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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
+\begin_preamble
+\input{../defs}
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style french
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Sean
+\begin_inset Formula $K$
+\end_inset
+
+ un cuerpo y
+\begin_inset Formula $M$
+\end_inset
+
+ el
+\begin_inset Formula $K[X]$
+\end_inset
+
+-módulo asociado a un par
+\begin_inset Formula $(V,f)$
+\end_inset
+
+ de un espacio vectorial y un
+\begin_inset Formula $K$
+\end_inset
+
+-endomorfismo
+\begin_inset Formula $V\to V$
+\end_inset
+
+,
+\begin_inset Formula $M$
+\end_inset
+
+ es de torsión finitamente generado si y sólo si
+\begin_inset Formula $_{K}V$
+\end_inset
+
+ es de dimensión finita, y si
+\begin_inset Formula $p\in K[X]$
+\end_inset
+
+ es irreducible,
+\begin_inset Formula $M$
+\end_inset
+
+ es finitamente generado de
+\begin_inset Formula $p$
+\end_inset
+
+-torsión si y sólo si
+\begin_inset Formula $_{K}V$
+\end_inset
+
+ es de dimensión finita y
+\begin_inset Formula $p(f)^{m}=0\in\text{End}_{K}(V)$
+\end_inset
+
+ para cierto
+\begin_inset Formula $m>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+En el resto de la sección, salvo que se indique lo contrario,
+\begin_inset Formula $K$
+\end_inset
+
+ es un cuerpo,
+\begin_inset Formula $V$
+\end_inset
+
+ un
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial de dimensión finita,
+\begin_inset Formula $f:V\to V$
+\end_inset
+
+ un
+\begin_inset Formula $K$
+\end_inset
+
+-endomorfismo y
+\begin_inset Formula $M$
+\end_inset
+
+ el
+\begin_inset Formula $K[X]$
+\end_inset
+
+-módulo asociado a
+\begin_inset Formula $(V,f)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teoremas de clasificación de endomorfismos de espacios vectoriales:
+\end_layout
+
+\begin_layout Enumerate
+Existen
+\begin_inset Formula $p_{1},\dots,p_{k}\in K[X]$
+\end_inset
+
+ mónicos irreducibles distintos y
+\begin_inset Formula $n_{ij}\in\mathbb{N}^{*}$
+\end_inset
+
+ para
+\begin_inset Formula $i\in\{1,\dots,k\}$
+\end_inset
+
+ y
+\begin_inset Formula $j\in\{1,\dots,r_{i}\}$
+\end_inset
+
+, unívocamente determinados, y vectores
+\begin_inset Formula $v_{ij}\in V$
+\end_inset
+
+, tales que
+\begin_inset Formula
+\[
+\bigoplus_{i=1}^{k}\bigoplus_{j=1}^{r_{i}}K\{f^{s}(v_{ij})\}_{s\geq0}
+\]
+
+\end_inset
+
+es una descomposición de
+\begin_inset Formula $V$
+\end_inset
+
+ en suma directa interna de subespacios vectoriales
+\begin_inset Formula $f$
+\end_inset
+
+-in
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+va
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+rian
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+tes y cada
+\begin_inset Formula $p_{i}(f)^{n_{ij}}(v_{ij})=0\neq p_{i}(f)^{n_{ij}-1}(v_{ij})$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean
+\begin_inset Formula $W\leq V$
+\end_inset
+
+ y
+\begin_inset Formula $N$
+\end_inset
+
+ el
+\begin_inset Formula $K[X]$
+\end_inset
+
+-submódulo de
+\begin_inset Formula $M$
+\end_inset
+
+ asociado a
+\begin_inset Formula $(W,f|_{W})$
+\end_inset
+
+, basta ver que
+\begin_inset Formula $N\cong\frac{K[X]}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+ si y sólo si existe
+\begin_inset Formula $v\in V$
+\end_inset
+
+ tal que
+\begin_inset Formula $W=K\{f^{s}(v)_{s\geq0}\}$
+\end_inset
+
+ y
+\begin_inset Formula $p_{i}(f)^{n_{ij}}(v)=0\neq p_{i}(f)^{n_{ij}-1}(v)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\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 $\phi:\frac{K[X]}{(p_{i}^{n_{ij}})}\to N$
+\end_inset
+
+ el isomorfismo y
+\begin_inset Formula $v\coloneqq\phi(\overline{1})$
+\end_inset
+
+,
+\begin_inset Formula $p_{i}^{n_{ij}}\overline{1}=0$
+\end_inset
+
+ y por tanto
+\begin_inset Formula $0=p_{i}^{n_{ij}}\phi(\overline{1})=p_{i}^{n_{ij}}v=p_{i}(f)^{n_{ij}}(v)$
+\end_inset
+
+ por la definición del
+\begin_inset Formula $K[X]$
+\end_inset
+
+-módulo, pero
+\begin_inset Formula $p_{i}^{n_{ij}-1}\overline{1}\neq0$
+\end_inset
+
+ y por tanto
+\begin_inset Formula $p_{i}(f)^{n_{ij}-1}(v_{ij})\neq0$
+\end_inset
+
+.
