible
+\series default
+ si existe 
+\begin_inset Formula $B\in M_{n}(K)$
+\end_inset
+
+ tal que 
+\begin_inset Formula $AB=BA=I_{n}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $B$
+\end_inset
+
+ es la 
+\series bold
+matriz inversa
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ y se representa 
+\begin_inset Formula $A^{-1}$
+\end_inset
+
+.
+ Supongamos que 
+\begin_inset Formula $B$
+\end_inset
+
+ y 
+\begin_inset Formula $C$
+\end_inset
+
+ son inversas de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $C=CI_{n}=C(AB)=(CA)B=I_{n}B=B$
+\end_inset
+
+, por lo que la inversa es única.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $A,B\in M_{n}(K)$
+\end_inset
+
+ son invertibles, entonces 
+\begin_inset Formula $AB$
+\end_inset
+
+ es también invertible, y 
+\begin_inset Formula $(AB)^{-1}=B^{-1}A^{-1}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+(B^{-1}A^{-1})(AB)=B^{-1}(A^{-1}A)B=B^{-1}I_{n}B=B^{-1}B=I_{n}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Las ecuaciones de cambio de base se pueden expresar como
+\begin_inset Formula 
+\[
+\left(\begin{array}{c}
+x_{1}\\
+x_{2}\\
+\vdots\\
+x_{n}
+\end{array}\right)=\left(\begin{array}{cccc}
+p_{11} & p_{12} & \cdots & p_{1n}\\
+p_{21} & p_{22} & \cdots & p_{2n}\\
+\vdots & \vdots & \ddots & \vdots\\
+p_{n1} & p_{n2} & \cdots & p_{nn}
+\end{array}\right)\left(\begin{array}{c}
+x'_{1}\\
+x'_{2}\\
+\vdots\\
+x'_{n}
+\end{array}\right)
+\]
+
+\end_inset
+
+donde las columnas de 
+\begin_inset Formula $(p_{ij})$
+\end_inset
+
+ son los vectores de 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ respecto a 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+.
+ Abreviadamente, 
+\begin_inset Formula $X_{{\cal B}}=M_{{\cal B}{\cal B}'}X'_{{\cal B}'}$
+\end_inset
+
+, donde a 
+\begin_inset Formula $M_{{\cal B}{\cal B}'}$
+\end_inset
+
+ la llamamos 
+\series bold
+matriz de cambio de base
+\series default
+ de 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ a 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+.
+ Podemos deducir que 
+\begin_inset Formula $M_{{\cal B}{\cal B}}=I_{n}$
+\end_inset
+
+, 
+\begin_inset Formula $M_{{\cal B}''{\cal B}}=M_{{\cal B}''{\cal B}'}M_{{\cal B}'{\cal B}}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $M_{{\cal B}'{\cal B}}M_{{\cal B}{\cal B}'}=M_{{\cal B}{\cal B}'}M_{{\cal B}'{\cal B}}=I_{n}$
+\end_inset
+
+.
+ Así, toda matriz de cambio de base es invertible, y viceversa.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+¿Buscar la demostración de que si es invertible también es de cambio de
+ base?
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Operaciones con subespacios
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $\{U_{i}\}_{i\in I}\neq\emptyset$
+\end_inset
+
+ es un conjunto de subespacios vectoriales de 
+\begin_inset Formula $V$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\bigcap_{i\in I}U_{i}$
+\end_inset
+
+ también es subespacio vectorial de 
+\begin_inset Formula $V$
+\end_inset
+
+, pero en general 
+\begin_inset Formula $\bigcup_{i\in I}U_{i}$
+\end_inset
+
+ no es subespacio vectorial.
+ Llamamos 
+\series bold
+suma
+\series default
+ de 
+\begin_inset Formula $U_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $U_{2}$
+\end_inset
+
+ al subespacio
+\begin_inset Formula 
+\[
+U_{1}+U_{2}=\{u_{1}+u_{2}|u_{1}\in U_{1},u_{2}\in U_{2}\}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Este es el menor subespacio vectorial 
+\begin_inset Formula $V$
+\end_inset
+
+ que contiene a 
+\begin_inset Formula $U_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $U_{2}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $u\in U_{1}$
+\end_inset
+
+, como 
+\begin_inset Formula $0\in U_{2}$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $u=u+0\in U_{1}+U_{2}$
+\end_inset
+
+, y viceversa, de modo que 
+\begin_inset Formula $U_{1}\cup U_{2}\subseteq U_{1}+U_{2}$
+\end_inset
+
+ y además 
+\begin_inset Formula $U_{1}+U_{2}\neq\emptyset$
+\end_inset
+
+.
+ Demostraremos ahora que es un espacio vectorial.
+ Si 
+\begin_inset Formula $v,v'\in U_{1}+U_{2}$
+\end_inset
+
+, existirán 
+\begin_inset Formula $u_{1},u'_{1}\in U$
+\end_inset
+
+ y 
+\begin_inset Formula $u_{2},u'_{2}\in U_{2}$
+\end_inset
+
+ con 
+\begin_inset Formula $v=u_{1}+u_{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $v'=u'_{1}+u'_{2}$
+\end_inset
+
+, y si 
+\begin_inset Formula $\alpha,\alpha'\in K$
+\end_inset
+
+, se tiene que
+\begin_inset Formula 
+\[
+\alpha v+\alpha'v'=\alpha u_{1}+\alpha u_{2}+\alpha'u'_{1}+\alpha'u'_{2}=(\alpha u_{1}+\alpha'u'_{1})+(\alpha u_{2}+\alpha'u'_{2})\in U_{1}+U_{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+La 
+\series bold
+fórmula de Grassmann
+\series default
+ o 
+\series bold
+teorema de Grassman
+\series default
+ nos dice que si 
+\begin_inset Formula $U_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $U_{2}$
+\end_inset
+
+ son subespacios de 
+\begin_inset Formula $V$
+\end_inset
+
+, y 
+\begin_inset Formula $V$
+\end_inset
+
+ tiene dimensión finita, entonces 
+\begin_inset Formula 
+\[
+\dim(U_{1})+\dim(U_{2})=\dim(U_{1}\cap U_{2})+\dim(U_{1}+U_{2})
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración.
+
+\series default
+ Sea 
+\begin_inset Formula $\{u_{1},\dots,u_{t}\}$
+\end_inset
+
+ base de 
+\begin_inset Formula $U_{1}\cap U_{2}$
+\end_inset
+
+, que completamos por un lado a la base 
+\begin_inset Formula $\{u_{1},\dots,u_{t},u_{t+1},\dots,u_{r}\}$
+\end_inset
+
+ de 
+\begin_inset Formula $U_{1}$
+\end_inset
+
+ y por otro a la base 
+\begin_inset Formula $\{u_{1},\dots,u_{t},v_{t+1},\dots,v_{s}\}$
+\end_inset
+
+ de 
+\begin_inset Formula $U_{2}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\{u_{1},\dots,u_{t},u_{t+1},\dots,u_{r},v_{t+1},\dots,v_{s}\}$
+\end_inset
+
+ es sistema de generadores de 
+\begin_inset Formula $U_{1}+U_{2}$
+\end_inset
+
+.
+ Queda ver que es además linealmente independiente.
+\end_layout
+
+\begin_layout Standard
+Supongamos 
+\begin_inset Formula $\alpha_{1}u_{1}+\dots+\alpha_{t}u_{t}+\alpha_{t+1}u_{t+1}+\dots+\alpha_{r}u_{r}+\beta_{t+1}v_{t+1}+\dots+\beta_{s}v_{s}=0$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\alpha_{1}u_{1}+\dots+\alpha_{r}u_{r}=-\beta_{t+1}v_{t+1}-\dots-\beta_{s}v_{s}\in U_{1}\cap U_{2}$
+\end_inset
+
+.
+ Pero como 
+\begin_inset Formula $\{u_{1},\dots,u_{t}\}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $U_{1}\cap U_{2}$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $-\beta_{t+1}v_{t+1}-\dots-\beta_{s}v_{s}=\gamma_{1}u_{1}+\dots+\gamma_{t}u_{t}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $\gamma_{1}u_{1}+\dots+\gamma_{t}u_{t}+\beta_{t+1}v_{t+1}+\dots+\beta_{s}v_{s}=0$
+\end_inset
+
+.
+ Al ser 
+\begin_inset Formula $\{u_{1},\dots,u_{t},v_{t+1},\dots,v_{s}\}$
+\end_inset
+
+ base de 
+\begin_inset Formula $U_{2}$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $\beta_{t+1}=\dots=\beta_{s}=0$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $\alpha_{1}u_{1}+\dots+\alpha_{r}u_{r}=0$
+\end_inset
+
+.
+ Pero como 
+\begin_inset Formula $\{u_{1},\dots,u_{r}\}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $U_{1}$
+\end_inset
+
+ se tiene que 
+\begin_inset Formula $\alpha_{1}=\dots=\alpha_{r}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así, se tiene que 
+\begin_inset Formula $\{u_{1},\dots,u_{t},u_{t+1},\dots,u_{r},v_{t+1},\dots,v_{s}\}$
+\end_inset
+
+ es una familia linealmente independiente y por tanto una base del subespacio
+ 
+\begin_inset Formula $U_{1}+U_{2}$
+\end_inset
+
+, que tendrá dimensión 
+\begin_inset Formula $\dim(U_{1}+U_{2})=r+s-t=\dim(U_{1})+\dim(U_{2})-\dim(U_{1}\cap U_{2})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $U_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $U_{2}$
+\end_inset
+
+ subespacios de 
+\begin_inset Formula $V$
+\end_inset
+
+ están en 
+\series bold
+suma directa
+\series default
+ si 
+\begin_inset Formula $U_{1}\cap U_{2}=\{0\}$
+\end_inset
+
+.
+ Se dice entonces que 
+\begin_inset Formula $U_{1}+U_{2}$
+\end_inset
+
+ es una suma directa y se representa 
+\begin_inset Formula $U_{1}\oplus U_{2}$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $\dim(U_{1}\oplus U_{2})=\dim(U_{1})+\dim(U_{2})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una suma de subespacios 
+\begin_inset Formula $U_{1}+U_{2}$
+\end_inset
+
+ es directa si y sólo si 
+\begin_inset Formula $\forall v\in U_{1}+U_{2},\exists!u_{1}\in U_{1},u_{2}\in U_{2}:v=u_{1}+u_{2}$
+\end_inset
+
+.
+ Llamamos a 
+\begin_inset Formula $u_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $u_{2}$
+\end_inset
+
+ las 
+\series bold
+componentes
+\series default
+ de 
+\begin_inset Formula $v$
+\end_inset
+
+.
+\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 $u_{1}+u_{2}=u'_{1}+u'_{2}$
+\end_inset
+
+ con 
+\begin_inset Formula $u_{1},u'_{1}\in U_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $u_{2},u'_{2}\in U_{2}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $u_{1}-u'_{1}=u'_{2}-u_{2}\in U_{1}\cap U_{2}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $u_{1}-u'_{1}=u'_{2}-u_{2}=0$
+\end_inset
+
+ y 
+\begin_inset Formula $u_{1}=u'_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $u_{2}=u'_{2}$
+\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 
+\begin_inset Formula $u\in U_{1}\cap U_{2}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $u=u+0=0+u$
+\end_inset
+
+, pero como 
+\begin_inset Formula $u$
+\end_inset
+
+ se expresa de modo único como suma de un vector de 
+\begin_inset Formula $U_{1}$
+\end_inset
+
+ y otro de 
+\begin_inset Formula $U_{2}$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $u=0$
+\end_inset
+
+, y por tanto 
+\begin_inset Formula $U_{1}\cap U_{2}=\{0\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dado 
+\begin_inset Formula $U$
+\end_inset
+
+ subespacio vectorial de 
+\begin_inset Formula $V$
+\end_inset
+
+ existe 
+\begin_inset Formula $U'$
+\end_inset
+
+, llamado 
+\series bold
+complementario
+\series default
+ de 
+\begin_inset Formula $U$
+\end_inset
+
+ en 
+\begin_inset Formula $V$
+\end_inset
+
+, tal que 
+\begin_inset Formula $V=U\oplus U'$
+\end_inset
+
+, pues si expandimos la base 
+\begin_inset Formula $\{u_{1},\dots,u_{r}\}$
+\end_inset
+
+ de 
+\begin_inset Formula $U$
+\end_inset
+
+ a una base 
+\begin_inset Formula $\{u_{1},\dots,u_{r},u_{r+1},\dots,u_{n}\}$
+\end_inset
+
+ de 
+\begin_inset Formula $V$
+\end_inset
+
+ entonces 
+\begin_inset Formula $U'=<u_{r+1},\dots,u_{n}>$
+\end_inset
+
+ satisface la condición.
+ El complementario no tiene por qué ser único.
+\end_layout
+
+\begin_layout Standard
+Una suma 
+\begin_inset Formula $U_{1}+\dots+U_{k}$
+\end_inset
+
+ es suma directa si todo vector de la suma se expresa de modo único como
+ suma de un vector de cada 
+\begin_inset Formula $U_{i}$
+\end_inset
+
+, lo que ocurre si y sólo si 
+\begin_inset Formula $\forall i\in\{1,\dots,k\},U_{i}\cap(\sum_{1\leq j\leq k,j\neq i}U_{j})=\{0\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $u_{1},\dots,u_{k}\in V$
+\end_inset
+
+, entonces 
+\begin_inset Formula $<u_{1},\dots,u_{k}>=<u_{1}>+\dots+<u_{k}>$
+\end_inset
+
+.
+ Si además son linealmente independientes, entonces 
+\begin_inset Formula $<u_{1},\dots,u_{k}>=<u_{1}>\oplus\dots\oplus<u_{n}>$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $\{u_{1},\dots,u_{n}\}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $V$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $V=<u_{1}>\oplus\dots\oplus<u_{n}>$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/algl/n2.lyx b/algl/n2.lyx
new file mode 100644
index 0000000..bd834d3
--- /dev/null
+++ b/algl/n2.lyx
@@ -0,0 +1,2102 @@
+#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
+\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 swiss
+\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
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ es una 
+\series bold
+aplicación lineal
+\series default
+ u 
+\series bold
+homomorfismo de espacios vectoriales
+\series default
+ si 
+\begin_inset Formula $f(u+u')=f(u)+f(u')\forall u,u'\in U$
+\end_inset
+
+ y 
+\begin_inset Formula $f(\alpha u)=\alpha f(u)\forall\alpha\in K,u\in U$
+\end_inset
+
+, es decir, si 
+\begin_inset Formula $f(\sum\alpha_{i}u_{i})=\sum\alpha_{i}f(u_{i})$
+\end_inset
+
+.
