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\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 unívocamente 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 independientes 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