+ Finalmente, como
+\begin_inset Formula $\frac{K[X]}{(p_{i}^{n_{ij}})}=K\{\overline{1},X\overline{1},\dots,X^{s}\overline{1},\dots\}$
+\end_inset
+
+,
+\begin_inset Formula $M=K\{f^{s}(v)\}_{s\geq0}$
+\end_inset
+
+ ya que
+\begin_inset Formula $\phi(X^{s}\overline{1})=X^{s}\phi(\overline{1})=f^{s}(v)$
+\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
+
+Por la hipótesis y la definición de
+\begin_inset Formula $N$
+\end_inset
+
+,
+\begin_inset Formula $N=(v)$
+\end_inset
+
+, pero
+\begin_inset Formula $v$
+\end_inset
+
+ es anulado por
+\begin_inset Formula $p_{i}(f)^{n_{ij}}$
+\end_inset
+
+ y por tanto hay un epimorfismo
+\begin_inset Formula $\psi:\frac{K[X]}{(p_{i}^{n_{ij}})}\twoheadrightarrow K[X]v=N$
+\end_inset
+
+ con
+\begin_inset Formula $\ker\psi\trianglelefteq\frac{K[X]}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+, pero los únicos ideales de
+\begin_inset Formula $\frac{K[X]}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+ son
+\begin_inset Formula $(\overline{p_{i}}^{k})$
+\end_inset
+
+ con
+\begin_inset Formula $k\in\{0,\dots,n_{ij}\}$
+\end_inset
+
+, y como
+\begin_inset Formula $p_{i}(f)^{n_{ij}-1}(v)\neq0$
+\end_inset
+
+,
+\begin_inset Formula $\overline{p_{i}}^{n_{ij}-1}\notin\ker\psi$
+\end_inset
+
+, con lo que
+\begin_inset Formula $\ker\psi=0$
+\end_inset
+
+ y
+\begin_inset Formula $\psi$
+\end_inset
+
+ es un isomorfismo.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Existen polinomios mónicos no constantes
+\begin_inset Formula $d_{1}\mid\dots\mid d_{t}$
+\end_inset
+
+ unívocamente determinados y vectores
+\begin_inset Formula $v_{j}\in V$
+\end_inset
+
+ tales que
+\begin_inset Formula $\bigoplus_{i=1}^{t}\text{span}\{f^{s}(v_{j})\}_{s\in\mathbb{N}_{\text{gr}(d_{j})}}$
+\end_inset
+
+ es una descomposición de
+\begin_inset Formula $V$
+\end_inset
+
+ en subespacios
+\begin_inset Formula $f$
+\end_inset
+
+-invariantes y cada
+\begin_inset Formula $d_{j}(f)(v_{j})=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean
+\begin_inset Formula $W\leq V$
+\end_inset
+
+ y
+\begin_inset Formula $N$
+\end_inset
+
+ el
+\begin_inset Formula $K[X]$
+\end_inset
+
+-submódulo de
+\begin_inset Formula $M$
+\end_inset
+
+ asociado a
+\begin_inset Formula $(W,f|_{W})$
+\end_inset
+
+, basta ver que
+\begin_inset Formula $N\cong\frac{K[X]}{(d_{j})}$
+\end_inset
+
+ si y sólo si existe
+\begin_inset Formula $v\in V$
+\end_inset
+
+ tal que
+\begin_inset Formula $\{f^{s}(v)\}{}_{s\in\mathbb{N}_{\text{gr}(d_{j})}}$
+\end_inset
+
+ es base de
+\begin_inset Formula $W$
+\end_inset
+
+ como
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial y
+\begin_inset Formula $d_{j}(f)(v)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\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 $\phi:\frac{K[X]}{(p_{i}^{n_{ij}})}\to N$
+\end_inset
+
+ el isomorfismo y
+\begin_inset Formula $v\coloneqq\phi(\overline{1})$
+\end_inset
+
+,
+\begin_inset Formula $d_{j}\overline{1}=0$
+\end_inset
+
+ y por tanto
+\begin_inset Formula $0=d_{j}\phi(\overline{1})=d_{j}v=d_{j}(f)(v)$
+\end_inset
+
+, y como
+\begin_inset Formula $\frac{K[X]}{(d_{j})}=K\{\overline{1},X\overline{1},\dots,X^{\text{gr}d_{j}-1}\overline{1}\}$
+\end_inset
+
+ con
+\begin_inset Formula $(X^{s}\overline{1})_{s\in\mathbb{N}_{\text{gr}(d_{j})}}$
+\end_inset
+
+ linealmente independiente,
+\begin_inset Formula $N=K\{f^{s}(v)\}_{s\in\mathbb{N}_{\text{gr}(d_{j})}}$
+\end_inset
+
+ con
+\begin_inset Formula $(f^{s}(v))_{s\in\mathbb{N}_{\text{gr}(d_{j})}}$
+\end_inset
+
+ linealmente independiente.
+\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
+
+
+\begin_inset Formula $v$
+\end_inset
+
+ es anulado por
+\begin_inset Formula $p_{i}(f)^{n_{ij}}$
+\end_inset
+
+ y por tanto hay un epimorfismo
+\begin_inset Formula $\psi:\frac{K[X]}{(d_{j})}\twoheadrightarrow K[X]v=K\{f^{s}(v)\}_{s\in\mathbb{N}}=K\{f^{s}(v)\}_{s\in\mathbb{N}_{\text{gr}(d_{j})}}=N$
+\end_inset
+
+, pero si
+\begin_inset Formula $p\in K[X]$
+\end_inset
+
+ con
+\begin_inset Formula $\text{gr}p<\text{gr}d_{j}$
+\end_inset
+
+ cumple
+\begin_inset Formula $\psi(\overline{p})=p(f)(v)=\sum_{i}p_{i}f^{i}(v)=0$
+\end_inset
+
+, como los
+\begin_inset Formula $f^{i}(v)$
+\end_inset
+
+ son linealmente independiente, cada
+\begin_inset Formula $p_{i}=0$
+\end_inset
+
+ y
+\begin_inset Formula $p=0$
+\end_inset
+
+, y como cada elemento de
+\begin_inset Formula $\frac{K[X]}{(d_{j})}$
+\end_inset
+
+ tiene un representante de grado menor que el de
+\begin_inset Formula $d_{j}$
+\end_inset
+
+,
+\begin_inset Formula $\ker\psi=0$
+\end_inset
+
+ y
+\begin_inset Formula $\psi$
+\end_inset
+
+ es un isomorfismo.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Polinomio mínimo
+\end_layout
+
+\begin_layout Standard
+Sea
+\begin_inset Formula $\varphi\in K[X]$
+\end_inset
+
+ el polinomio característico de
+\begin_inset Formula $f$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Teorema de Cayley-Hamilton:
+\series default
+
+\begin_inset Formula $\varphi(f)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean
+\begin_inset Formula $C\in{\cal M}_{n}(K)$
+\end_inset
+
+ la matriz asociada a
+\begin_inset Formula $f$
+\end_inset
+
+ bajo cualquier base de
+\begin_inset Formula $V$
+\end_inset
+
+ e
+\begin_inset Formula $I\coloneqq I_{n}$
+\end_inset
+
+, queremos ver que
+\begin_inset Formula $\varphi=\det(XI-C)$
+\end_inset
+
+ cumple
+\begin_inset Formula $\sum_{i=0}^{n}\varphi_{i}C^{i}=0$
+\end_inset
+
+.