+ Ejemplos:
+\end_layout
+
+\begin_layout Itemize
+La 
+\series bold
+aplicación identidad:
+\series default
+ 
+\begin_inset Formula $Id_{V}:V\rightarrow V$
+\end_inset
+
+ con 
+\begin_inset Formula $Id_{V}(v)=v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+La 
+\series bold
+aplicación inclusión:
+\series default
+ 
+\begin_inset Formula $i:U\subseteq V\rightarrow V$
+\end_inset
+
+ con 
+\begin_inset Formula $i(u)=u$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+La 
+\series bold
+aplicación lineal nula:
+\series default
+ 
+\begin_inset Formula $0:U\rightarrow V$
+\end_inset
+
+ con 
+\begin_inset Formula $0(u)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+La 
+\series bold
+homotecia de razón 
+\begin_inset Formula $\alpha$
+\end_inset
+
+:
+\series default
+ 
+\begin_inset Formula $h_{\alpha}:V\rightarrow V$
+\end_inset
+
+ con 
+\begin_inset Formula $h_{\alpha}(v)=\alpha v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Las 
+\series bold
+proyecciones de 
+\begin_inset Formula $V$
+\end_inset
+
+ sobre 
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula $W$
+\end_inset
+
+,
+\series default
+ con 
+\begin_inset Formula $V=U\oplus W$
+\end_inset
+
+: 
+\begin_inset Formula $p_{U}:V\rightarrow U$
+\end_inset
+
+ y 
+\begin_inset Formula $p_{W}:V\rightarrow W$
+\end_inset
+
+, tales que si 
+\begin_inset Formula $v=u+w$
+\end_inset
+
+ con 
+\begin_inset Formula $v\in V$
+\end_inset
+
+, 
+\begin_inset Formula $u\in U$
+\end_inset
+
+ y 
+\begin_inset Formula $w\in W$
+\end_inset
+
+, entonces 
+\begin_inset Formula $p_{U}(v)=u$
+\end_inset
+
+ y 
+\begin_inset Formula $p_{W}(v)=w$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+La aplicación 
+\begin_inset Formula $f_{A}:K^{n}\rightarrow K^{m}$
+\end_inset
+
+ con 
+\begin_inset Formula $A\in M_{m,n}(K)$
+\end_inset
+
+, dada por 
+\begin_inset Formula 
+\[
+f_{A}(v)=A\left(\begin{array}{c}
+|\\
+v\\
+|
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Tenemos que 
+\begin_inset Formula $f(0_{U})=0_{V}$
+\end_inset
+
+, y que 
+\begin_inset Formula $f(-u)=-f(u)$
+\end_inset
+
+.
+ Además, si 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ y 
+\begin_inset Formula $g:V\rightarrow W$
+\end_inset
+
+ son aplicaciones lineales, 
+\begin_inset Formula $g\circ f:U\rightarrow W$
+\end_inset
+
+ también lo es.
+\end_layout
+
+\begin_layout Section
+Aplicaciones lineales y subespacios.
+ Núcleo e Imagen
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+núcleo
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ se define como 
+\begin_inset Formula $\text{Nuc}(f)=\ker(f)=f^{-1}(\{0\})$
+\end_inset
+
+, y la 
+\series bold
+imagen
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ como 
+\begin_inset Formula $\text{Im}(f)=\{f(u)\}_{u\in U}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $U'\leq U$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f(U')\leq V$
+\end_inset
+
+, y si 
+\begin_inset Formula $V'\leq V$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\text{Nuc}(f)\leq f^{-1}(V')\leq U$
+\end_inset
+
+.
+ En particular, 
+\begin_inset Formula $\text{Nuc}(f)$
+\end_inset
+
+ e 
+\begin_inset Formula $\text{Im}(f)$
+\end_inset
+
+ son espacios vectoriales, y si 
+\begin_inset Formula $U'=<u_{1},\dots,u_{r}>\leq U$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f(U')=<f(u_{1}),\dots,f(u_{r})>$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $u_{1},u_{2}\in U,\alpha_{1},\alpha_{2}\in K$
+\end_inset
+
+, y sean 
+\begin_inset Formula $v_{1}=f(u_{1}),v_{2}=f(u_{2})\in f(U')$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\alpha_{1}v_{1}+\alpha_{2}v_{2}=\alpha_{1}f(u_{1})+\alpha_{2}f(u_{2})=f(\alpha_{1}u_{1}+\alpha_{2}u_{2})\in f(U')$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $f(U')$
+\end_inset
+
+ es un espacio vectorial.
+ Ahora bien, si 
+\begin_inset Formula $V'$
+\end_inset
+
+ es un subespacio de 
+\begin_inset Formula $V$
+\end_inset
+
+ entonces 
+\begin_inset Formula $\{0\}\subseteq V'$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $f^{-1}(\{0\})=\text{Nuc}(f)\subseteq f^{-1}(V')$
+\end_inset
+
+.
+ Entonces si 
+\begin_inset Formula $u_{1},u_{2}\in f^{-1}(V')$
+\end_inset
+
+ y 
+\begin_inset Formula $\alpha_{1},\alpha_{2}\in K$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f(\alpha_{1}u_{1}+\alpha_{2}u_{2})=\alpha_{1}f(u_{1})+\alpha_{2}f(u_{2})\in V'$
+\end_inset
+
+, y por lo tanto 
+\begin_inset Formula $\alpha_{1}u_{1}+\alpha_{2}u_{2}\in f^{-1}(V')$
+\end_inset
+
+ y 
+\begin_inset Formula $f^{-1}(V')$
+\end_inset
+
+ es un espacio vectorial.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema:
+\series default
+ Para 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ con 
+\begin_inset Formula $\dim(U)$
+\end_inset
+
+ finita, entonces 
+\begin_inset Formula $\dim(U)=\dim(\text{Nuc}(f))+\dim(\text{Im}(f))$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $\{v_{1},\dots,v_{n}\}$
+\end_inset
+
+ base de 
+\begin_inset Formula $U$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\text{Im}(f)=<f(v_{1}),\dots f(v_{n})>$
+\end_inset
+
+ es de dimensión finita.
+ Ahora sea 
+\begin_inset Formula $\{v_{1},\dots,v_{k}\}$
+\end_inset
+
+ base de 
+\begin_inset Formula $\text{Nuc}(f)\leq U$
+\end_inset
+
+, con 
+\begin_inset Formula $k\leq n$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $f(v_{1})=\dots=f(v_{k})=0$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $\text{Im}(f)=<f(v_{1}),\dots,f(v_{k}),f(v_{k+1}),\dots,f(v_{n})>=<f(v_{k+1}),\dots,f(v_{n})>$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $\{f(v_{k+1}),\dots,f(v_{n})\}$
+\end_inset
+
+ es sistema de generadores de 
+\begin_inset Formula $\text{Im}(f)$
+\end_inset
+
+.
+ A continuación mostramos que es linealmente independiente.
+ Sea 
+\begin_inset Formula $0=\alpha_{k+1}f(v_{k+1})+\dots+\alpha_{n}f(v_{n})=f(\alpha_{k+1}v_{k+1}+\dots+\alpha_{n}v_{n})$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\alpha_{k+1}v_{k+1}+\dots+\alpha_{n}v_{n}\in\text{Nuc}(f)$
+\end_inset
+
+, por lo que existen 
+\begin_inset Formula $\beta_{1},\dots,\beta_{k}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $\alpha_{k+1}v_{k+1}+\dots+\alpha_{n}v_{n}=\beta_{1}v_{1}+\dots+\beta_{k}v_{k}$
+\end_inset
+
+.
+ Pero entonces 
+\begin_inset Formula $\beta_{1}v_{1}+\dots+\beta_{k}v_{k}-\alpha_{k+1}v_{k+1}-\dots-\alpha_{n}v_{n}=0$
+\end_inset
+
+, y como 
+\begin_inset Formula $\{v_{1},\dots,v_{n}\}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $U$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $\beta_{1}=\dots=\beta_{k}=\alpha_{k+1}=\dots=\alpha_{n}=0$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $\{v_{k+1},\dots,v_{n}\}$
+\end_inset
+
+ es linealmente independiente y por ello 
+\begin_inset Formula $\{f(v_{k+1}),\dots,f(v_{n})\}$
+\end_inset
+
+ también, por lo que es base de 
+\begin_inset Formula $\text{Im}(f)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+rango
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ a la dimensión de la imagen: 
+\begin_inset Formula $\text{rang}(f)=\dim(\text{Im}(f))$
+\end_inset
+
+.
+ Así, dada 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ y 
+\begin_inset Formula $\{u_{1},\dots,u_{n}\}$
+\end_inset
+
+ base de 
+\begin_inset Formula $U$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f(U)=<f(u_{1}),\dots,f(u_{n})>$
+\end_inset
+
+ y por tanto
+\begin_inset Formula 
+\[
+\text{rang}(f)=\dim(\text{Im}(f))=\dim(<f(u_{1}),\dots,f(u_{n})>)=\text{rang}(\{f(u_{1}),\dots,f(u_{n})\})
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ es una aplicación lineal y 
+\begin_inset Formula $\dim(U)=\dim(V)<\infty$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+f\text{ inyectiva}\iff f\text{ suprayectiva}\iff f\text{ biyectiva}\iff\text{rang}(f)=\dim(U)\iff\text{Nuc}(f)=\{0\}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $1\iff2\iff3]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Equivalen al hecho de que, para 
+\begin_inset Formula $f:A\rightarrow B$
+\end_inset
+
+ con 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ conjuntos finitos, es lo mismo decir que 
+\begin_inset Formula $f$
+\end_inset
+
+ sea inyectiva, suprayectiva o biyectiva.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $3\iff4]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\text{rang}(f)=\dim(\text{Im}(U))\overset{\text{(supray.)}}{=}\text{dim}(V)$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $f$
+\end_inset
+
+ no fuera suprayectiva, entonces 
+\begin_inset Formula $\dim(\text{Im}(U))<\dim(V)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $1\implies5]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $u\in\text{Nuc}(f)\implies f(u)=0_{V}=f(0_{U})\implies u=0_{U}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $5\implies1]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\text{Nuc}(f)=\{0\}\implies\left(f(u)=f(u')\implies0=f(u-u')\implies u-u'\in\text{Nuc}(f)\implies u=u'\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+El homomorfismo 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ es un 
+\series bold
+isomorfismo de espacios vectoriales
+\series default
+ si es biyectivo, un 
+\series bold
+endomorfismo
+\series default
+ de 
+\begin_inset Formula $U$
+\end_inset
+
+ si 
+\begin_inset Formula $U=V$
+\end_inset
+
+ y un 
+\series bold
+automorfismo
+\series default
+ es un endomorfismo biyectivo.
+ Ahora, dado el isomorfismo 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+, 
+\begin_inset Formula $f^{-1}:V\rightarrow U$
+\end_inset
+
+ es una aplicación lineal y por tanto un isomorfismo.
+ 
+\series bold
+Demostración:
+\series default
+ Consideramos 
+\begin_inset Formula $u=f^{-1}(v)$
+\end_inset
+
+ y 
+\begin_inset Formula $u'=f^{-1}(v')$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $f(u+u')=f(u)+f(u')=v+v'$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $f^{-1}(v+v')=u+u'=f^{-1}(v)+f^{-1}(v')$
+\end_inset
+
+.
+ Del mismo modo, 
+\begin_inset Formula $f(\alpha u)=\alpha f(u)=\alpha v$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $f^{-1}(\alpha v)=\alpha u=\alpha f^{-1}(v)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula $V$
+\end_inset
+
+ son 
+\series bold
+isomorfos
+\series default
+ (
+\begin_inset Formula $U\cong V$
+\end_inset
+
+) si existe un isomorfismo 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+.
+ Podemos comprobar que la relación es de equivalencia, y si 
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula $V$
+\end_inset
+
+ son 
+\begin_inset Formula $K$
+\end_inset
+
+-espacios vectoriales, entonces 
+\begin_inset Formula $U\cong V\iff\dim(U)=\dim(V)$
+\end_inset
+
+.
+\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
+
+
+\begin_inset Formula $U\cong V\implies\dim(U)=\dim(\text{Nuc}(f))+\dim(\text{Im}(f))=0+\dim(V)$
+\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
+
+Sean 
+\begin_inset Formula $f:U\rightarrow K^{n}$
+\end_inset
+
+ y 
+\begin_inset Formula $g:V\rightarrow K^{n}$
+\end_inset
+
+ isomorfismos con 
+\begin_inset Formula $f(u)=[u]_{{\cal B}}$
+\end_inset
+
+ y 
+\begin_inset Formula $g(v)=[v]_{\beta'}$
+\end_inset
+
+ ; entonces 
+\begin_inset Formula $g^{-1}\circ f:U\rightarrow V$
+\end_inset
+
+ también es un isomorfismo.
+\end_layout
+
+\begin_layout Section
+Determinación de una aplicación lineal
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula $V$
+\end_inset
+
+ son 
+\begin_inset Formula $K$
+\end_inset
+
+-espacios vectoriales con 
+\begin_inset Formula ${\cal B}=\{u_{1},\dots,u_{n}\}$
+\end_inset
+
+ base de 
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ vectores cualesquiera de 
+\begin_inset Formula $V$
+\end_inset
+
+, existe una única 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ con 
+\begin_inset Formula $f(u_{i})=v_{i}\forall i$
+\end_inset
+
+, pues es aquella dada por 
+\begin_inset Formula $f(\alpha_{1}u_{1}+\dots+\alpha_{n}u_{n})=\alpha_{1}v_{1}+\dots+\alpha_{n}v_{n}$
+\end_inset
+
+.
+ Esto también se cumple para espacios de dimensión infinita.
+\end_layout
+
+\begin_layout Standard
+También, si 
+\begin_inset Formula ${\cal B}=\{u_{i}\}_{i\in I}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $U$
+\end_inset
+
+ (la cual puede ser infinita) y 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ lineal entonces:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f$
+\end_inset
+
+ es inyectiva si y sólo si 
+\begin_inset Formula $\{f(u_{i})\}_{i\in I}$
+\end_inset
+
+ es linealmente independiente.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $0=\alpha_{1}f(u_{1})+\dots+\alpha_{k}f(u_{k})=f(\alpha_{1}u_{1}+\dots+\alpha_{k}u_{k})\implies\alpha_{1}u_{1}+\dots+\alpha_{k}u_{k}\in\text{Nuc}(f)=\{0\}\implies\alpha_{1}u_{1}+\dots+\alpha_{k}u_{k}=0\implies\alpha_{1},\dots,\alpha_{k}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Partimos de que 
+\begin_inset Formula $\{f(u_{i})\}_{i\in I}$
+\end_inset
+
+ es linealmente independiente.
+ Sea 
+\begin_inset Formula $u\in\text{Nuc}(f)$
+\end_inset
+
+, si 
+\begin_inset Formula $u=\alpha_{1}u_{1}+\dots+\alpha_{k}u_{k}$
+\end_inset
+
+ entonces 
+\begin_inset Formula $0=f(u)=\alpha_{1}f(u_{1})+\dots+\alpha_{k}f(u_{k})\implies\alpha_{i}=0\forall i\implies u=0$
+\end_inset
+
+, por lo que entonces 
+\begin_inset Formula $\text{Nuc}(f)=\{0\}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f$
+\end_inset
+
+ es suprayectiva si y sólo si 
+\begin_inset Formula $\{f(u_{i})\}_{i\in I}$
+\end_inset
+
+ es una familia de generadores de 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+f\text{ suprayectiva}\iff f(U)=V\iff<\{f(u_{i})\}_{i\in I}>=V
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f$
+\end_inset
+
+ es biyectiva, y por tanto isomorfismo, si y sólo si 
+\begin_inset Formula $\{f(u_{i})\}_{i\in I}$
+\end_inset
+
+ es
+\begin_inset space \space{}
+\end_inset
+
+ base de 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Representación matricial de una aplicación lineal.
+ Rango de una matriz.
