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+#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
generated by cgit v1.2.3 (git 2.46.0) at 2025-10-30 14:22:21 +0000