+ Por la prueba de la fórmula de la matriz inversa, para toda matriz
+\begin_inset Formula $A$
+\end_inset
+
+ es
+\begin_inset Formula $A\cdot\text{adj}(A)^{\intercal}=|A|I$
+\end_inset
+
+, por lo que viendo
+\begin_inset Formula $XI-C\in{\cal M}_{n}(K[X])$
+\end_inset
+
+ es
+\begin_inset Formula $(XI-C)\text{adj}(XI-C)^{\intercal}=\varphi I$
+\end_inset
+
+.
+ Como las entradas de
+\begin_inset Formula $\text{adj}(XI-C)^{\intercal}$
+\end_inset
+
+ son polinomios de grado máximo
+\begin_inset Formula $n-1$
+\end_inset
+
+, podemos escribir
+\begin_inset Formula $\text{adj}(XI-C)^{t}\eqqcolon\sum_{i=0}^{n-1}B_{i}X^{i}$
+\end_inset
+
+ con cada
+\begin_inset Formula $B_{i}\in{\cal M}_{n}(K)$
+\end_inset
+
+ y entonces
+\begin_inset Formula $(XI-C)\sum_{i=0}^{n-1}B_{i}X^{i}=\sum_{i=0}^{n}\varphi_{i}I$
+\end_inset
+
+.
+ Viendo esta igualdad en
+\begin_inset Formula ${\cal M}_{n}(K)[X]$
+\end_inset
+
+, igualando coeficientes,
+\begin_inset Formula
+\begin{align*}
+B_{n-1} & =\varphi_{n}I, & B_{n-2}-CB_{n-1} & =\varphi_{n-1}I, & & \cdots, & B_{0}-B_{1}C & =\varphi_{1}I, & -B_{0}C & =\varphi_{0}I,
+\end{align*}
+
+\end_inset
+
+y multiplicando la primera igualdad por
+\begin_inset Formula $C^{n}$
+\end_inset
+
+, la segunda por
+\begin_inset Formula $C^{n-1}$
+\end_inset
+
+, etc.,
+\begin_inset Formula
+\begin{align*}
+C^{n}B_{n-1} & =\varphi_{n}C^{n}, & C^{n-1}B_{n-2}-C^{n}B_{n-1} & =\varphi_{n-1}C^{n-1}, & & \dots,\\
+CB_{0}-C^{2}B_{1} & =\varphi_{1}C, & -CB_{0} & =\varphi_{0}I,
+\end{align*}
+
+\end_inset
+
+luego sumando es
+\begin_inset Formula $0=\varphi_{n}C^{n}+\dots+\varphi_{1}C+\varphi_{0}=\varphi_{C}(C)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Los divisores irreducibles de
+\begin_inset Formula $M$
+\end_inset
+
+ son precisamente los divisores irreducibles de
+\begin_inset Formula $\varphi$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $p\in K[X]$
+\end_inset
+
+ irreducible es divisor irreducible de
+\begin_inset Formula $M$
+\end_inset
+
+ si y sólo si existe
+\begin_inset Formula $v\in M\setminus\{0\}$
+\end_inset
+
+ con
+\begin_inset Formula $pv=p(f)(v)=0$
+\end_inset
+
+, si y sólo si
+\begin_inset Formula $\ker(p(f))\neq0$
+\end_inset
+
+, si y sólo si
+\begin_inset Formula $p(f):V\to V$
+\end_inset
+
+ como endomorfismo no es un isomorfismo, si y sólo si
+\begin_inset Formula $\det(p(f))=0$
+\end_inset
+
+.
+ Sea
+\begin_inset Formula $\overline{K}$
+\end_inset
+
+ la clausura algebraica de
+\begin_inset Formula $K$
+\end_inset
+
+,
+\begin_inset Formula $p=(X-\lambda_{1})\cdots(X-\lambda_{t})\in\overline{K}[X]$
+\end_inset
+
+.
+ Si
+\begin_inset Formula $p\mid\varphi$
+\end_inset
+
+, sea
+\begin_inset Formula $C$
+\end_inset
+
+ la matriz asociada a
+\begin_inset Formula $f$
+\end_inset
+
+ bajo cualquier base, los
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ son valores propios de
+\begin_inset Formula $C$
+\end_inset
+
+ en
+\begin_inset Formula $\overline{K}$
+\end_inset
+
+ y por tanto existen
+\begin_inset Formula $v_{i}\in\overline{K}^{n}\setminus\{0\}$
+\end_inset
+
+ con
+\begin_inset Formula $Cv_{i}=\lambda v_{i}$
+\end_inset
+
+ y
+\begin_inset Formula $(C-\lambda_{i}I)=0$
+\end_inset
+
+.
+ Pero
+\begin_inset Formula $(C-\lambda_{i}I)(C-\lambda_{j}I)=C^{2}-\lambda_{i}I-\lambda_{j}I+\lambda_{i}\lambda_{j}I=(C-\lambda_{j}I)(C-\lambda_{i}I)$
+\end_inset
+
+, por lo que
+\begin_inset Formula $(C-\lambda_{i}I)(C-\lambda_{j}I)(v_{i})=0$
+\end_inset
+
+ y
+\begin_inset Formula $p(C)(v)=\left(\prod_{j}(C-\lambda_{j}I)\right)(v_{i})=0$
+\end_inset
+
+, de modo que
+\begin_inset Formula $\ker_{\overline{K}}(p(C))\neq0$
+\end_inset
+
+ y
+\begin_inset Formula $\det(p(C))=0$
+\end_inset
+
+, lo que no depende de si consideramos
+\begin_inset Formula $p(C)$
+\end_inset
+
+ sobre
+\begin_inset Formula $K$
+\end_inset
+
+ o sobre
+\begin_inset Formula $\overline{K}$
+\end_inset
+
+ y por tanto
+\begin_inset Formula $p$
+\end_inset
+
+ es divisor irreducible de
+\begin_inset Formula $M$
+\end_inset
+
+.