+ Matrices equivalentes
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula $V$
+\end_inset
+
+ son 
+\begin_inset Formula $K$
+\end_inset
+
+-espacios vectoriales de dimensión finita, 
+\begin_inset Formula ${\cal B}=\{u_{1},\dots,u_{n}\}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'=\{v_{1},\dots,v_{m}\}$
+\end_inset
+
+ base de 
+\begin_inset Formula $V$
+\end_inset
+
+, y 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ es un homomorfismo, entonces para cada 
+\begin_inset Formula $j$
+\end_inset
+
+ existirán 
+\begin_inset Formula $a_{ij}$
+\end_inset
+
+ tales que
+\begin_inset Formula 
+\[
+f(u_{j})=\sum_{i=1}^{m}a_{ij}v_{i}
+\]
+
+\end_inset
+
+Así, si 
+\begin_inset Formula $[u]_{{\cal B}}=(x_{1},\dots,x_{n})$
+\end_inset
+
+, entonces 
+\begin_inset Formula $u=\sum_{j=1}^{n}x_{j}u_{j}\in U$
+\end_inset
+
+ y
+\begin_inset Formula 
+\[
+f(u)=f\left(\sum_{j=1}^{n}x_{j}u_{j}\right)=\sum_{j=1}^{n}x_{j}f(u_{j})=\sum_{j=1}^{n}x_{j}\left(\sum_{i=1}^{m}a_{ij}v_{i}\right)=\sum_{j=1}^{n}\sum_{i=1}^{m}(x_{j}a_{ij})v_{i}=\sum_{i=1}^{m}\left(\sum_{j=1}^{n}a_{ij}x_{j}\right)v_{i}
+\]
+
+\end_inset
+
+por lo que 
+\begin_inset Formula $[f(u)]_{{\cal B}'}=(\sum_{j=1}^{n}a_{1j}x_{j},\dots,\sum_{j=1}^{n}a_{mj}x_{j})$
+\end_inset
+
+, de modo que, si 
+\begin_inset Formula $[f(u)]_{{\cal B}'}=(y_{1},\dots,y_{m})$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+\left(\begin{array}{c}
+y_{1}\\
+\vdots\\
+y_{m}
+\end{array}\right)=\left(\begin{array}{ccc}
+a_{11} & \cdots & a_{1n}\\
+\vdots & \ddots & \vdots\\
+a_{m1} & \cdots & a_{mn}
+\end{array}\right)\left(\begin{array}{c}
+x_{1}\\
+\vdots\\
+x_{n}
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Siendo las columnas de 
+\begin_inset Formula $(a_{ij})$
+\end_inset
+
+ los 
+\begin_inset Formula $[f(u_{j})]_{{\cal B}'}$
+\end_inset
+
+, es decir, las imágenes de los elementos de 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ respecto de 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $[f(u)]_{{\cal B}'}=A[u]_{{\cal B}}$
+\end_inset
+
+, lo que se conoce como 
+\series bold
+representación matricial de 
+\begin_inset Formula $f$
+\end_inset
+
+ respecto a las bases 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+
+\series default
+.
+ A la matriz 
+\begin_inset Formula $(a_{ij})$
+\end_inset
+
+ se le llama 
+\series bold
+matriz de 
+\begin_inset Formula $f$
+\end_inset
+
+
+\series default
+ o 
+\series bold
+matriz asociada a 
+\begin_inset Formula $f$
+\end_inset
+
+ respecto a las bases 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+
+\series default
+, y se denomina 
+\begin_inset Formula $M_{{\cal B}',{\cal B}}(f)$
+\end_inset
+
+.
+ Así,
+\begin_inset Formula 
+\[
+[f(u)]_{{\cal B}'}=M_{{\cal B}',{\cal B}}(f)[u]_{{\cal B}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Tenemos que 
+\begin_inset Formula $M_{{\cal B}',{\cal B}}(f)$
+\end_inset
+
+ está completamente determinada por 
+\begin_inset Formula $f$
+\end_inset
+
+, y de igual modo, 
+\begin_inset Formula $f$
+\end_inset
+
+ está univocamente determinada por 
+\begin_inset Formula $M_{{\cal B}',{\cal B}}(f)$
+\end_inset
+
+.
+ Además, si 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ y 
+\begin_inset Formula $g:V\rightarrow W$
+\end_inset
+
+ son aplicaciones lineales y 
+\begin_inset Formula ${\cal B}_{1}$
+\end_inset
+
+, 
+\begin_inset Formula ${\cal B}_{2}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}_{3}$
+\end_inset
+
+ son bases respectivas de 
+\begin_inset Formula $U$
+\end_inset
+
+, 
+\begin_inset Formula $V$
+\end_inset
+
+ y 
+\begin_inset Formula $W$
+\end_inset
+
+, entonces 
+\begin_inset Formula $M_{{\cal B}_{3},{\cal B}_{1}}(g\circ f)=M_{{\cal B}_{3},{\cal B}_{2}}(g)M_{{\cal B}_{2},{\cal B}_{1}}(f)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Para cada 
+\begin_inset Formula $u\in U,v\in V$
+\end_inset
+
+, tenemos que 
+\begin_inset Formula $M_{{\cal B}_{2},{\cal B}_{1}}(f)[u]_{{\cal B}_{1}}=[f(u)]_{{\cal B}_{2}}$
+\end_inset
+
+ y 
+\begin_inset Formula $M_{{\cal B}_{3},{\cal B}_{2}}(g)[v]_{{\cal B}_{2}}=[g(v)]_{{\cal B}_{3}}$
+\end_inset
+
+, por lo que
+\begin_inset Formula 
+\[
+M_{{\cal B}_{3},{\cal B}_{2}}(g)M_{{\cal B}_{2},{\cal B}_{1}}(f)[u]_{{\cal B}_{1}}=M_{{\cal B}_{3},{\cal B}_{2}}(g)[f(u)]_{{\cal B}_{2}}=[g(f(u))]_{{\cal B}_{3}}=[(g\circ f)(u)]_{{\cal B}_{3}}
+\]
+
+\end_inset
+
+y por la unicidad de la matriz de una aplicación lineal respecto a dos bases,
+ se tiene que 
+\begin_inset Formula $M_{{\cal B}_{3},{\cal B}_{1}}(g\circ f)=M_{{\cal B}_{3},{\cal B}_{2}}(g)M_{{\cal B}_{2},{\cal B}_{1}}(f)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula $V$
+\end_inset
+
+ 
+\begin_inset Formula $K$
+\end_inset
+
+-espacios vectoriales con bases respectivas 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+, 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+ una aplicación lineal y 
+\begin_inset Formula $A=M_{{\cal B}',{\cal B}}(f)$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es un isomorfismo si y sólo si 
+\begin_inset Formula $A$
+\end_inset
+
+ es invertible, y entonces 
+\begin_inset Formula $A^{-1}=M_{{\cal B},{\cal B}'}(f^{-1})$
+\end_inset
+
+.
+\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
+
+Sea 
+\begin_inset Formula $B=M_{{\cal B},{\cal B}'}(f^{-1})$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+AB=M_{{\cal B}',{\cal B}}(f)M_{{\cal B},{\cal B}'}(f^{-1})=M_{{\cal B}',{\cal B}'}(f\circ f^{-1})=M_{{\cal B}',{\cal B}'}(Id_{V})=I_{n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+BA=M_{{\cal B},{\cal B}'}(f^{-1})M_{{\cal B}',{\cal B}}(f)=M_{{\cal B},{\cal B}}(f^{-1}\circ f)=M_{{\cal B},{\cal B}}(Id_{U})=I_{n}
+\]
+
+\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
+
+Al ser invertible es cuadrada, por lo que 
+\begin_inset Formula $\dim(U)=\dim(V)=n$
+\end_inset
+
+, y si consideramos 
+\begin_inset Formula $g:V\rightarrow U$
+\end_inset
+
+ tal que 
+\begin_inset Formula $M_{{\cal B},{\cal B}'}(g)=A^{-1}$
+\end_inset
+
+, se tiene que
+\begin_inset Formula 
+\[
+M_{{\cal B},{\cal B}}(g\circ f)=M_{{\cal B},{\cal B}'}(g)M_{{\cal B}',{\cal B}}(f)=A^{-1}A=I_{n}=M_{{\cal B},{\cal B}}(Id_{U})\implies g\circ f=Id_{U}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+M_{{\cal B}',{\cal B}'}(f\circ g)=M_{{\cal B}',{\cal B}}(f)M_{{\cal B},{\cal B}'}(g)=AA^{-1}=I_{n}=M_{{\cal B}',{\cal B}'}(Id_{V})\implies f\circ g=Id_{V}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Así, si 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ son bases de 
+\begin_inset Formula $V$
+\end_inset
+
+, como 
+\begin_inset Formula $M_{{\cal B},{\cal B}'}=M_{{\cal B},{\cal B}'}(Id_{V})$
+\end_inset
+
+, se tiene que
+\begin_inset Formula 
+\[
+M_{{\cal B}',{\cal B}}^{-1}=(M_{{\cal B}',{\cal B}}(Id_{V}))^{-1}=M_{{\cal B},{\cal B}'}(Id_{V}^{-1})=M_{{\cal B},{\cal B}'}(Id_{V})=M_{{\cal B},{\cal B}'}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+También se tiene que si 
+\begin_inset Formula ${\cal B}_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}_{2}$
+\end_inset
+
+ son bases de 
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'_{2}$
+\end_inset
+
+ son bases de 
+\begin_inset Formula $V$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+M_{{\cal B}'_{2},{\cal B}_{2}}(f)=M_{{\cal B}'_{2},{\cal B}_{2}}(Id_{V}\circ f\circ Id_{U})=M_{{\cal B}'_{2},{\cal B}'_{1}}\cdot M_{{\cal B}'_{1}{\cal B}_{1}}(f)\cdot M_{{\cal B}_{1},{\cal B}_{2}}
+\]
+
+\end_inset
+
+Para 
+\begin_inset Formula $A,B\in M_{m,n}(K)$
+\end_inset
+
+, 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ son 
+\series bold
+equivalentes
+\series default
+ si existen matrices invertibles 
+\begin_inset Formula $P\in M_{m}(K)$
+\end_inset
+
+ y 
+\begin_inset Formula $Q\in M_{n}(K)$
+\end_inset
+
+ tales que 
+\begin_inset Formula $B=PAQ$
+\end_inset
+
+.
+ Esta es una relación de equivalencia.
+\end_layout
+
+\begin_layout Standard
+Se llama 
+\series bold
+rango
+\series default
+ de 
+\begin_inset Formula $A\in M_{m,n}(K)$
+\end_inset
+
+ al máximo de columnas linealmente independientres consideradas como vectores
+ de 
+\begin_inset Formula $K^{m}$
+\end_inset
+
+, es decir, 
+\begin_inset Formula $\text{rang}(A)=\dim(<C_{1},\dots,C_{n}>)$
+\end_inset
+
+, y por tanto 
+\begin_inset Formula $\text{rang}(A)\leq m,n$
+\end_inset
+
+.
+ Dado que las columnas de 
+\begin_inset Formula $M_{{\cal B}',{\cal B}}(f)$
+\end_inset
+
+ son las imágenes en 
+\begin_inset Formula $f$
+\end_inset
+
+ de los elementos de 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ sobre 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $\text{rang}(M_{{\cal B}',{\cal B}}(f))=\text{rang}(f)$
+\end_inset
+
+, y como las matrices invertibles corresponden a isomorfismos, se tiene
+ que 
+\begin_inset Formula $A\in M_{n}(K)$
+\end_inset
+
+ es invertible si y sólo si 
+\begin_inset Formula $\text{rang}(A)=n$
+\end_inset
+
+, para lo que basta con considerar el homomorfismo 
+\begin_inset Formula $f:K^{n}\rightarrow K^{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $M_{{\cal C},{\cal C}}(f)=A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $K$
+\end_inset
+
+-espacios vectoriales 
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula $V$
+\end_inset
+
+ con dimensiones respectivas 
+\begin_inset Formula $n$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+, y el homomorfismo 
+\begin_inset Formula $f:U\rightarrow V$
+\end_inset
+
+, existen bases 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ de 
+\begin_inset Formula $U$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ de 
+\begin_inset Formula $V$
+\end_inset
+
+ tales que
+\begin_inset Formula 
+\[
+M_{{\cal B}',{\cal B}}(f)=\left(\begin{array}{c|c}
+I_{r} & 0\\
+\hline 0 & 0
+\end{array}\right)\in M_{m,n}(K)
+\]
+
+\end_inset
+
+con 
+\begin_inset Formula $r=\text{rang}(f)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $r=\text{rang}(f)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $\dim(\text{Nuc}(f))=n-r$
+\end_inset
+
+.
+ Además, si 
+\begin_inset Formula $\{u_{r+1},\dots,u_{n}\}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $\text{Nuc}(f)$
+\end_inset
+
+ que se extiende a la base 
+\begin_inset Formula ${\cal B}=\{u_{1},\dots,u_{r},u_{r+1},\dots,u_{n}\}$
+\end_inset
+
+ de 
+\begin_inset Formula $U$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f(u_{r+1})=\dots=f(u_{n})=0$
+\end_inset
+
+, y 
+\begin_inset Formula $f(u_{1}),\dots,f(u_{r})$
+\end_inset
+
+ son linealmente dependientes, dado que si 
+\begin_inset Formula $\alpha_{1}f(u_{1})+\dots+\alpha_{r}f(u_{r})=0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $\alpha_{1}u_{1}+\dots+\alpha_{r}u_{r}\in\text{Nuc}(f)=<u_{r+1},\dots,u_{n}>$
+\end_inset
+
+ y como 
+\begin_inset Formula $<u_{1},\dots,u_{r}>\cap<u_{r+1},\dots,u_{n}>=\{0\}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\alpha_{1}u_{1}+\dots+\alpha_{r}u_{r}=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\alpha_{1}=\dots=\alpha_{r}=0$
+\end_inset
+
+.
+ Si extendemos este conjunto a la base 
+\begin_inset Formula ${\cal B}'=\{f(u_{1}),\dots,f(u_{r}),v_{r+1},\dots,v_{m}\}$
+\end_inset
+
+, se tiene la 
+\begin_inset Formula $M_{{\cal B}',{\cal B}}(f)$
+\end_inset
+
+ buscada.
+\end_layout
+
+\begin_layout Standard
+Toda 
+\begin_inset Formula $A\in M_{m,n}(K)$
+\end_inset
+
+ es equivalente a una de esta forma, con 
+\begin_inset Formula $r=\text{rang}(A)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $f:K^{n}\rightarrow K^{m}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $M_{{\cal C}',{\cal C}}(f)=A$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $M_{{\cal B}',{\cal B}}(f)=M_{{\cal B}',{\cal {\cal C}}'}\cdot A\cdot M_{{\cal C},{\cal B}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+matriz traspuesta
+\series default
+ de 
+\begin_inset Formula $A=(a_{ij})\in M_{m,n}(K)$
+\end_inset
+
+ a la matriz 
+\begin_inset Formula $A^{t}=(b_{ij})\in M_{n,m}(K)$
+\end_inset
+
+ con 
+\begin_inset Formula $b_{ij}=a_{ji}$
+\end_inset
+
+.
+ Se verifica que 
+\begin_inset Formula $(A^{t})^{t}=A$
+\end_inset
+
+, 
+\begin_inset Formula $(\alpha A)^{t}=\alpha A^{t}$
+\end_inset
+
+, 
+\begin_inset Formula $(A+B)^{t}=A^{t}+B^{t}$
+\end_inset
+
+, 
+\begin_inset Formula $(AC)^{t}=C^{t}A^{t}$
+\end_inset
+
+, y si 
+\begin_inset Formula $A$
+\end_inset
+
+ es invertible entonces 
+\begin_inset Formula $A^{t}$
+\end_inset
+
+ también lo es y 
+\begin_inset Formula $(A^{t})^{-1}=(A^{-1})^{t}$
+\end_inset
+
+.
+ Así, si
+\begin_inset Formula 
+\[
+B=\left(\begin{array}{c|c}
+I_{r} & 0\\
+\hline 0 & 0
+\end{array}\right)\in M_{m,n}(K)
+\]
+
+\end_inset
+
+con 
+\begin_inset Formula $r=\text{rang}(A)$
+\end_inset
+
+ y 
+\begin_inset Formula $A=M_{m,n}(K)$
+\end_inset
+
+, existirán 
+\begin_inset Formula $P\in M_{m}(K)$
+\end_inset
+
+ y 
+\begin_inset Formula $Q\in M_{n}(K)$
+\end_inset
+
+ tales que 
+\begin_inset Formula $B=PAQ$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $Q^{t}A^{t}P^{t}=(PAQ)^{t}=B^{t}$
+\end_inset
+
+ con 
+\begin_inset Formula $\text{rang}(B^{t})=\text{rang}(B)$
+\end_inset
+
+, y por tanto 
+\begin_inset Formula $\text{rang}(A)=\text{rang}(A^{t})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Matrices elementales.