+ Si
+\begin_inset Formula $p$
+\end_inset
+
+ es divisor irreducible de
+\begin_inset Formula $M$
+\end_inset
+
+, divide al mayor factor invariante de
+\begin_inset Formula $M$
+\end_inset
+
+,
+\begin_inset Formula $d_{t}$
+\end_inset
+
+, pero para
+\begin_inset Formula $v\in M$
+\end_inset
+
+,
+\begin_inset Formula $\varphi v=\varphi(f)(v)=0$
+\end_inset
+
+, con lo que
+\begin_inset Formula $\varphi\in\text{ann}_{A}(M)=(d_{t})$
+\end_inset
+
+ y
+\begin_inset Formula $p\mid d_{t}\mid\varphi$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{reminder}{AlgL}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $A,B\in{\cal M}_{n}(K)$
+\end_inset
+
+ son
+\series bold
+semejantes
+\series default
+ si
+\begin_inset Formula $\exists P\in{\cal M}_{n}(K):B=P^{-1}AP$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{reminder}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sean
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ una base de
+\begin_inset Formula $V$
+\end_inset
+
+,
+\begin_inset Formula $C\coloneqq M_{{\cal B}}(f)$
+\end_inset
+
+ y
+\begin_inset Formula $f_{C}:K^{n}\to K^{n}$
+\end_inset
+
+ dado por
+\begin_inset Formula $f_{C}(y)\coloneqq Cy$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+El isomorfismo
+\begin_inset Formula $\phi:V\to K^{n}$
+\end_inset
+
+ que lleva
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ a la base canónica induce un isomorfismo entre el
+\begin_inset Formula $K[X]$
+\end_inset
+
+-módulo asociado a
+\begin_inset Formula $(V,f)$
+\end_inset
+
+ y el asociado a
+\begin_inset Formula $(K^{n},f_{C})$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Claramente la biyección
+\begin_inset Formula $\hat{\phi}$
+\end_inset
+
+ inducida conserva la suma y el producto por escalares de
+\begin_inset Formula $K$
+\end_inset
+
+, y
+\begin_inset Formula $\hat{\phi}(Xv)=\phi(f(v))=\phi((\phi^{-1}\circ f_{C}\circ\phi)(v))=f_{C}(\phi(v))=X\phi(v)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sean
+\begin_inset Formula $W$
+\end_inset
+
+ otro
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial,
+\begin_inset Formula $g:W\to W$
+\end_inset
+
+ un
+\begin_inset Formula $K$
+\end_inset
+
+-endomorfismo,
+\begin_inset Formula $\phi:V\to W$
+\end_inset
+
+ un
+\begin_inset Formula $K$
+\end_inset
+
+-isomorfismo con
+\begin_inset Formula $\phi\circ f=g\circ\phi$
+\end_inset
+
+,
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ una base de
+\begin_inset Formula $V$
+\end_inset
+
+ y
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ la base correspondiente de
+\begin_inset Formula $W$
+\end_inset
+
+ por
+\begin_inset Formula $\phi$
+\end_inset
+
+, se tiene
+\begin_inset Formula $M_{{\cal B}}(f)=M_{{\cal B}'}(g)$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si
+\begin_inset Formula ${\cal B}\eqqcolon(b_{i})_{i}$
+\end_inset
+
+,
+\begin_inset Formula ${\cal B}'=(\phi(b_{i}))_{i}$
+\end_inset
+
+, pero
+\begin_inset Formula $M_{{\cal B}}(f)$
+\end_inset
+
+ tiene como columnas los
+\begin_inset Formula $f(b_{i})$
+\end_inset
+
+ respecto de
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ y
+\begin_inset Formula $M_{{\cal B}'}(g)$
+\end_inset
+
+ tiene como columnas los
+\begin_inset Formula $g(\phi(b_{i}))=\phi(f(b_{i}))$
+\end_inset
+
+ respecto de
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+, por lo que
+\begin_inset Formula $M_{{\cal B}}(f)=M_{{\cal B}'}(g)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si
+\begin_inset Formula $W$
+\end_inset
+
+ es otro
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial de dimensión finita y
+\begin_inset Formula $g:W\to W$
+\end_inset
+
+ un
+\begin_inset Formula $K$
+\end_inset
+
+-endomorfismo, los
+\begin_inset Formula $K[X]$
+\end_inset
+
+-módulos asociados a
+\begin_inset Formula $(V,f)$
+\end_inset
+
+ y
+\begin_inset Formula $(W,g)$
+\end_inset
+
+ son isomorfos si y sólo si
+\begin_inset Formula $\dim V=\dim W$
+\end_inset
+
+ y existen bases respectivas
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ y
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ de
+\begin_inset Formula $V$
+\end_inset
+
+ y
+\begin_inset Formula $W$
+\end_inset
+
+ tales que
+\begin_inset Formula $M_{{\cal B}}(f)$
+\end_inset
+
+ y
+\begin_inset Formula $M_{{\cal B}'}(g)$
+\end_inset
+
+ son semejantes.
+\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
+
+Sea
+\begin_inset Formula $\phi:M\to N$
+\end_inset
+
+ el isomorfismo, claramente
+\begin_inset Formula $\phi:V\to W$
+\end_inset
+
+ es un
+\begin_inset Formula $K$
+\end_inset
+
+-isomorfismo y por tanto
+\begin_inset Formula $\dim_{K}V=\dim_{K}W$
+\end_inset
+
+, y basta tomar una base
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ de
+\begin_inset Formula $V$
+\end_inset
+
+ y, como
+\begin_inset Formula $\phi(f(v))=\phi(Xv)=X\phi(v)=g(\phi(v))$
+\end_inset
+
+, estamos en las condiciones del anterior apartado.
+\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
+
+Por cambio de base podemos suponer
+\begin_inset Formula $M_{{\cal B}}(f)=M_{{\cal B}'}(g)\eqqcolon(a_{ij})_{1\leq i,j\leq n}$
+\end_inset
+
+, y si
+\begin_inset Formula ${\cal B}=(b_{1},\dots,b_{n})$
+\end_inset
+
+ y
+\begin_inset Formula ${\cal B}'=(b'_{1},\dots,b'_{n})$
+\end_inset
+
+, tomando el isomorfismo vectorial
+\begin_inset Formula $\phi:V\to W$
+\end_inset
+
+ que lleva cada
+\begin_inset Formula $b_{i}$
+\end_inset
+
+ a
+\begin_inset Formula $b'_{i}$
+\end_inset
+
+ y viéndolo como un
+\begin_inset Formula $K[X]$
+\end_inset
+
+-isomorfismo
+\begin_inset Formula $\phi:M\to N$
+\end_inset
+
+,
+\begin_inset Formula $\phi(Xb_{i})=\phi(f(b_{i}))=\phi(\sum_{j}a_{ji}b_{j})=\sum_{j}a_{ji}b'_{j}=g(b'_{i})=X\phi(b_{i})$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Si
+\begin_inset Formula $A$
+\end_inset
+
+ es una matriz cuadrada, llamamos
+\begin_inset Formula $\text{rk}A$
+\end_inset
+
+ al rango de
+\begin_inset Formula $A$
+\end_inset
+
+, y si
+\begin_inset Formula $f:V\to V$
+\end_inset
+
+ es un endomorfismo,
+\begin_inset Formula $\text{rk}f\coloneqq\text{rk}M_{{\cal B}}(f)$
+\end_inset
+
+ para cualquier base
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ de
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Newpage pagebreak
+\end_inset
+
+Llamamos
+\series bold
+polinomio mínimo
+\series default
+ de
+\begin_inset Formula $M$
+\end_inset
+
+ a su mayor factor invariante, elegido mónico.
+\end_layout
+
+\begin_layout Enumerate
+Para
+\begin_inset Formula $G\in K[X]$
+\end_inset
+
+ y
+\begin_inset Formula $j\in\mathbb{N}$
+\end_inset
+
+,
+\begin_inset Formula $\text{ann}_{M}(G^{j})=\ker(G^{j}(f))$
+\end_inset
+
+, y
+\begin_inset Formula $G^{j}\in\text{ann}_{K[X]}(M)\iff G^{j}(f)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $G^{j}\in\text{ann}_{K[X]}(M)\iff\text{ann}_{M}(G^{j})=\ker(G^{j}(f))=M\iff G^{j}(f)=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+El polinomio mínimo de
+\begin_inset Formula $M$
+\end_inset
+
+ es el menor
+\begin_inset Formula $d_{t}\in K[X]$
+\end_inset
+
+ (por divisibilidad) con
+\begin_inset Formula $d_{t}(f)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si este es
+\begin_inset Formula $d_{t}$
+\end_inset
+
+,
+\begin_inset Formula $(d_{t})=\text{ann}_{K[X]}(M)$
+\end_inset
+
+, y basta aplicar el apartado anterior.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si
+\begin_inset Formula $\varphi$
+\end_inset
+
+ es el polinomio característico de
+\begin_inset Formula $f$
+\end_inset
+
+,
+\begin_inset Formula $d_{t}\mid\varphi$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\varphi(f)=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si
+\begin_inset Formula $p$
+\end_inset
+
+ es divisor irreducible de
+\begin_inset Formula $M$
+\end_inset
+
+ y
+\begin_inset Formula $n\coloneqq\min\{s\in\mathbb{N}\mid\ker(p(f)^{s})=\ker(p(f)^{s+1})\}=\min\{s\in\mathbb{N}\mid\text{rk}(p(f)^{s})=\text{rk}(p(f)^{s+1})\}$
+\end_inset
+
+, entonces
+\begin_inset Formula $M(p)=\ker(p(f)^{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\ker(p(f)^{s})=\ker(p(f)^{s+1})$
+\end_inset
+
+ implica
+\begin_inset Formula $\text{rk}(p(f)^{s})=\text{rk}(p(f)^{s+1})$
+\end_inset
+
+, y el recíproco se cumple porque entonces
+\begin_inset Formula $\dim\ker(p(f)^{s})=\dim\ker(p(f)^{s+1})$
+\end_inset
+
+ con
+\begin_inset Formula $p(f)^{s}\subseteq p(f)^{s+1}$
+\end_inset
+
+.