+ Aplicaciones
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+matriz elemental
+\series default
+ de tamaño 
+\begin_inset Formula $n$
+\end_inset
+
+ a toda matriz obtenida al efectuar una operación elemental (por filas o
+ columnas) en 
+\begin_inset Formula $I_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $E_{n}(i,j)$
+\end_inset
+
+ es la resultante de intercambiar las filas 
+\begin_inset Formula $i$
+\end_inset
+
+ y 
+\begin_inset Formula $j$
+\end_inset
+
+, o las columnas 
+\begin_inset Formula $i$
+\end_inset
+
+ y 
+\begin_inset Formula $j$
+\end_inset
+
+, en 
+\begin_inset Formula $I_{n}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $E_{n}(i,j)^{-1}=E_{n}(i,j)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $E_{n}(r[i])$
+\end_inset
+
+ es la resultante de multiplicar por 
+\begin_inset Formula $r$
+\end_inset
+
+ la fila 
+\begin_inset Formula $i$
+\end_inset
+
+, o la columna 
+\begin_inset Formula $i$
+\end_inset
+
+, en 
+\begin_inset Formula $I_{n}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $E_{n}(r[i])^{-1}=E_{n}(r^{-1}[i])$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $E_{n}([i]+r[j])$
+\end_inset
+
+ es la resultante de añadir a la fila 
+\begin_inset Formula $i$
+\end_inset
+
+ la fila 
+\begin_inset Formula $j$
+\end_inset
+
+ multiplicada por 
+\begin_inset Formula $r$
+\end_inset
+
+, o a la columna 
+\begin_inset Formula $j$
+\end_inset
+
+ la columna 
+\begin_inset Formula $i$
+\end_inset
+
+ multiplicada por 
+\begin_inset Formula $r$
+\end_inset
+
+, en 
+\begin_inset Formula $I_{n}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $E_{n}([i]+r[j])^{-1}=E_{n}([i]-r[j])$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $B$
+\end_inset
+
+ se obtiene al realizar una operación elemental por filas en 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $E$
+\end_inset
+
+ al realizar la misma en 
+\begin_inset Formula $I_{m}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $B=EA$
+\end_inset
+
+.
+ Del mismo modo, si 
+\begin_inset Formula $B$
+\end_inset
+
+ se obtiene de aplicar una operación elemental por columnas en 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $E$
+\end_inset
+
+ al aplicarla a 
+\begin_inset Formula $I_{n}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $B=AE$
+\end_inset
+
+.
+ Así, realizar una serie de estas operaciones en una matriz equivale a multiplic
+arla por uno o ambos lados por un producto de matrices elementales, el cual
+ es invertible.
+ Dada una matriz 
+\begin_inset Formula $A$
+\end_inset
+
+, para obtener las matrices 
+\begin_inset Formula $P$
+\end_inset
+
+ y 
+\begin_inset Formula $Q$
+\end_inset
+
+ tales que
+\begin_inset Formula 
+\[
+PAQ=\left(\begin{array}{c|c}
+I_{r} & 0\\
+\hline 0 & 0
+\end{array}\right)
+\]
+
+\end_inset
+
+podemos partir de
+\begin_inset Formula 
+\[
+\left(\begin{array}{c|c}
+A & I_{m}\\
+\hline I_{n}
+\end{array}\right)
+\]
+
+\end_inset
+
+y realizar operaciones elementales hasta llegar a una matriz de la forma
+\begin_inset Formula 
+\[
+\left(\begin{array}{c|c}
+\begin{array}{c|c}
+I_{r} & 0\\
+\hline 0 & 0
+\end{array} & P\\
+\hline Q
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula $A\in M_{n}(K)$
+\end_inset
+
+ es invertible cuando tiene rango precisamente 
+\begin_inset Formula $n$
+\end_inset
+
+, por lo que al hacer operaciones elementales por filas para obtener una
+ matriz escalonada reducida, esta será precisamente 
+\begin_inset Formula $I_{n}$
+\end_inset
+
+, de forma que 
+\begin_inset Formula $(E_{k}\cdots E_{1})A=I_{n}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $A^{-1}=E_{k}\cdots E_{1}$
+\end_inset
+
+, de forma que 
+\begin_inset Formula $A^{-1}$
+\end_inset
+
+ es la matriz resultante de efectuar en 
+\begin_inset Formula $I_{n}$
+\end_inset
+
+ las mismas operaciones elementales fila que se hacen en 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ A efectos prácticos, formamos la matriz 
+\begin_inset Formula $\left(\begin{array}{c|c}
+A & I_{n}\end{array}\right)$
+\end_inset
+
+ y hacemos operaciones elementales por filas hasta llegar a 
+\begin_inset Formula $\left(\begin{array}{c|c}
+I_{n} & A^{-1}\end{array}\right)$
+\end_inset
+
+.
+ Por otro lado, si 
+\begin_inset Formula $A^{-1}=E_{k}\cdots E_{1}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $A=(A^{-1})^{-1}=(E_{k}\cdots E_{1})^{-1}=E_{1}^{-1}\cdots E_{k}^{-1}$
+\end_inset
+
+, de forma que toda matriz invertible es producto de matrices elementales.
+\end_layout
+
+\end_body
+\end_document
diff --git a/algl/n3.lyx b/algl/n3.lyx
new file mode 100644
index 0000000..99baa10
--- /dev/null
+++ b/algl/n3.lyx
@@ -0,0 +1,652 @@
+#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
+\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 swiss
+\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
+ecuación lineal
+\series default
+ en las 
+\begin_inset Formula $n$
+\end_inset
+
+ 
+\series bold
+incógnitas
+\series default
+ 
+\begin_inset Formula $x_{1},\dots,x_{n}$
+\end_inset
+
+ sobre el cuerpo 
+\begin_inset Formula $K$
+\end_inset
+
+ es una expresión de la forma 
+\begin_inset Formula $a_{1}x_{1}+\dots+a_{n}x_{n}=b$
+\end_inset
+
+, con los 
+\begin_inset Formula $a_{i}\in K$
+\end_inset
+
+ (
+\series bold
+coeficientes
+\series default
+) y 
+\begin_inset Formula $b\in K$
+\end_inset
+
+ (
+\series bold
+término independiente
+\series default
+).
+ Un 
+\series bold
+sistema de 
+\begin_inset Formula $m$
+\end_inset
+
+ ecuaciones lineales
+\series default
+ con 
+\begin_inset Formula $n$
+\end_inset
+
+ incógnitas sobre el cuerpo 
+\series bold
+
+\begin_inset Formula $K$
+\end_inset
+
+
+\series default
+ tiene la forma
+\begin_inset Formula 
+\[
+\left.\begin{array}{ccccccc}
+a_{11}x_{1} & + & \dots & + & a_{1n}x_{n} & = & b_{1}\\
+ &  &  &  &  & \vdots\\
+a_{m1}x_{1} & + & \dots & + & a_{mn}x_{n} & = & b_{m}
+\end{array}\right\} 
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Puede expresarse matricialmente de la forma 
+\begin_inset Formula $AX=B$
+\end_inset
+
+, donde 
+\begin_inset Formula $A=(a_{ij})\in M_{m,n}(K)$
+\end_inset
+
+ es la 
+\series bold
+matriz de los coeficientes
+\series default
+, 
+\begin_inset Formula $B=(b_{i})\in M_{m,1}(K)$
+\end_inset
+
+ es la 
+\series bold
+matriz de los términos independientes
+\series default
+ y 
+\begin_inset Formula $X=(x_{i})\in M_{n,1}(K)$
+\end_inset
+
+ es la matriz de incógnitas.
+ A la matriz 
+\begin_inset Formula $(A|B)\in M_{m,n+1}(K)$
+\end_inset
+
+ se le llama 
+\series bold
+matriz ampliada
+\series default
+ del sistema.
+ Un sistema es 
+\series bold
+homogéneo
+\series default
+ si 
+\begin_inset Formula $B=0$
+\end_inset
+
+, y a cada sistema 
+\begin_inset Formula $AX=B$
+\end_inset
+
+ se le puede asociar el sistema homogéneo 
+\begin_inset Formula $AX=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Se llama 
+\series bold
+solución
+\series default
+ a toda 
+\begin_inset Formula $n$
+\end_inset
+
+-upla 
+\begin_inset Formula $(r_{1},\dots,r_{n})\in K^{n}$
+\end_inset
+
+ tal que si 
+\begin_inset Formula $R=(r_{i})\in M_{n,1}(K)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $AR=B$
+\end_inset
+
+.
+ Un sistema es 
+\series bold
+compatible
+\series default
+ si tiene alguna solución, 
+\series bold
+determinado
+\series default
+ si tiene solo una e 
+\series bold
+indeterminado
+\series default
+ si tiene más; o 
+\series bold
+incompatible
+\series default
+ si no tiene ninguna.
+ 
+\series bold
+Discutir
+\series default
+ un sistema es determinar su compatibilidad, y 
+\series bold
+resolverlo
+\series default
+ es encontrar las soluciones.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema:
+\series default
+ Si un sistema 
+\begin_inset Formula $AX=B$
+\end_inset
+
+ tiene una solución 
+\begin_inset Formula $P$
+\end_inset
+
+, todas las soluciones son de la forma 
+\begin_inset Formula $P+M$
+\end_inset
+
+, donde 
+\begin_inset Formula $M$
+\end_inset
+
+ es solución de 
+\begin_inset Formula $AX=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Teorema de Rouché-Frobenius
+\end_layout
+
+\begin_layout Standard
+Un sistema 
+\begin_inset Formula $AX=B$
+\end_inset
+
+ es compatible si y sólo si 
+\begin_inset Formula $\text{rang}(A)=\text{rang}(A|B)$
+\end_inset
+
+, en cuyo caso es determinado si 
+\begin_inset Formula $\text{rang}(A)=n$
+\end_inset
+
+.
+ En concreto, si 
+\begin_inset Formula $k=n-\text{rang}(A)>0$
+\end_inset
+
+, existen 
+\begin_inset Formula $u_{1},\dots,u_{k}$
+\end_inset
+
+ soluciones linealmente independientes de 
+\begin_inset Formula $AX=0$
+\end_inset
+
+ tales que cualquier solución del sistema es de la forma 
+\begin_inset Formula $x_{0}+\lambda_{1}u_{1}+\dots+\lambda_{k}u_{k}$
+\end_inset
+
+.
+ Decimos del sistema que 
+\series bold
+depende de 
+\begin_inset Formula $k$
+\end_inset
+
+ parámetros
+\series default
+ 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ o que tiene 
+\begin_inset Formula $k$
+\end_inset
+
+ 
+\series bold
+grados de libertad
+\series default
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Si tenemos la aplicación 
+\begin_inset Formula $f_{A}:K^{n}\rightarrow K^{m}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $A=M_{{\cal C^{0}},{\cal C}}(f)$
+\end_inset
+
+, entonces si 
+\begin_inset Formula $K^{n}=<u_{1},\dots,u_{n}>$
+\end_inset
+
+ y 
+\begin_inset Formula $v$
+\end_inset
+
+ es el vector definido por 
+\begin_inset Formula $B$
+\end_inset
+
+, el conjunto de soluciones es 
+\begin_inset Formula $f^{-1}(v)=\{x\in K^{n}|f(x)=v\}$
+\end_inset
+
+.
+ Entonces,
+\begin_inset Formula 
+\begin{multline*}
+\exists x\in U:f(x)=v\iff v\in\text{Im}(f)\iff<f(u_{1}),\dots,f(u_{n}),v>=<f(u_{1}),\dots,f(u_{n})>\iff\\
+\iff\dim(<f(u_{1}),\dots,f(u_{n}),v>)=\dim(<f(u_{1}),\dots,f(u_{n})>)=\dim(\text{Im}(f))=\text{rang}(f)
+\end{multline*}
+
+\end_inset
+
+Por tanto 
+\begin_inset Formula $AX=B$
+\end_inset
+
+ es compatible si y sólo si 
+\begin_inset Formula $\text{rang}(A)=\text{rang}(A|B)$
+\end_inset
+
+.
+ Por otro lado, si 
+\begin_inset Formula $f(x_{0})=v$
+\end_inset
+
+, las soluciones serán 
+\begin_inset Formula $f^{-1}(v)=\{u\in U|f(u)=v\}=x_{0}+\text{Nuc}(f)$
+\end_inset
+
+.
+ Como además 
+\begin_inset Formula $\dim(K^{n})=\text{rang}(f)+\dim(\text{Nuc}(f))$
+\end_inset
+
+, entonces 
+\begin_inset Formula $k:=\dim(\text{Nuc}(f))=n-\text{rang}(f)$
+\end_inset
+
+, por lo que existen 
+\begin_inset Formula $k$
+\end_inset
+
+ soluciones linealmente independientes de 
+\begin_inset Formula $AX=0$
+\end_inset
+
+ (una base de 
+\begin_inset Formula $\text{Nuc}(f)$
+\end_inset
+
+).
+ Por tanto las soluciones del sistema homogéneo serán combinaciones lineales
+ de dicha base.
+\end_layout
+
+\begin_layout Section
+Resolución de sistemas de ecuaciones lineales.
+ Método de Gauss
+\end_layout
+
+\begin_layout Standard
+Dos sistemas de 
+\begin_inset Formula $m$
+\end_inset
+
+ ecuaciones lineales con 
+\begin_inset Formula $n$
+\end_inset
+
+ incógnitas sobre un mismo cuerpo son 
+\series bold
+e
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+\series bold
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+qui
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+\series bold
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+va
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+\series bold
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+len
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+\series bold
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+tes
+\series default
+ si tienen las mismas soluciones.
+ Si 
+\begin_inset Formula $P\in M_{m}(K)$
+\end_inset
+
+, 
+\begin_inset Formula $(PA)X=PB$
+\end_inset
+
+ es equivalente a 
+\begin_inset Formula $AX=B$
+\end_inset
+
+, y en particular, si 
+\begin_inset Formula $E\in M_{m}(K)$
+\end_inset
+
+ es una matriz elemental, 
+\begin_inset Formula $(EA)X=EB$
+\end_inset
+
+ también lo es, por lo que al hacer operaciones elementales por filas sobre
+ 
+\begin_inset Formula $(A|B)$
+\end_inset
+
+ se obtiene un sistema con las mismas soluciones.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+método de Gauss
+\series default
+ comienza por convertir la matriz ampliada a una escalonada reducida por
+ filas 
+\begin_inset Formula $(A'|B')$
+\end_inset
+
+.
+ Si obtenemos que 
+\begin_inset Formula $\text{rang}(A|B)>\text{rang}(A)$
+\end_inset
+
+, el sistema es incompatible.
+ Si 
+\begin_inset Formula $r=\text{rang}(A)=\text{rang}(A|B)$
+\end_inset
+
+, las filas nulas de 
+\begin_inset Formula $A'$
+\end_inset
+
+ lo son de 
+\begin_inset Formula $B'$
+\end_inset
+
+.
+ Reordenamos las incógnitas, lo que equivale a reordenar las columnas de
+ 
+\begin_inset Formula $A'$
+\end_inset
+
+, para conseguir un sistema de la forma
+\begin_inset Formula 
+\[
+\left(\begin{array}{c|c}
+I_{r} & C\\
+\hline 0 & 0
+\end{array}\right)\left(\begin{array}{c}
+y_{1}\\
+\vdots\\
+y_{r}\\
+\hline y_{r+1}\\
+\vdots\\
+y_{n}
+\end{array}\right)=\left(\begin{array}{c}
+b_{1}^{\prime}\\
+\vdots\\
+b_{r}^{\prime}\\
+\hline 0\\
+\vdots\\
+0
+\end{array}\right)
+\]
+
+\end_inset
+
+Donde 
+\begin_inset Formula $y_{1},\dots,y_{n}$
+\end_inset
+
+ son los 
+\begin_inset Formula $x_{1},\dots,x_{n}$
+\end_inset
+
+ reordenados de la misma forma que las columnas.
+ Esto equivale a
+\begin_inset Formula 
+\[
+I_{r}\left(\begin{array}{c}
+y_{1}\\
+\vdots\\
+y_{r}
+\end{array}\right)+C\left(\begin{array}{c}
+y_{r+1}\\
+\vdots\\
+y_{n}
+\end{array}\right)=\left(\begin{array}{c}
+b_{1}^{\prime}\\
+\vdots\\
+b_{r}^{\prime}
+\end{array}\right)\implies\left(\begin{array}{c}
+y_{1}\\
+\vdots\\
+y_{r}
+\end{array}\right)=\left(\begin{array}{c}
+b_{1}^{\prime}\\
+\vdots\\
+b_{r}^{\prime}
+\end{array}\right)-C\left(\begin{array}{c}
+y_{r+1}\\
+\vdots\\
+y_{n}
+\end{array}\right)
+\]
+
+\end_inset
+
+De modo que al asignar valores arbitrarios a 
+\begin_inset Formula $y_{r+1},\dots,y_{n}$
+\end_inset
+
+, que llamamos 
+\series bold
+incógnitas libres
+\series default
+, obtenemos valores de 
+\begin_inset Formula $y_{1},\dots,y_{r}$
+\end_inset
+
+, que llamamos 
+\series bold
+incógnitas principales
+\series default
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/algl/n4.lyx b/algl/n4.lyx
new file mode 100644
index 0000000..cf26416
--- /dev/null
+++ b/algl/n4.lyx
@@ -0,0 +1,1759 @@
+#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
+\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 swiss
+\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 Section
+Determinante de una matriz.
+ Propiedades
+\end_layout
+
+\begin_layout Standard
+Una aplicación 
+\begin_inset Formula $f:U_{1}\times\dots\times U_{n}\rightarrow V$
+\end_inset
+
+ es una 
+\series bold
+aplicación multilineal
+\series default
+ si es lineal en cada una de las 
+\begin_inset Formula $n$
+\end_inset
+
+ variables, es decir, si 
+\begin_inset Formula 
+\[
+f(u_{1},\dots,\alpha u_{i}+\beta u_{i}^{\prime},\dots,u_{n})=\alpha f(u_{1},\dots,u_{i},\dots,u_{n})+\beta f(u_{1},\dots,u_{i}^{\prime},\dots,u_{n})
+\]
+
+\end_inset
+
+Una aplicación multilineal 
+\begin_inset Formula $f:U^{n}\rightarrow V$
+\end_inset
+
+ se llama 
+\series bold
+aplicación 
+\begin_inset Formula $n$
+\end_inset
+
+-lineal
+\series default
+.
+ Si además 
+\begin_inset Formula $V=K$
+\end_inset
+
+ es una 
+\series bold
+forma 
+\begin_inset Formula $n$
+\end_inset
+
+-lineal
+\series default
+.
+ Una forma 
+\begin_inset Formula $n$
+\end_inset
+
+-lineal 
+\begin_inset Formula $f:U^{n}\rightarrow K$
+\end_inset
+
+ es 
+\series bold
+alternada
+\series default
+ si se anula en cada 
+\begin_inset Formula $n$
+\end_inset
+
+-upla con dos componentes iguales, es decir, tal que 
+\begin_inset Formula $f(u_{1},\dots,u_{k},\dots,u_{l},\dots,u_{n})=0$
+\end_inset
+
+ cuando 
+\begin_inset Formula $u_{k}=u_{l}$
+\end_inset
+
+ (con 
+\begin_inset Formula $k\neq l$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Standard
+Una 
+\series bold
+aplicación determinante
+\series default
+ 
+\begin_inset Formula $\det:M_{n}(K)\rightarrow K$
+\end_inset
+
+ es una forma 
+\begin_inset Formula $n$
+\end_inset
+
+-lineal alternada que a cada matriz cuadrada 
+\begin_inset Formula $A$
+\end_inset
+
+ le asigna un escalar, llamado 
+\series bold
+determinante
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+, que denotamos 
+\begin_inset Formula $\det(A)$
+\end_inset
+
+, 
+\begin_inset Formula $|A|$
+\end_inset
+
+ o 
+\begin_inset Formula $\det(A_{1},\dots,A_{n})$
+\end_inset
+
+ (donde 
+\begin_inset Formula $A_{i}$
+\end_inset
+
+ son las columnas de 
+\begin_inset Formula $A$
+\end_inset
+
+), tal que 
+\begin_inset Formula $|I_{n}|=1$
+\end_inset
+
+.
+ Algunas aplicaciones determinantes son:
+\end_layout
+
+\begin_layout Enumerate
+La aplicación 
+\begin_inset Formula $||:M_{2}(K)\rightarrow K$
+\end_inset
+
+ dada por
+\begin_inset Formula 
+\[
+\left|\begin{array}{cc}
+a_{11} & a_{12}\\
+a_{21} & a_{22}
+\end{array}\right|=a_{11}a_{22}-a_{12}a_{21}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+La 
+\series bold
+regla de Sarrus
+\series default
+, aplicación 
+\begin_inset Formula $||:M_{3}(K)\rightarrow K$
+\end_inset
+
+ dada por
+\begin_inset Formula 
+\[
+\left|\begin{array}{ccc}
+a_{11} & a_{12} & a_{13}\\
+a_{21} & a_{22} & a_{23}\\
+a_{31} & a_{32} & a_{33}
+\end{array}\right|=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{11}a_{23}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Las aplicaciones determinantes verifican que:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula 
+\[
+\left|\begin{array}{cccc}
+a_{1} & 0 & \cdots & 0\\
+0 & a_{2} & \cdots & 0\\
+\vdots & \vdots & \ddots & \vdots\\
+0 & 0 & \cdots & a_{n}
+\end{array}\right|=a_{1}a_{2}\cdots a_{n}
+\]
+
+\end_inset
+
+Si 
+\begin_inset Formula $\{e_{1},\dots,e_{n}\}$
+\end_inset
+
+ es la base canónica de 
+\begin_inset Formula $K^{n}$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+\left|\begin{array}{cccc}
+a_{1} & 0 & \cdots & 0\\
+0 & a_{2} & \cdots & 0\\
+\vdots & \vdots & \ddots & \vdots\\
+0 & 0 & \cdots & a_{n}
+\end{array}\right|=\det(a_{1}e_{1},\dots,a_{n}e_{n})=a_{1}\cdots a_{n}\det(e_{1},\dots,e_{n})=a_{1}\cdots a_{n}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ tiene una columna nula entonces 
+\begin_inset Formula $\det(A)=0$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Si 
+\begin_inset Formula $A_{i}=0$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+\begin{array}{c}
+\det(A_{1},\dots,A_{i-1},0,A_{i+1},\dots,A_{n})=\det(A_{1},\dots,A_{i-1},0+0,A_{i+1},\dots,A_{n})=\\
+=\det(A_{1},\dots,A_{i-1},0,A_{i+1},\dots,A_{n})+\det(A_{1},\dots,A_{i-1},0,A_{i+1},\dots,A_{n})
+\end{array}
+\]
+
+\end_inset
+
+luego 
+\begin_inset Formula $\det A=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Al intercambiar dos columnas, el determinante cambia de signo.
+\begin_inset Formula 
+\[
+\begin{array}{c}
+0=\det(A_{1},\dots,A_{i}+A_{j},\dots,A_{i}+A_{j},\dots,A_{n})=\\
+=\det(A_{1},\dots,A_{i},\dots,A_{i},\dots,A_{n})+\det(A_{1},\dots,A_{i},\dots,A_{j},\dots,A_{n})+\\
++\det(A_{1},\dots,A_{j},\dots,A_{i},\dots,A_{n})+\det(A_{1},\dots,A_{j},\dots,A_{j},\dots,A_{n})=\\
+=\det(A_{1},\dots,A_{i},\dots,A_{j},\dots,A_{n})+\det(A_{1},\dots,A_{j},\dots,A_{i},\dots,A_{n})
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Si a una columna se le añade otra multiplicada por un escalar, el determinante
+ no varía.
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\begin{array}{c}
+\det(A_{1},\dots,A_{i}+\alpha A_{j},\dots,A_{j},\dots,A_{n})=\\
+=\det(A_{1},\dots,A_{i},\dots,A_{j},\dots,A_{n})+\alpha\det(A_{1},\dots,A_{j},\dots,A_{j},\dots,A_{n})=\\
+=\det(A_{1},\dots,A_{i},\dots,A_{j},\dots,A_{n})
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Si las columnas de una matriz cuadrada son linealmente dependientes, su
+ determinante es 0.
+ Por tanto una matriz no invertible tiene determinante 0.
+\begin_inset Newline newline
+\end_inset
+
+Habrá una columna que será combinación lineal del resto: 
+\begin_inset Formula $A_{k}=\sum_{j\neq k}\alpha_{j}A_{j}$
+\end_inset
+
+.
+ Así,
+\begin_inset Formula 
+\[
+\begin{array}{c}
+\det(A_{1},\dots,A_{k},\dots,A_{n})=\det(A_{1},\dots,\sum_{j\neq k}\alpha_{j}A_{j},\dots,A_{n})=\\
+=\sum_{j\neq k}\alpha_{j}\det(A_{1},\dots,A_{j},\dots,A_{n})=0
+\end{array}
+\]
+
+\end_inset
+
+Ya que cada matriz del último sumatorio tiene dos columnas iguales.
+\end_layout
+
+\begin_layout Standard
+De aquí podemos deducir que 
+\begin_inset Formula $|E_{n}(i,j)|=-1$
+\end_inset
+
+, 
+\begin_inset Formula $|E_{n}(\alpha[i])|=\alpha$
+\end_inset
+
+ y 
+\begin_inset Formula $|E_{n}([i]+\alpha[j])|=1$
+\end_inset
+
+, y que si 
+\begin_inset Formula $A,E\in M_{n}(K)$
+\end_inset
+
+, siendo 
+\begin_inset Formula $E$
+\end_inset
+
+ una matriz elemental, entonces 
+\begin_inset Formula $|AE|=|A||E|$
+\end_inset
+
+.
+ Se deducen los siguientes teoremas:
+\end_layout
+
+\begin_layout Enumerate
+Una matriz cuadrada 
+\begin_inset Formula $A$
+\end_inset
+
+ es invertible si y sólo si 
+\begin_inset Formula $|A|\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Toda matriz invertible es producto de matrices elementales, y por lo anterior,
+ 
+\begin_inset Formula $|A|=|I_{n}E_{1}\cdots E_{k}|=|I_{n}||E_{1}|\cdots|E_{k}|=|E_{1}|\cdots|E_{k}|$
+\end_inset
+
+.
+ Como ninguno de los factores es nulo, se tiene que 
+\begin_inset Formula $|A|\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Inmediato de la última propiedad.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A,B\in M_{n}(K)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $|AB|=|A||B|$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Si alguna de las dos no es invertible, su producto tampoco (pues si lo fuera,
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ serían invertibles).
+ En tal caso, 
+\begin_inset Formula $|AB|=0=|A||B|$
+\end_inset
+
+.
+ Si son ambas invertibles, existen matrices elementales 
+\begin_inset Formula $E_{1},\dots,E_{k}$
+\end_inset
+
+ con 
+\begin_inset Formula $B=E_{1}\cdots E_{k}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $|AB|=|AE_{1}\cdots E_{k}|=|A||E_{1}|\cdots|E_{k}|=|A||B|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+De aquí tenemos que 
+\begin_inset Formula $|A^{-1}|=|A|^{-1}$
+\end_inset
+
+, pues 
+\begin_inset Formula $1=|I_{n}|=|AA^{-1}|=|A||A^{-1}|$
+\end_inset
+
+.
+ Tenemos también que la aplicación determinante es única, pues 
+\begin_inset Formula $\det(A)=0$
+\end_inset
+
+ para matrices no invertibles y 
+\begin_inset Formula $\det(A)=|E_{1}|\cdots|E_{k}|$
+\end_inset
+
+ para aquellas que sí lo son, y podemos entonces comprobar que esta operación
+ está bien definida.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema:
+\series default
+ 
+\begin_inset Formula $|A^{t}|=|A|$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $A$
+\end_inset
+
+ no es invertible, 
+\begin_inset Formula $A^{t}$
+\end_inset
+
+ tampoco, por lo que 
+\begin_inset Formula $|A^{t}|=0=|A|$
+\end_inset
+
+.
+ Si lo es, existen 
+\begin_inset Formula $E_{1},\dots,E_{k}$
+\end_inset
+
+ con 
+\begin_inset Formula $A=E_{1}\cdots E_{k}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $|A^{t}|=|(E_{1}\cdots E_{k})^{t}|=|E_{k}^{t}\cdots E_{1}^{t}|=|E_{k}^{t}|\cdots|E_{1}^{t}|=|E_{1}|\cdots|E_{k}|=|E_{1}\cdots E_{k}|=|A|$
+\end_inset
+
+.
+ Esto significa que todo lo relativo a determinantes que se diga para columnas
+ también es válido para filas.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $A=(a_{ij})\in M_{n}(K)$
+\end_inset
+
+ e 
+\begin_inset Formula $i,j\in\{1,\dots,n\}$
+\end_inset
+
+, llamamos 
+\series bold
+menor complementario
+\series default
+ del elemento 
+\begin_inset Formula $a_{ij}$
+\end_inset
+
+ al determinante 
+\begin_inset Formula $|A_{ij}|$
+\end_inset
+
+ de la matriz 
+\begin_inset Formula $A_{ij}\in M_{n-1}(K)$
+\end_inset
+
+ resultado de eliminar la fila 
+\begin_inset Formula $i$
+\end_inset
+
+ y la columna 
+\begin_inset Formula $j$
+\end_inset
+
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+adjunto
+\series default
+ de 
+\begin_inset Formula $a_{ij}$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+ al escalar 
+\begin_inset Formula $\Delta_{ij}:=(-1)^{i+j}|A_{ij}|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema:
+\series default
+ Las aplicaciones 
+\begin_inset Formula $||:M_{n}(K)\rightarrow K$
+\end_inset
+
+ definidas para 
+\begin_inset Formula $n=1$
+\end_inset
+
+ como 
+\begin_inset Formula $|(a)|=a$
+\end_inset
+
+ y para 
+\begin_inset Formula $n>1$
+\end_inset
+
+ como 
+\begin_inset Formula $|(a_{ij})|=a_{11}\Delta_{11}+\dots+a_{1n}\Delta_{1n}$
+\end_inset
+
+ son aplicaciones determinante.
+ 
+\series bold
+Demostración:
+\series default
+ Para 
+\begin_inset Formula $n=1$
+\end_inset
+
+ es trivial.
+ Ahora supongamos que la aplicación determinante está definida para 
+\begin_inset Formula $n-1$
+\end_inset
+
+ y probamos que se cumplen las condiciones para 
+\begin_inset Formula $n-1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Multilineal: Sea 
+\begin_inset Formula $A=(a_{ij})=(A_{1},\dots,A_{n})\in M_{n}(K)$
+\end_inset
+
+ y 
+\begin_inset Formula $A^{\prime}=(a_{ij}^{\prime})=(A_{1},\dots,\alpha A_{k},\dots,A_{n})$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $a_{ik}^{\prime}=\alpha a_{ik}$
+\end_inset
+
+ y para 
+\begin_inset Formula $j\neq k$
+\end_inset
+
+, 
+\begin_inset Formula $a_{ij}^{\prime}=a_{ij}$
+\end_inset
+
+.