+ Pero sabemos que
+\begin_inset Formula $M(p)=\text{ann}_{M}(p^{n_{r}})=\ker(p(f)^{n_{r}})$
+\end_inset
+
+ siendo
+\begin_inset Formula $n_{r}=\min\{s\in\mathbb{N}\mid\text{ann}_{M}(p^{s})=\text{ann}_{M}(p^{s+1})\}=n$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+La multiplicidad de
+\begin_inset Formula $p$
+\end_inset
+
+ como factor irreducible de
+\begin_inset Formula $\varphi$
+\end_inset
+
+ es
+\begin_inset Formula $m\geq n$
+\end_inset
+
+ y cumple
+\begin_inset Formula $M(p)=\ker(p(f)^{m})$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sea
+\begin_inset Formula $\varphi\eqqcolon p^{m}G$
+\end_inset
+
+ con
+\begin_inset Formula $p\nmid G$
+\end_inset
+
+, la identidad de Bézout
+\begin_inset Formula $1=p^{m}R+GS$
+\end_inset
+
+ implica, evaluando en
+\begin_inset Formula $f$
+\end_inset
+
+ sobre un
+\begin_inset Formula $v\in V$
+\end_inset
+
+, que
+\begin_inset Formula
+\[
+v=p(f)^{m}(R(f)(v))+G(f)(S(f)(v))=R(f)(p(f)^{m}(v))+S(f)(G(f)(v)),
+\]
+
+\end_inset
+
+y por el teorema de Cayley-Hamilton,
+\begin_inset Formula $(p^{m}G)(f)=p^{m}(f)\circ G(f)=G(f)\circ p^{m}(f)=0$
+\end_inset
+
+ y entonces
+\begin_inset Formula $p(f)^{m}(R(f)(v))\in\ker(G(f))$
+\end_inset
+
+ y
+\begin_inset Formula $G(f)(S(f)(v))\in\ker(p(f)^{m})$
+\end_inset
+
+, luego
+\begin_inset Formula $V=\ker(p(f)^{m})+\ker(G(f))$
+\end_inset
+
+ y si
+\begin_inset Formula $v\in\ker(p(f)^{m})\cap\ker(G(f))$
+\end_inset
+
+ la igualdad anterior nos da
+\begin_inset Formula $v=0+0=0$
+\end_inset
+
+, con lo que la suma es directa y
+\begin_inset Formula $V=\text{ann}_{M}(p^{m})\oplus\text{ann}_{M}(G)$
+\end_inset
+
+, de donde
+\begin_inset Formula $M(p)=\text{ann}_{M}(p^{m})=\ker(p(f)^{m})$
+\end_inset
+
+ y, por la afirmación anterior,
+\begin_inset Formula $m\geq n$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sea
+\begin_inset Formula $V=V_{1}\oplus\dots\oplus V_{t}$
+\end_inset
+
+ con los
+\begin_inset Formula $V_{i}$
+\end_inset
+
+
+\begin_inset Formula $f$
+\end_inset
+
+-invariantes, el polinomio mínimo de
+\begin_inset Formula $f$
+\end_inset
+
+ es el mínimo común múltiplo de los polinomios mínimos de los
+\begin_inset Formula $f|_{V_{i}}:V_{i}\to V_{i}$
+\end_inset
+
+.
+
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean
+\begin_inset Formula $\hat{f}_{i}\coloneqq f|_{V_{i}}:V_{i}\to V_{i}$
+\end_inset
+
+,
+\begin_inset Formula $P$
+\end_inset
+
+ el polinomio mínimo de
+\begin_inset Formula $f$
+\end_inset
+
+ y
+\begin_inset Formula $Q_{i}$
+\end_inset
+
+ el de
+\begin_inset Formula $\hat{f}_{i}$
+\end_inset
+
+, como
+\begin_inset Formula $P(\hat{f}_{i})=P(f)|_{V_{i}}=0$
+\end_inset
+
+,
+\begin_inset Formula $Q_{i}\mid P$
+\end_inset
+
+, y si
+\begin_inset Formula $F\in K[X]$
+\end_inset
+
+ es tal que
+\begin_inset Formula $Q_{1},\dots,Q_{t}\mid F$
+\end_inset
+
+, para
+\begin_inset Formula $v\in V$
+\end_inset
+
+, sea
+\begin_inset Formula $v\eqqcolon v_{1}+\dots+v_{t}$
+\end_inset
+
+ con cada
+\begin_inset Formula $v_{i}\in V_{i}$
+\end_inset
+
+, entonces
+\begin_inset Formula $f(v)=f(v_{1})+\dots+f(v_{t})=\hat{f}_{1}(v_{1})+\dots+\hat{f}_{t}(v_{t})=0$
+\end_inset
+
+, luego
+\begin_inset Formula $F(f)=0$
+\end_inset
+
+ y
+\begin_inset Formula $P\mid F$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{exinfo}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+7.
+\end_layout
+
+\end_inset
+
+Si
+\begin_inset Formula $f$
+\end_inset
+
+ es nilpotente, su polinomio característico es
+\begin_inset Formula $X^{n}$
+\end_inset
+
+ con
+\begin_inset Formula $n\coloneqq\dim V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+8.
+\end_layout
+
+\end_inset
+
+Dados
+\begin_inset Formula $f,g\in\text{End}_{K}V$
+\end_inset
+
+, las matrices asociadas a
+\begin_inset Formula $f$
+\end_inset
+
+ y
+\begin_inset Formula $g$
+\end_inset
+
+ son semejantes si y solo si
+\begin_inset Formula $f$
+\end_inset
+
+ y
+\begin_inset Formula $g$
+\end_inset
+
+ tienen el mismo polinomio característico con factorización irreducible
+
+\begin_inset Formula $\varphi=p_{1}^{m_{1}}\cdots p_{k}^{m_{k}}$
+\end_inset
+
+ y
+\begin_inset Formula $\text{rk}(p_{i}(f)^{s})=\text{rk}(p_{i}(g)^{s})$
+\end_inset
+
+ para todo
+\begin_inset Formula $i$
+\end_inset
+
+ y
+\begin_inset Formula $s\in\mathbb{N}^{*}$
+\end_inset
+
+, si y sólo si tienen el mismo polinomio mínimo con factorización irreducible
+
+\begin_inset Formula $d=p_{1}^{n_{1}}\cdots p_{k}^{n_{k}}$
+\end_inset
+
+ y
+\begin_inset Formula $\text{rk}(p_{i}(f)^{s})=\text{rk}(p_{i}(g)^{s})$
+\end_inset
+
+ para todo
+\begin_inset Formula $i$
+\end_inset
+
+ y
+\begin_inset Formula $s\in\mathbb{N}^{*}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Que dos endomorfismos tengan el mismo polinomio característico y el mismo
+ polinomio mínimo no implica que sus matrices asociadas bajo alguna base
+ sean semejantes.