+ Si llamamos 
+\begin_inset Formula $\Delta_{ij}$
+\end_inset
+
+ y 
+\begin_inset Formula $\Delta_{ij}^{\prime}$
+\end_inset
+
+ a los correspondientes adjuntos, 
+\begin_inset Formula $\Delta_{ik}^{\prime}=\Delta_{ik}$
+\end_inset
+
+ y para 
+\begin_inset Formula $j\neq k$
+\end_inset
+
+, 
+\begin_inset Formula $\Delta_{ij}^{\prime}=\alpha\Delta_{ij}$
+\end_inset
+
+.
+ Así,
+\begin_inset Formula 
+\[
+\begin{array}{c}
+|A^{\prime}|=a_{11}^{\prime}\Delta_{11}^{\prime}+\dots+a_{1n}^{\prime}\Delta_{1n}^{\prime}=\\
+=a_{11}\alpha\Delta_{11}+\dots+a_{1(k-1)}\alpha\Delta_{1(k-1)}+\alpha a_{1k}\Delta_{ik}+a_{1(k+1)}\alpha\Delta_{i(k+1)}+\dots+a_{1n}\alpha\Delta_{in}=\\
+=\alpha(a_{11}\Delta_{11}+\dots+a_{1n}\Delta_{1n})=\alpha|A|
+\end{array}
+\]
+
+\end_inset
+
+Del mismo modo, sea 
+\begin_inset Formula $A=(a_{ij})=(A_{1},\dots,A_{k}^{\prime}+A_{k}^{\prime\prime},\dots,A_{n})$
+\end_inset
+
+ y sean 
+\begin_inset Formula $A^{\prime}=(a_{ij}^{\prime})=(A_{1},\dots,A_{k}^{\prime},\dots,A_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $A^{\prime\prime}=(a_{ij}^{\prime\prime})=(A_{1},\dots,A_{k}^{\prime\prime},\dots,A_{n})$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $a_{ik}=a_{ik}^{\prime}+a_{ik}^{\prime\prime}$
+\end_inset
+
+ y si 
+\begin_inset Formula $j\neq k$
+\end_inset
+
+, 
+\begin_inset Formula $a_{ij}=a_{ij}^{\prime}=a_{ij}^{\prime\prime}$
+\end_inset
+
+.
+ Del mismo modo, 
+\begin_inset Formula $\Delta_{ik}=\Delta_{ik}^{\prime}=\Delta_{ik}^{\prime\prime}$
+\end_inset
+
+ y si 
+\begin_inset Formula $j\neq k$
+\end_inset
+
+, 
+\begin_inset Formula $\Delta_{ij}=\Delta_{ij}^{\prime}+\Delta_{ij}^{\prime\prime}$
+\end_inset
+
+, por lo que
+\begin_inset Formula 
+\[
+\begin{array}{c}
+|A|=a_{11}\Delta_{11}+\dots+a_{1n}\Delta_{1n}=\\
+=a_{11}(\Delta_{11}^{\prime}+\Delta_{11}^{\prime\prime})+\dots+a_{1(k-1)}(\Delta_{1(k-1)}^{\prime}+\Delta_{1(k-1)}^{\prime\prime})+(a_{1k}^{\prime}+a_{1k}^{\prime\prime})\Delta_{1k}+\\
++a_{1(k+1)}(\Delta_{1(k+1)}^{\prime}+\Delta_{1(k+1)}^{\prime\prime})+\dots+a_{1n}(\Delta_{1n}^{\prime}+\Delta_{1n}^{\prime\prime})=\\
+=a_{11}^{\prime}\Delta_{11}^{\prime}+\dots+a_{1n}\Delta_{1n}^{\prime}+a_{11}^{\prime\prime}\Delta_{11}^{\prime\prime}+\dots+a_{1n}\Delta_{1n}^{\prime\prime}=|A^{\prime}|+|A^{\prime\prime}|
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Alternada: Sea 
+\begin_inset Formula $A=(a_{ij})=(A_{1},\dots,A_{n})\in M_{n}(K)$
+\end_inset
+
+.
+ Si para 
+\begin_inset Formula $r<s$
+\end_inset
+
+ se tiene que 
+\begin_inset Formula $A_{r}=A_{s}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a_{r}=a_{s}$
+\end_inset
+
+ y para 
+\begin_inset Formula $j\neq r,s$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $\Delta_{1j}=0$
+\end_inset
+
+, pues el menor complementario posee dos columnas iguales, por lo que 
+\begin_inset Formula $|A|=a_{11}\Delta_{11}+\dots+a_{1n}\Delta_{1n}=a_{1r}\Delta_{1r}+a_{1s}\Delta_{1s}$
+\end_inset
+
+.
+ Por otro lado, si llamamos 
+\begin_inset Formula $A_{j}^{\prime}$
+\end_inset
+
+ al elemento de 
+\begin_inset Formula $A_{j}$
+\end_inset
+
+ resultado de eliminar 
+\begin_inset Formula $a_{1j}$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+\begin{array}{c}
+|A_{1s}|=|(A_{1}^{\prime},\dots,A_{r-1}^{\prime},A_{r}^{\prime},A_{r+1}^{\prime},\dots,A_{s-1}^{\prime},A_{s+1}^{\prime},\dots,A_{n}^{\prime})|=\\
+=-|(A_{1}^{\prime},\dots,A_{r-1}^{\prime},A_{r+1}^{\prime},A_{r}^{\prime},\dots,A_{s-1}^{\prime},A_{s+1}^{\prime},\dots,A_{n}^{\prime})|=\dots=\\
+=(-1)^{s-r-1}|(A_{1}^{\prime},\dots,A_{r-1}^{\prime},A_{r+1}^{\prime},\dots,A_{s-1}^{\prime},A_{r}^{\prime},A_{s+1}^{\prime},\dots,A_{n})|=(-1)^{s-r-1}|A_{1r}|
+\end{array}
+\]
+
+\end_inset
+
+pues 
+\begin_inset Formula $A_{r}^{\prime}=A_{s}^{\prime}$
+\end_inset
+
+.
+ Por tanto
+\begin_inset Formula 
+\[
+\begin{array}{c}
+|A|=a_{1r}(-1)^{1+r}|A_{1r}|+a_{1s}(-1)^{1+s}(-1)^{s-r-1}|A_{1r}|=\\
+=a_{1r}|A_{1r}|((-1)^{1+r}+(-1)^{1+2s-r-1})=a_{1r}|A_{1r}|((-1)^{1+r}+(-1)^{-r})=0
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $|I_{n}|=\delta_{11}\Delta_{11}+\dots+\delta_{1n}\Delta_{1n}=\Delta_{11}=(-1)^{1+1}|I_{n-1}|=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Se puede probar que, para 
+\begin_inset Formula $1\leq i\leq n$
+\end_inset
+
+, cada aplicación dada por 
+\begin_inset Formula $|A|=a_{i1}\Delta_{i1}+\dots+a_{in}\Delta_{in}$
+\end_inset
+
+, que llamamos 
+\series bold
+desarrollo del determinante
+\series default
+ de la matriz 
+\begin_inset Formula $A$
+\end_inset
+
+ por la 
+\begin_inset Formula $i$
+\end_inset
+
+-ésima fila, también cumple las condiciones.
+ Por otro lado, como 
+\begin_inset Formula $|A|=|A^{t}|$
+\end_inset
+
+, también se puede desarrollar por filas.
+ En la práctica se pueden hacer operaciones elementales para obtener ceros
+ en una fila o columna y luego desarrollar por ella.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Determinantes de Vandermonde:
+\series default
+ Restando a cada fila la anterior por 
+\begin_inset Formula $x_{1}$
+\end_inset
+
+, desarrollando, dividiendo lo resultante por cada elemento de la primera
+ fila y repitiendo el proceso, se tiene que:
+\begin_inset Formula 
+\[
+\begin{array}{c}
+\left|\begin{array}{cccc}
+1 & 1 & \cdots & 1\\
+x_{1} & x_{2} & \cdots & x_{n}\\
+\vdots & \vdots &  & \vdots\\
+x_{1}^{n-1} & x_{2}^{n-1} & \cdots & x_{n}^{n-1}
+\end{array}\right|=\left|\begin{array}{cccc}
+1 & 1 & \cdots & 1\\
+0 & x_{2}-x_{1} & \cdots & x_{n}-x_{1}\\
+\vdots & \vdots &  & \vdots\\
+0 & x_{2}^{n-1}-x_{1}x_{2}^{n-2} & \cdots & x_{n}^{n-1}-x_{1}x_{n}^{n-2}
+\end{array}\right|=\\
+=(x_{2}-x_{1})\cdots(x_{n}-x_{1})\left|\begin{array}{ccc}
+1 & \cdots & 1\\
+x_{2} & \cdots & x_{n}\\
+\vdots &  & \vdots\\
+x_{2}^{n-2} & \cdots & x_{n}^{n-2}
+\end{array}\right|=\dots=\prod_{1\leq j<i\leq n}(x_{i}-x_{j})
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Desarrollo de un determinante por menores, de
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+sa
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+rro
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+llo de Laplace
+\end_layout
+
+\begin_layout Standard
+Se llama 
+\series bold
+submatriz
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ a la obtenida al eliminar determinadas filas y columnas de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Toda matriz es submatriz de sí misma.
+ Un 
+\series bold
+menor de orden 
+\begin_inset Formula $n$
+\end_inset
+
+
+\series default
+ es el determinante de una submatriz de tamaño 
+\begin_inset Formula $n\times n$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es cuadrada, el 
+\series bold
+menor complementario
+\series default
+ 
+\begin_inset Formula $M^{\prime}$
+\end_inset
+
+ del menor 
+\begin_inset Formula $M$
+\end_inset
+
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ es el determinante de la matriz formada por las filas y columnas restantes.
+ Un menor es de 
+\series bold
+clase par
+\series default
+ o de 
+\series bold
+clase impar
+\series default
+ según lo sea la suma de los índices de sus filas (
+\begin_inset Formula $i_{1},\dots,i_{p}$
+\end_inset
+
+) y columnas (
+\begin_inset Formula $j_{1},\dots,j_{p}$
+\end_inset
+
+).
+ La 
+\series bold
+signatura
+\series default
+ de un menor 
+\begin_inset Formula $M$
+\end_inset
+
+ es 
+\begin_inset Formula $\varepsilon(M)=(-1)^{(i_{1}+\dots+i_{p})+(j_{1}+\dots+j_{p})}$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $(1+\dots+n)+(1+\dots+n)$
+\end_inset
+
+ es par, todo menor tiene la misma signatura que su complementario.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\begin_inset Formula $\chi_{r}$
+\end_inset
+
+ al conjunto de combinaciones de 
+\begin_inset Formula $r$
+\end_inset
+
+ filas o columnas:
+\begin_inset Formula 
+\[
+\chi_{r}=\{(i_{1},\dots,i_{r}):1\leq i_{1}<\dots<i_{r}\leq n\}
+\]
+
+\end_inset
+
+Si 
+\begin_inset Formula $I,J\in\chi_{r}$
+\end_inset
+
+, llamamos 
+\begin_inset Formula $A_{IJ}$
+\end_inset
+
+ al menor determinado por las filas 
+\begin_inset Formula $I$
+\end_inset
+
+ y las columnas 
+\begin_inset Formula $J$
+\end_inset
+
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Entonces:
+\end_layout
+
+\begin_layout Itemize
+Dado 
+\begin_inset Formula $I\in\chi_{r}$
+\end_inset
+
+, 
+\begin_inset Formula $|A|=\sum_{J\in\chi_{r}}\varepsilon(A_{IJ})A_{IJ}A_{IJ}^{\prime}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Dado 
+\begin_inset Formula $J\in\chi_{r}$
+\end_inset
+
+, 
+\begin_inset Formula $|A|=\sum_{I\in\chi_{r}}\varepsilon(A_{IJ})A_{IJ}A_{IJ}^{\prime}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Esto es útil cuando 
+\begin_inset Formula $A$
+\end_inset
+
+ está formada por bloques de tamaño adecuado alguno de los cuales es nulo.
+ De aquí se tiene que 
+\begin_inset Formula $\left|\left(\begin{array}{c|c}
+P & Q\\
+\hline 0 & R
+\end{array}\right)\right|=|P||R|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Determinante de un endomorfismo
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}^{\prime}$
+\end_inset
+
+ son bases de 
+\begin_inset Formula $V$
+\end_inset
+
+ y 
+\begin_inset Formula $f\in\text{End}(V)$
+\end_inset
+
+, entonces 
+\begin_inset Formula 
+\[
+M_{{\cal B}}(f)=M_{{\cal B}{\cal B}^{\prime}}M_{{\cal B}^{\prime}}(f)M_{{\cal B}^{\prime}{\cal B}}=P^{-1}M_{{\cal B}^{\prime}}(f)P
+\]
+
+\end_inset
+
+por lo que 
+\begin_inset Formula $|M_{{\cal B}}(f)|=|P|^{-1}|M_{{\cal B}^{\prime}}(f)||P|=|M_{{\cal B}^{\prime}}(f)|$
+\end_inset
+
+.
+ Así, llamamos 
+\series bold
+determinante del endomorfismo 
+\begin_inset Formula $f$
+\end_inset
+
+
+\series default
+ al de la matriz asociada a 
+\begin_inset Formula $f$
+\end_inset
+
+ respecto de cualquier base de 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Matriz adjunta.
+ Aplicación al cálculo de la inversa
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+matriz adjunta
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ a la matriz 
+\begin_inset Formula $\hat{A}=(\Delta_{ij})\in M_{n}(K)$
+\end_inset
+
+.
+ 
+\series bold
+Teorema:
+\series default
+ Si 
+\begin_inset Formula $A\in M_{n}(K)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $A\cdot\hat{A}^{t}=\hat{A}^{t}\cdot A=|A|I_{n}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $A=(a_{ij})$
+\end_inset
+
+, 
+\begin_inset Formula $\hat{A}=(\Delta_{ij})$
+\end_inset
+
+ y 
+\begin_inset Formula $\hat{A}^{t}=(b_{ij})$
+\end_inset
+
+, entonces 
+\begin_inset Formula $b_{ij}=\Delta_{ji}$
+\end_inset
+
+.
+ Sea entonces 
+\begin_inset Formula $C=(c_{ij})=A\cdot\hat{A}^{t}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $c_{ij}=\sum_{k=1}^{n}a_{ik}b_{kj}=\sum_{k=1}^{n}a_{ik}\Delta_{jk}$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $i\neq j$
+\end_inset
+
+, esto corresponde al desarrollo por la fila 
+\begin_inset Formula $j$
+\end_inset
+
+-ésima del determinante de la matriz que se diferencia de 
+\begin_inset Formula $A$
+\end_inset
+
+ en que tiene la fila 
+\begin_inset Formula $i$
+\end_inset
+
+-ésima copiada en la 
+\begin_inset Formula $j$
+\end_inset
+
+-ésima, por lo que entonces 
+\begin_inset Formula $c_{ij}=0$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $i\neq j$
+\end_inset
+
+, este es el desarrollo por la fila 
+\begin_inset Formula $j$
+\end_inset
+
+-ésima de 
+\begin_inset Formula $A$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $c_{ii}=|A|$
+\end_inset
+
+ y 
+\begin_inset Formula $C=|A|I_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Como consecuencia, se tiene el 
+\series bold
+teorema
+\series default
+ de que 
+\begin_inset Formula $A$
+\end_inset
+
+ es invertible si y sólo si 
+\begin_inset Formula $|A|\neq0$
+\end_inset
+
+ y entonces
+\begin_inset Formula 
+\[
+A^{-1}=\frac{1}{|A|}\hat{A}^{t}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Cálculo del rango de una matriz por determinantes
+\end_layout
+
+\begin_layout Standard
+El rango de 
+\begin_inset Formula $A$
+\end_inset
+
+ es el mayor de los órdenes de los menores no nulos de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $A=(a_{ij})\in M_{m,n}(K)$
+\end_inset
+
+ con 
+\begin_inset Formula $A\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $r=\text{rang}(A)$
+\end_inset
+
+ y 
+\begin_inset Formula $p$
+\end_inset
+
+ el mayor de los tamaños de los menores no nulos, que existe si 
+\begin_inset Formula $A\neq0$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $A^{\prime}$
+\end_inset
+
+ es una submatriz cuadrada de 
+\begin_inset Formula $A$
+\end_inset
+
+ de tamaño 
+\begin_inset Formula $p\times p$
+\end_inset
+
+ con 
+\begin_inset Formula $|A^{\prime}|\neq0$
+\end_inset
+
+, entonces la submatriz 
+\begin_inset Formula $B$
+\end_inset
+
+ formada por las filas de 
+\begin_inset Formula $A^{\prime}$
+\end_inset
+
+ pero con todas las columnas de 
+\begin_inset Formula $A$
+\end_inset
+
+ tiene 
+\begin_inset Formula $p$
+\end_inset
+
+ columnas linealmente independientes (las de 
+\begin_inset Formula $A^{\prime}$
+\end_inset
+
+) y por tanto también tiene 
+\begin_inset Formula $p$
+\end_inset
+
+ filas linealmente independientes, pero entonces 
+\begin_inset Formula $A$
+\end_inset
+
+ tiene al menos 
+\begin_inset Formula $p$
+\end_inset
+
+ filas linealmente independientes y 
+\begin_inset Formula $r\geq p$
+\end_inset
+
+.