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{exinfo}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Formas canónicas
+\end_layout
+
+\begin_layout Standard
+Para
+\begin_inset Formula $F\in K[X]$
+\end_inset
+
+ mónico de grado
+\begin_inset Formula $n>0$
+\end_inset
+
+, llamamos
+\series bold
+matriz compañera
+\series default
+ de
+\begin_inset Formula $F$
+\end_inset
+
+ a
+\begin_inset Formula
+\[
+C(F)\coloneqq\begin{pmatrix} & & & -F_{0}\\
+1 & & & -F_{1}\\
+ & \ddots & & \vdots\\
+ & & 1 & -F_{n-1}
+\end{pmatrix}\in{\cal M}_{n}(K),
+\]
+
+\end_inset
+
+y para
+\begin_inset Formula $r>0$
+\end_inset
+
+ escribimos
+\begin_inset Formula
+\[
+C_{r}(F)=\begin{pmatrix}\boxed{C(F)} & \boxed{U}\\
+ & \ddots & \ddots\\
+ & & \ddots & \boxed{U}\\
+ & & & \boxed{C(F)}
+\end{pmatrix}\in{\cal M}_{rn}(K),
+\]
+
+\end_inset
+
+donde
+\begin_inset Formula
+\[
+U\coloneqq\begin{pmatrix} & & 1\\
+\\
+\\
+\end{pmatrix}\in{\cal M}_{n}(K).
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+El polinomio característico de un
+\begin_inset Formula $C_{r}(F)$
+\end_inset
+
+ es
+\begin_inset Formula $F^{r}$
+\end_inset
+
+.
+
+\series bold
+Demostración:
+\series default
+ Primero vemos que el de
+\begin_inset Formula $C(F)$
+\end_inset
+
+ es
+\begin_inset Formula $F$
+\end_inset
+
+.
+ Para
+\begin_inset Formula $n\coloneqq\text{gr}F=1$
+\end_inset
+
+,
+\begin_inset Formula $C(F)=(-F_{0})\in{\cal M}_{1}(K)$
+\end_inset
+
+ y
+\begin_inset Formula $\det(XI-C(F))=X+F_{0}=F$
+\end_inset
+
+.
+ Para
+\begin_inset Formula $n>1$
+\end_inset
+
+,
+\begin_inset Formula
+\begin{align*}
+\det(XI-C(F)) & =\begin{vmatrix}X & & & F_{0}\\
+-1 & \ddots & & \vdots\\
+ & \ddots & X & F_{n-2}\\
+ & & -1 & X+F_{n-1}
+\end{vmatrix}=\\
+ & =X\begin{vmatrix}X & & & F_{1}\\
+-1 & \ddots & & \vdots\\
+ & \ddots & X & F_{n-2}\\
+ & & -1 & X+F_{n-1}
+\end{vmatrix}+(-1)^{n+1}F_{0}\begin{vmatrix}-1 & X\\
+ & \ddots & \ddots\\
+ & & \ddots & X\\
+ & & & -1
+\end{vmatrix}=\\
+ & =X(F_{1}+XF_{2}+\dots+X^{n-2}F_{n-1}+X^{n-1}F_{n})+(-1)^{n+1}(-1)^{n-1}F_{0}=F,
+\end{align*}
+
+\end_inset
+
+donde para el primer sumando hemos usado la hipótesis de inducción.
+ Para
+\begin_inset Formula $C_{r}F$
+\end_inset
+
+, el caso
+\begin_inset Formula $r=1$
+\end_inset
+
+ está hecho, y para
+\begin_inset Formula $r>1$
+\end_inset
+
+,
+\begin_inset Formula
+\[
+\det(XI-C_{r}(F))=\begin{vmatrix}\boxed{C(F)} & \boxed{U}\\
+ & \ddots & \ddots\\
+ & & \ddots & \boxed{U}\\
+ & & & \boxed{C(F)}
+\end{vmatrix}=\det(C(F))\det(C_{r-1}(F))=FF^{r-1}=F^{r}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{exinfo}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sean
+\begin_inset Formula $p\in K[X]$
+\end_inset
+
+ un divisor irreducible del polinomio característico de
+\begin_inset Formula $f$
+\end_inset
+
+,
+\begin_inset Formula $h\in\mathbb{N}^{*}$
+\end_inset
+
+ y
+\begin_inset Formula $\{v_{1},\dots,v_{t}\}\subseteq\ker(p(f)^{h})$
+\end_inset
+
+,
+\begin_inset Formula $(\overline{v_{1}},\dots,\overline{v_{t}})$
+\end_inset
+
+ es base de
+\begin_inset Formula $\frac{\ker(p(f)^{h})}{\ker(p(f)^{h-1})}$
+\end_inset
+
+ como
+\begin_inset Formula $\frac{K[X]}{(p)}$
+\end_inset
+
+-espacio vectorial si y sólo si
+\begin_inset Formula $\left(\overline{f^{i}(v_{j})}\right)_{0\leq i<d}^{1\leq j\leq t}$
+\end_inset
+
+ es base de
+\begin_inset Formula $\frac{\ker(p(f)^{h})}{\ker(p(f)^{h-1})}$
+\end_inset
+
+ como
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial.