+ Por otro lado, si 
+\begin_inset Formula $A_{i_{1}},\dots,A_{i_{r}}$
+\end_inset
+
+ son filas linealmente independientes de 
+\begin_inset Formula $A$
+\end_inset
+
+ y tomamos la submatriz 
+\begin_inset Formula $B\in M_{r,n}(K)$
+\end_inset
+
+ formada por estas filas y todas las columnas, 
+\begin_inset Formula $B$
+\end_inset
+
+ tendrá rango 
+\begin_inset Formula $r$
+\end_inset
+
+, luego tendrá 
+\begin_inset Formula $r$
+\end_inset
+
+ columnas 
+\begin_inset Formula $j_{1},\dots,j_{r}$
+\end_inset
+
+ linealmente independientes.
+ Si tomamos la submatriz 
+\begin_inset Formula $A^{\prime}\in M_{r}(K)$
+\end_inset
+
+ formada por estas columnas, al ser linealmente independientes, 
+\begin_inset Formula $|A^{\prime}|\neq0$
+\end_inset
+
+, luego 
+\begin_inset Formula $p\geq r$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $p=r$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $A_{i}=(a_{1i},\dots,a_{ni})\in K^{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $A_{1},\dots,A_{r}$
+\end_inset
+
+ linealmente independientes y
+\begin_inset Formula 
+\[
+\left|\begin{array}{ccc}
+a_{i_{1}1} & \cdots & a_{i_{1}r}\\
+\vdots & \ddots & \vdots\\
+a_{i_{r}1} & \cdots & a_{i_{r}r}
+\end{array}\right|\neq0
+\]
+
+\end_inset
+
+
+\begin_inset Formula $B=(b_{1},\dots,b_{n})$
+\end_inset
+
+ es combinación lineal de 
+\begin_inset Formula $A_{1},\dots,A_{r}$
+\end_inset
+
+ si y sólo si para todo 
+\begin_inset Formula $j$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+\left|\begin{array}{cccc}
+a_{i_{1}1} & \dots & a_{i_{1}r} & b_{i_{1}}\\
+\vdots & \ddots & \vdots & \vdots\\
+a_{i_{r}1} & \cdots & a_{i_{r}r} & b_{i_{r}}\\
+a_{j1} & \cdots & a_{jr} & b_{j}
+\end{array}\right|=0
+\]
+
+\end_inset
+
+
+\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 son linealmente dependientes, los determinantes son nulos.
+\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 todos son nulos, desarrollando por la última fila, se obtiene, para cada
+ 
+\begin_inset Formula $j$
+\end_inset
+
+, que
+\begin_inset Formula 
+\[
+a_{j1}\left|\begin{array}{cccc}
+a_{i_{1}2} & \cdots & a_{i_{1}r} & b_{i_{1}}\\
+\vdots & \ddots & \vdots & \vdots\\
+a_{i_{r}2} & \cdots & a_{i_{r}r} & b_{i_{r}}
+\end{array}\right|\pm\dots\pm b_{j}\left|\begin{array}{ccc}
+a_{i_{1}1} & \cdots & a_{i_{1}r}\\
+\vdots & \ddots & \vdots\\
+a_{i_{r}1} & \cdots & a_{i_{r}r}
+\end{array}\right|=0
+\]
+
+\end_inset
+
+Por lo que
+\begin_inset Formula 
+\[
+b_{j}=\frac{1}{\left|\begin{array}{ccc}
+a_{i_{1}1} & \cdots & a_{i_{1}r}\\
+\vdots & \ddots & \vdots\\
+a_{i_{r}1} & \cdots & a_{i_{r}r}
+\end{array}\right|}\left(\pm a_{j1}\left|\begin{array}{ccc}
+a_{i_{1}2} & \cdots & b_{i_{1}}\\
+\vdots & \ddots & \vdots\\
+a_{i_{r}2} & \cdots & b_{i_{r}}
+\end{array}\right|\pm\dots\pm a_{j_{r}}\left|\begin{array}{ccc}
+a_{i_{1}1} & \cdots & b_{i1}\\
+\vdots & \ddots & \vdots\\
+a_{i_{r}1} & \cdots & b_{i_{r}}
+\end{array}\right|\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+A efectos prácticos, esto significa que, una vez encontrado un menor no
+ nulo de orden 
+\begin_inset Formula $k$
+\end_inset
+
+ en una matriz 
+\begin_inset Formula $A$
+\end_inset
+
+, podemos 
+\emph on
+orlarlo
+\emph default
+ (obtener otro añadiendo una fila y una columna a la submatriz) de todas
+ las formas posibles y, si todos los menores resultantes son nulos, entonces
+ 
+\begin_inset Formula $\text{rang}(A)=k$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Regla de Cramer
+\end_layout
+
+\begin_layout Standard
+Un sistema de ecuaciones lineales 
+\begin_inset Formula $AX=B$
+\end_inset
+
+ es un 
+\series bold
+sistema de Cramer
+\series default
+ si 
+\begin_inset Formula $A$
+\end_inset
+
+ es invertible.
+ En tal caso tiene solución única 
+\begin_inset Formula $X=A^{-1}B$
+\end_inset
+
+.
+ 
+\series bold
+Regla de Cramer:
+\series default
+ si las columnas de 
+\begin_inset Formula $A$
+\end_inset
+
+ son 
+\begin_inset Formula $(A_{1},\dots,A_{n})$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+x_{i}=\frac{\det(A_{1},\dots,A_{i-1},B,A_{i+1},\dots,A_{n})}{\det(A)}
+\]
+
+\end_inset
+
+
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $A^{-1}=\frac{1}{|A|}\hat{A}^{t}$
+\end_inset
+
+, y si 
+\begin_inset Formula $X=(x_{i})$
+\end_inset
+
+, 
+\begin_inset Formula $A=(a_{ij})$
+\end_inset
+
+ y 
+\begin_inset Formula $\hat{A}=(\Delta_{ij})$
+\end_inset
+
+, entonces 
+\begin_inset Formula $x_{i}=\sum_{j=1}^{n}\frac{1}{|A|}\Delta_{ji}b_{j}=\frac{1}{|A|}\sum_{j=1}^{n}\Delta_{ji}b_{j}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $A\in M_{m,n}(K)$
+\end_inset
+
+ con 
+\begin_inset Formula $\text{rang}(A)=r$
+\end_inset
+
+, habrá un menor 
+\begin_inset Formula $M\neq0$
+\end_inset
+
+ de orden 
+\begin_inset Formula $r$
+\end_inset
+
+, por lo que las 
+\begin_inset Formula $n-r$
+\end_inset
+
+ últimas filas serán combinaciones lineales de las 
+\begin_inset Formula $r$
+\end_inset
+
+ primeras, y moviendo al lado derecho los 
+\begin_inset Formula $m-r$
+\end_inset
+
+ coeficientes que no están en la submatriz de 
+\begin_inset Formula $M$
+\end_inset
+
+, nos queda el sistema
+\begin_inset Formula 
+\[
+\left.\begin{array}{ccc}
+a_{11}x_{1}+\dots+a_{1r}x_{r} & = & b_{1}-(a_{1r+1}x_{r+1}+\dots+a_{1n}x_{n})\\
+ & \vdots\\
+a_{r1}x_{1}+\dots+a_{rr}x_{r} & = & b_{r}-(a_{rr+1}x_{r+1}+\dots+a_{rn}x_{n})
+\end{array}\right\} 
+\]
+
+\end_inset
+
+que podemos resolver por Cramer.
+\end_layout
+
+\end_body
+\end_document
diff --git a/algl/n5.lyx b/algl/n5.lyx
new file mode 100644
index 0000000..bb844d5
--- /dev/null
+++ b/algl/n5.lyx
@@ -0,0 +1,1323 @@
+#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
+\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 swiss
+\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 Section
+Semejanza de matrices
+\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
+
+.
+ Esta relación es de equivalencia, y si dos matrices son semejantes también
+ son equivalentes y por tanto tienen el mismo rango, si bien el recíproco
+ no se cumple.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $A\in M_{n}(K)$
+\end_inset
+
+ una matriz formada por 
+\series bold
+bloques
+\series default
+ cuadrados en la diagonal y ceros en el resto:
+\begin_inset Formula 
+\[
+A=\left(\begin{array}{ccc}
+\boxed{A_{1}} &  & 0\\
+ & \ddots\\
+0 &  & \boxed{A_{t}}
+\end{array}\right)
+\]
+
+\end_inset
+
+con 
+\begin_inset Formula $A_{i}\in M_{n_{i}}(K)$
+\end_inset
+
+ y 
+\begin_inset Formula $n_{1}+\dots+n_{t}=n$
+\end_inset
+
+.
+ Por el desarrollo de Laplace, su determinante es 
+\begin_inset Formula $|A|=|A_{1}|\cdots|A_{t}|$
+\end_inset
+
+; su rango es la suma de los rangos de los 
+\begin_inset Formula $A_{i}$
+\end_inset
+
+, y su potencia 
+\begin_inset Formula $k$
+\end_inset
+
+-ésima es
+\begin_inset Formula 
+\[
+A^{k}=\left(\begin{array}{ccc}
+\boxed{A_{1}^{k}} &  & 0\\
+ & \ddots\\
+0 &  & \boxed{A_{t}^{k}}
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Entonces, si 
+\begin_inset Formula $B$
+\end_inset
+
+ es semejante a 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $|B|=|P^{-1}AP|=|P|^{-1}|A||P|=|A|$
+\end_inset
+
+, su rango es el de 
+\begin_inset Formula $A$
+\end_inset
+
+ y
+\begin_inset Formula 
+\[
+B^{k}=\underset{k\text{ veces}}{(P^{-1}AP)\cdots(P^{-1}AP)}=P^{-1}A^{k}P
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Subespacios invariantes
+\end_layout
+
+\begin_layout Standard
+Dado 
+\begin_inset Formula $f\in\text{End}_{K}(V)$
+\end_inset
+
+ y 
+\begin_inset Formula $W\leq V$
+\end_inset
+
+, 
+\begin_inset Formula $W$
+\end_inset
+
+ es 
+\series bold
+invariante
+\series default
+ por 
+\begin_inset Formula $f$
+\end_inset
+
+ si 
+\begin_inset Formula $f(W)\subseteq W$
+\end_inset
+
+.
+ Entonces la restricción de 
+\begin_inset Formula $f$
+\end_inset
+
+ a 
+\begin_inset Formula $W$
+\end_inset
+
+ es 
+\begin_inset Formula $f|_{W}\in\text{End}_{K}(W)$
+\end_inset
+
+.
+ También se tiene que 
+\begin_inset Formula $\{0\}$
+\end_inset
+
+ y 
+\begin_inset Formula $V$
+\end_inset
+
+ son invariantes de cada 
+\begin_inset Formula $f\in\text{End}_{K}(V)$
+\end_inset
+
+, y la suma e intersección de subespacios invariantes por 
+\begin_inset Formula $f$
+\end_inset
+
+ también son subespacios invariantes por 
+\begin_inset Formula $f$
+\end_inset
+
+.
+ Ahora supongamos que 
+\begin_inset Formula $V=W_{1}\oplus\dots\oplus W_{t}$
+\end_inset
+
+, donde cada 
+\begin_inset Formula $W_{i}$
+\end_inset
+
+ es un subespacio invariante no nulo por 
+\begin_inset Formula $f\in\text{End}(V)$
+\end_inset
+
+.
+ Si tomamos 
+\begin_inset Formula ${\cal B}={\cal B}_{1}\cup\dots\cup{\cal B}_{t}$
+\end_inset
+
+, siendo cada 
+\begin_inset Formula ${\cal B}_{i}$
+\end_inset
+
+ una base de 
+\begin_inset Formula $W_{i}$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+M_{{\cal B}}(f)=\left(\begin{array}{ccc}
+\boxed{A_{1}} &  & 0\\
+ & \ddots\\
+0 &  & \boxed{A_{t}}
+\end{array}\right)
+\]
+
+\end_inset
+
+donde 
+\begin_inset Formula $A_{i}=M_{{\cal B}_{i}}(f|_{W})\in M_{n_{i}}(K)$
+\end_inset
+
+ para 
+\begin_inset Formula $n_{i}=\dim(W_{i})$
+\end_inset
+
+.
+ Recíprocamente, si 
+\begin_inset Formula $M_{{\cal B}}(f)$
+\end_inset
+
+ tiene dicha forma y
+\begin_inset Formula 
+\[
+{\cal B}=\{v_{1},\dots,v_{n_{1}},v_{n_{1}+1},\dots,v_{n_{1}+n_{2}},\dots,v_{n_{1}+\dots+n_{t-1}+1},\dots,v_{n_{1}+\dots+n_{t}}\}
+\]
+
+\end_inset
+
+entonces 
+\begin_inset Formula $W_{1}=<v_{1},\dots,v_{n_{1}}>$
+\end_inset
+
+, 
+\begin_inset Formula $W_{2}=<v_{n_{1}+1},\dots,v_{n_{1}+n_{2}}>$
+\end_inset
+
+, etc.
+ son subespacios vectoriales invariantes por 
+\begin_inset Formula $f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Endomorfismos y matrices diagonalizables
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f\in\text{End}_{K}(V)$
+\end_inset
+
+ es 
+\series bold
+diagonalizable
+\series default
+ si existe una base 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ de 
+\begin_inset Formula $V$
+\end_inset
+
+ tal que 
+\begin_inset Formula $M_{{\cal B}}(f)$
+\end_inset
+
+ es diagonal, y una matriz cuadrada es diagonalizable si lo es el endomorfismo
+ de 
+\begin_inset Formula $K^{n}$
+\end_inset
+
+ que cuya matriz respecto a la base canónica es 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Equivalentemente, una matriz cuadrada es diagonalizable si y sólo si es
+ semejante a una matriz diagonal, y un endomorfismo es diagonalizable si
+ su matriz asociada respecto a cualquier base lo es.
+ Denotamos las matrices diagonales como
+\begin_inset Formula 
+\[
+\left(\begin{array}{ccc}
+\lambda_{1} &  & 0\\
+ & \ddots\\
+0 &  & \lambda_{n}
+\end{array}\right)=\text{Diag}(\lambda_{1},\dots,\lambda_{n})
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Vectores y valores propios
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula ${\cal B}=\{u_{1,}\dots,u_{n}\}$
+\end_inset
+
+ y 
+\begin_inset Formula $M_{{\cal B}}(f)=\text{Diag}(\lambda_{1},\dots,\lambda_{n})$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $f(u_{i})=\lambda_{1}u_{1}$
+\end_inset
+
+.
+ Así:
+\end_layout
+
+\begin_layout Itemize
+Un 
+\series bold
+vector propio
+\series default
+, 
+\series bold
+autovector
+\series default
+ o 
+\series bold
+vector característico
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ es un vector 
+\begin_inset Formula $v\neq0$
+\end_inset
+
+ para el que existe un 
+\begin_inset Formula $\lambda\in K$
+\end_inset
+
+ con 
+\begin_inset Formula $f(v)=\lambda v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Un 
+\series bold
+valor propio
+\series default
+, 
+\series bold
+autovalor
+\series default
+ o 
+\series bold
+valor característico
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ es un escalar 
+\begin_inset Formula $\lambda\in K$
+\end_inset
+
+ para el que existe un 
+\begin_inset Formula $v\in V\backslash\{0\}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f(v)=\lambda v$
+\end_inset
+
+.
+ Decimos que 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ es 
+\series bold
+\emph on
+el
+\emph default
+ valor propio asociado al vector propio 
+\begin_inset Formula $v$
+\end_inset
+
+
+\series default
+, o que 
+\begin_inset Formula $v$
+\end_inset
+
+ es 
+\series bold
+\emph on
+un
+\emph default
+ vector propio asociado al valor propio 
+\begin_inset Formula $\lambda$
+\end_inset
+
+
+\series default
+.
+\end_layout
+
+\begin_layout Standard
+Así, 
+\begin_inset Formula $f\in\text{End}(V)$
+\end_inset
+
+ es diagonalizable si y sólo si existe una base de 
+\begin_inset Formula $V$
+\end_inset
+
+ formada por vectores propios de 
+\begin_inset Formula $f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Subespacios propios.
+ Polinomio característico
+\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
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $\lambda\in K$
+\end_inset
+
+ es valor propio si y sólo si existe 
+\begin_inset Formula $0\neq v\in V_{\lambda}$
+\end_inset
+
+, es decir, si 
+\begin_inset Formula $\text{Nuc}(f-\lambda Id)\neq\{0\}$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $0<\dim(\text{Nuc}(\lambda Id-f))=\dim(V)-\text{rang}(\lambda Id-f)$
+\end_inset
+
+, es decir, 
+\begin_inset Formula $\text{rang}(\lambda Id-f)<\dim(V)$
+\end_inset
+
+ o, equivalentemente, 
+\begin_inset Formula $\det(\lambda Id-f)=0$
+\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
+
+donde 
+\begin_inset Formula $\text{tr}(A)$
+\end_inset
+
+ es la 
+\series bold
+traza
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+, la suma de los elementos de su diagonal.
+ Obtenemos como resultado que los valores propios de 
+\begin_inset Formula $f\in\text{End}(V)$
+\end_inset
+
+ son las raíces de 
+\begin_inset Formula $P_{f}(x)$
+\end_inset
+
+, y que 
+\begin_inset Formula $f$
+\end_inset
+
+ tiene a lo sumo 
+\begin_inset Formula $\dim(V)$
+\end_inset
+
+ valores propios distintos.
+\end_layout
+
+\begin_layout Section
+Independencia de los subespacios propios
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{s}$
+\end_inset
+
+ son valores propios de 
+\begin_inset Formula $f$
+\end_inset
+
+ distintos dos a dos, entonces 
+\begin_inset Formula $\text{Nuc}(\lambda_{1}Id-f)+\dots+\text{Nuc}(\lambda_{s}Id-f)$
+\end_inset
+
+ es suma directa, y en particular, vectores propios correspondientes a valores
+ propios distintos dos a dos son linealmente independientes.
+ 
+\series bold
+Demostración:
+\series default
+ Para 
+\begin_inset Formula $s=2$
+\end_inset
+
+, sean 
+\begin_inset Formula $v_{1}\in\text{Nuc}(\lambda_{1}Id-f)$
+\end_inset
+
+ y 
+\begin_inset Formula $v_{2}\in\text{Nuc}(\lambda_{2}Id-f)$
+\end_inset
+
+ con 
+\begin_inset Formula $0=v_{1}+v_{2}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $0=f(0)=f(v_{1}+v_{2})=f(v_{1})+f(v_{2})=\lambda_{1}v_{1}+\lambda_{2}v_{2}$
+\end_inset
+
+, pero también 
+\begin_inset Formula $0=\lambda_{2}(v_{1}+v_{2})=\lambda_{2}v_{1}+\lambda_{2}v_{2}$
+\end_inset
+
+.
+ Restando, 
+\begin_inset Formula $0=(\lambda_{2}-\lambda_{1})v_{1}$
+\end_inset
+
+, pero como 
+\begin_inset Formula $\lambda_{2}\neq\lambda_{1}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $v_{1}=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $v_{2}$
+\end_inset
+
+ también.
+ Ahora sea 
+\begin_inset Formula $s>2$
+\end_inset
+
+ y supongamos el resultado cierto para 
+\begin_inset Formula $s-1$
+\end_inset
+
+.
+ Sean ahora 
+\begin_inset Formula $v_{1}\in\text{Nuc}(\lambda_{1}Id-f),\dots,v_{s}\in\text{Nuc}(\lambda_{s}Id-f)$
+\end_inset
+
+ con 
+\begin_inset Formula $0=v_{1}+\dots+v_{s}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $0=f(0)=f(v_{1}+\dots+v_{s})=f(v_{1})+\dots+f(v_{s})=\lambda_{1}v_{1}+\dots+\lambda_{s}v_{s}$
+\end_inset
+
+, pero también 
+\begin_inset Formula $0=\lambda_{s}v_{1}+\dots+\lambda_{s}v_{k-1}+\lambda_{s}v_{s}$
+\end_inset
+
+.
+ Restando, 
+\begin_inset Formula $0=(\lambda_{s}-\lambda_{1})v_{1}+\dots+(\lambda_{s}-\lambda_{s-1})v_{s-1}$
+\end_inset
+
+.
+ Aplicando la hipótesis de inducción, queda que 
+\begin_inset Formula $v_{1}=\dots=v_{s-1}=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $v_{s}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+También, si 
+\begin_inset Formula $f\in\text{End}(V)$
+\end_inset
+
+ tiene 
+\begin_inset Formula $\dim(V)$
+\end_inset
+
+ autovalores, entonces es diagonalizable.
+\series bold
+
+\begin_inset Newline newline
+\end_inset
+
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{n}$
+\end_inset
+
+ son valores propios de 
+\begin_inset Formula $f$
+\end_inset
+
+ distintos dos a dos y 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ son vectores propios asociados a cada uno, entonces 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ son 
+\begin_inset Formula $n$
+\end_inset
+
+ vectores linealmente independientes en un espacio de dimensión 
+\begin_inset Formula $n$
+\end_inset
+
+, por lo que constituyen una base formada por vectores propios de 
+\begin_inset Formula $f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Caracterización de los endomorfismos diagonalizables
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $P(x)$
+\end_inset
+
+ es un polinomio con coeficientes en 
+\begin_inset Formula $K$
+\end_inset
+
+ y 
+\begin_inset Formula $\lambda\in K$
+\end_inset
+
+ es una raíz de 
+\begin_inset Formula $P(x)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ tiene 
+\series bold
+multiplicidad
+\series default
+ 
+\begin_inset Formula $m$
+\end_inset
+
+ en 
+\begin_inset Formula $P(x)$
+\end_inset
+
+ si 
+\begin_inset Formula $(x-\lambda)^{m}|P(x)$
+\end_inset
+
+ pero 
+\begin_inset Formula $\neg((x-\lambda)^{m+1}|P(x))$
+\end_inset
+
+.
+ Si una raíz tiene multiplicidad 1, es una raíz 
+\series bold
+simple
+\series default
+\SpecialChar endofsentence
+ De lo contrario es una raíz
+\series bold
+ múltiple
+\series default
+\SpecialChar endofsentence
+
+\end_layout
+
+\begin_layout Standard
+Dado 
+\begin_inset Formula $f\in\text{End}_{K}(V)$
+\end_inset
+
+ y 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ un valor propio de 
+\begin_inset Formula $f$
+\end_inset
+
+, si 
+\begin_inset Formula $d=\dim(\text{Nuc}(\lambda Id-f))$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+ es la multiplicidad de 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ en 
+\begin_inset Formula $P_{f}(x)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $d\leq m$
+\end_inset
+
+.
+ En particular, si el valor propio es una raíz simple, entonces 
+\begin_inset Formula $d=m=1$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $\{v_{1},\dots,v_{d}\}$
+\end_inset
+
+ una base de 
+\begin_inset Formula $\text{Nuc}(\lambda Id-f)$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}=\{v_{1},\dots,v_{d},v_{d+1},\dots,v_{n}\}$
+\end_inset
+
+ una base de 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $M_{{\cal B}}(f)$
+\end_inset
+
+ tiene forma
+\begin_inset Formula 
+\[
+\left(\begin{array}{ccc|c}
+\lambda &  & 0\\
+ & \ddots &  & C\\
+0 &  & \lambda\\
+\hline  &  & \\
+ & 0 &  & D\\
+ &  & \\
+\end{array}\right)
+\]
+
+\end_inset
+
+por lo que 
+\begin_inset Formula $P_{f}(x)=(x-\lambda)^{d}\det(xI_{n-d}-D)=(x-\lambda)^{d}Q(x)$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $d\leq m$
+\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 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 $\lambda_{1},\dots,\lambda_{r}$
+\end_inset
+
+ los valores propios de 
+\begin_inset Formula $f$
+\end_inset
+
+, existirá una base 
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ de vectores propios en los que cada vector tendrá asociado un valor propio
+ y pertenecerá por tanto al subespacio propio correspondiente.
+ Agrupando, 
+\begin_inset Formula $M_{{\cal B}}(f)$
+\end_inset
+
+ tendrá forma
+\begin_inset Formula 
+\[
+\left(\begin{array}{ccccccc}
+\lambda_{1}\\
+ & \ddots &  &  &  & 0\\
+ &  & \lambda_{1}\\
+ &  &  & \ddots\\
+ &  &  &  & \lambda_{r}\\
+ & 0 &  &  &  & \ddots\\
+ &  &  &  &  &  & \lambda_{r}
+\end{array}\right)
+\]
+
+\end_inset
+
+donde cada 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ se repetirá 
+\begin_inset Formula $m_{i}$
+\end_inset
+
+ veces, el número de vectores propios de la base del subespacio.
+ Por tanto, 
+\begin_inset Formula $P_{f}(x)=(x-\lambda_{1})^{m_{1}}\cdots(x-\lambda_{r})^{m_{r}}$
+\end_inset
+
+ tiene todas sus raíces en 
+\begin_inset Formula $K$
+\end_inset
+
+.
+ Además, si 
+\begin_inset Formula $d_{i}=\dim(\text{Nuc}(\lambda_{i}Id-f))$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $\sum d_{i}=\dim(\text{Nuc}(\lambda_{1}Id-f)\oplus\cdots\oplus\text{Nuc}(\lambda_{r}Id-f))\leq\dim(V)=n$
+\end_inset
+
+, y como en la base hay 
+\begin_inset Formula $m_{i}$
+\end_inset
+
+ vectores linealmente independientes de 
+\begin_inset Formula $\text{Nuc}(\lambda_{i}Id-f)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $m_{i}\leq d_{i}$
+\end_inset
+
+, luego 
+\begin_inset Formula $n=\text{gr}(P_{f}(x))=\sum m_{i}\leq\sum d_{i}\leq n$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $m_{i}=d_{i}$
+\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 
+\begin_inset Formula $P_{f}(x)=(x-\lambda_{1})^{d_{1}}\cdots(x-\lambda_{r})^{d_{r}}$
+\end_inset
+
+ con 
+\begin_inset Formula $d_{i}=\dim(\text{Nuc}(f-\lambda Id))$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\dim(\text{Nuc}(\lambda_{1}Id-f)\oplus\cdots\oplus\text{Nuc}(\lambda_{r}Id-f))=d_{1}+\dots+d_{r}=\text{gr}(P_{f}(x))=\dim(V)$
+\end_inset
+
+, luego 
+\begin_inset Formula $V=\text{Nuc}(f-\lambda_{1}Id)\oplus\cdots\oplus\text{Nuc}(f-\lambda_{r}Id)$
+\end_inset
+
+ y la unión de las bases de cada subespacio será una base de 
+\begin_inset Formula $V$
+\end_inset
+
+ formada por vectores propios.
+\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 Section
+Aplicaciones
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $(x_{n})_{n}\subseteq K$
+\end_inset
+
+ verifica una 
+\series bold
+ecuación en diferencias lineales con coeficientes constantes
+\series default
+ (homogénea) si para todo 
+\begin_inset Formula $n$
+\end_inset
+
+ satisface que 
+\begin_inset Formula $x_{n+r}+a_{1}x_{n+r-1}+\dots+a_{r}x_{n}=0$
+\end_inset
+
+.
+ Llamamos a 
+\begin_inset Formula $r$
+\end_inset
+
+ el 
+\series bold
+orden
+\series default
+ de la ecuación.
+ Podemos definir entonces una sucesión auxiliar 
+\begin_inset Formula $(Y_{n})_{n}\subseteq M_{r,1}(K)$
+\end_inset
+
+ con 
+\begin_inset Formula $(Y_{n})_{i}=x_{n+r-i}$
+\end_inset
+
+.
+ Se tiene entonces que 
+\begin_inset Formula $x_{n+r}=-a_{1}x_{n+r-1}-\dots-a_{r}x_{n}$
+\end_inset
+
+, luego
+\begin_inset Formula 
+\[
+\begin{array}{c}
+Y_{n+1}=\left(\begin{array}{c}
+x_{n+r}\\
+\vdots\\
+x_{n+1}
+\end{array}\right)=\left(\begin{array}{c}
+-a_{1}x_{n+r-1}-\dots-a_{r}x_{n}\\
+x_{n+r-1}\\
+\vdots\\
+x_{n+1}
+\end{array}\right)=\\
+=\left(\begin{array}{cccc}
+-a_{1} & -a_{2} & \cdots & -a_{r}\\
+1 &  & 0 & 0\\
+ & \ddots &  & \vdots\\
+0 &  & 1 & 0
+\end{array}\right)\left(\begin{array}{c}
+x_{n+r-1}\\
+\vdots\\
+x_{n}
+\end{array}\right)=AY_{n}
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+sistema de ecuaciones en diferencias lineales de primer orden con coeficientes
+ constantes
+\series default
+ (homogéneo) es una relación entre los términos de unas sucesiones y sus
+ términos inmediatamente anteriores:
+\begin_inset Formula 
+\[
+\left.\begin{array}{ccc}
+x_{n+1} & = & a_{11}x_{n}+a_{12}y_{n}+a_{13}z_{n}\\
+y_{n+1} & = & a_{21}x_{n}+a_{22}y_{n}+a_{23}z_{n}\\
+z_{n+1} & = & a_{31}x_{n}+a_{32}y_{n}+a_{33}z_{n}
+\end{array}\right\} 
+\]
+
+\end_inset
+
+Estos pueden expresarme matricialmente de la forma 
+\begin_inset Formula $Y_{n+1}=AY_{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $A=(a_{ij})$
+\end_inset
+
+.
+ Por re
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+cu
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+rren
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+cia, en ambos casos se tiene que 
+\begin_inset Formula $Y_{n}=A^{n-1}Y_{1}=A^{n}Y_{0}$
+\end_inset
+
+.
+ Entonces es útil diagonalizar 
+\begin_inset Formula $A$
+\end_inset
+
+, si es posible, para poder calcular las potencias rápidamente.
+\end_layout
+
+\end_body
+\end_document