+ En particular, si
+\begin_inset Formula $p\in K[X]$
+\end_inset
+
+ es mónico irreducible con
+\begin_inset Formula $p(f)=0$
+\end_inset
+
+ y
+\begin_inset Formula $\{v_{1},\dots,v_{t}\}\subseteq V$
+\end_inset
+
+,
+\begin_inset Formula $(v_{1},\dots,v_{t})$
+\end_inset
+
+ es base de
+\begin_inset Formula $M$
+\end_inset
+
+ como
+\begin_inset Formula $\frac{K[X]}{(p)}$
+\end_inset
+
+-espacio vectorial si y sólo si
+\begin_inset Formula $(f^{i}(v_{j}))_{0\leq i<d}^{1\leq j\leq t}$
+\end_inset
+
+ es base del
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{exinfo}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sean
+\begin_inset Formula $F\in K[X]$
+\end_inset
+
+ un polinomio mónico de grado
+\begin_inset Formula $n>0$
+\end_inset
+
+ y
+\begin_inset Formula $r\in\mathbb{N}^{*}$
+\end_inset
+
+,
+\begin_inset Formula $M\cong\frac{K[X]}{(F^{r})}$
+\end_inset
+
+ si y sólo si existe
+\begin_inset Formula $v\in V$
+\end_inset
+
+ tal que
+\begin_inset Formula $(f^{s}(v))_{s=0}^{rn-1}$
+\end_inset
+
+ es base de
+\begin_inset Formula $v$
+\end_inset
+
+ y
+\begin_inset Formula $F(f)^{r}(v)=0$
+\end_inset
+
+, si y sólo si existe una base
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ de
+\begin_inset Formula $V$
+\end_inset
+
+ con
+\begin_inset Formula $M_{{\cal B}}(f)=C_{r}(F)$
+\end_inset
+
+, en cuyo caso el polinomio mínimo de
+\begin_inset Formula $M$
+\end_inset
+
+ coincide con el polinomio característico de
+\begin_inset Formula $f$
+\end_inset
+
+ y es
+\begin_inset Formula $F^{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\implies3]$
+\end_inset
+
+ Sea
+\begin_inset Formula ${\cal \tilde{B}}_{j}\coloneqq(\overline{F^{j}},\overline{XF^{j}},\dots,\overline{X^{n-1}F^{j}})$
+\end_inset
+
+ para
+\begin_inset Formula $j\in\{0,\dots,r-1\}$
+\end_inset
+
+ y
+\begin_inset Quotes cld
+\end_inset
+
+
+\begin_inset Formula $\star$
+\end_inset
+
+
+\begin_inset Quotes crd
+\end_inset
+
+ la concatenación de secuencias,
+\begin_inset Formula $\tilde{{\cal B}}\coloneqq\tilde{{\cal B}}_{r-1}\star\dots\star\tilde{{\cal B}}_{1}\star\tilde{{\cal B}}_{0}$
+\end_inset
+
+ es base de
+\begin_inset Formula $\frac{K[X]}{(F^{r})}$
+\end_inset
+
+ como
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial.
+ Para verlo, como
+\begin_inset Formula $|\tilde{{\cal B}}|=rn=\dim\frac{K[X]}{(F^{r})}$
+\end_inset
+
+, basta ver que
+\begin_inset Formula $\tilde{{\cal B}}$
+\end_inset
+
+ es linealmente independiente.
+ Si
+\begin_inset Formula $r=1$
+\end_inset
+
+,
+\begin_inset Formula $\tilde{{\cal B}}=(\overline{1},\overline{X},\dots,\overline{X}^{n-1})$
+\end_inset
+
+ y el resultado es claro.
+ Si
+\begin_inset Formula $r>1$
+\end_inset
+
+, sea
+\begin_inset Formula $\sum_{i=0}^{n-1}\sum_{j=0}^{r-1}\lambda_{ij}X^{i}F^{j}=0\in\frac{K[X]}{(F^{r})}$
+\end_inset
+
+ para ciertos
+\begin_inset Formula $\lambda_{ij}\in K$
+\end_inset
+
+, entonces
+\begin_inset Formula $\sum_{ij}\lambda_{ij}X^{i}F^{j}=F^{r}G\in K[X]$
+\end_inset
+
+ para cierto
+\begin_inset Formula $G\in K[X]$
+\end_inset
+
+, pero
+\begin_inset Formula $\sum_{ij}\lambda_{ij}X^{i}F^{j}=\sum_{i=0}^{n-1}\lambda_{i0}X^{i}+F(\sum_{i=0}^{n-1}\sum_{j=1}^{r-1}\lambda_{ij}X^{i}F^{j})$
+\end_inset
+
+, luego debe ser
+\begin_inset Formula $F\mid\sum_{i=0}^{n-1}\lambda_{i0}X^{i}$
+\end_inset
+
+ y, como
+\begin_inset Formula $\text{gr}F=n$
+\end_inset
+
+,
+\begin_inset Formula $\sum_{i=0}^{n-1}\lambda_{i0}X^{i}=0$
+\end_inset
+
+ y cada
+\begin_inset Formula $\lambda_{i0}=0$
+\end_inset
+
+.
+ Pero entonces, dividiendo por
+\begin_inset Formula $F$
+\end_inset
+
+,
+\begin_inset Formula $\sum_{i=0}^{n-1}\sum_{j=1}^{r-1}\lambda_{ij}X^{i}F^{j-1}=F^{r-1}G$
+\end_inset
+
+ y por hipótesis de inducción todos los
+\begin_inset Formula $\lambda_{ij}=0$
+\end_inset
+
+.
+ Sea
+\begin_inset Formula $g:\frac{K[X]}{(F^{r})}\to\frac{K[X]}{(F^{r})}$
+\end_inset
+
+ el endomorfismo
+\begin_inset Formula $G\mapsto XG$
+\end_inset
+
+, queremos ver que
+\begin_inset Formula $C\coloneqq M_{{\cal B}}(g)=C_{r}(F)$
+\end_inset
+
+.
+ Para
+\begin_inset Formula $j\in\{0,\dots,r-1\}$
+\end_inset
+
+,
+\begin_inset Formula $g(\tilde{{\cal B}}_{j})=(\overline{XF^{j}},\overline{X^{2}F^{j}},\dots,\overline{X^{n}F^{j}})$
+\end_inset
+
+, pero
+\begin_inset Formula
+\[
+\overline{F^{j+1}}-\overline{X^{n}F^{j}}=\overline{(F-X^{n})F^{j}}=\left(\sum_{i=0}^{n-1}F_{i}\overline{X^{i}}\right)\overline{F^{j}}=\sum_{i=0}^{n-1}F_{i}\overline{X^{i}F^{j}}
+\]
+
+\end_inset
+
+y por tanto
+\begin_inset Formula
+\[
+\overline{X^{n}F^{j}}=\overline{F^{j+1}}-\sum_{i=0}^{n-1}F_{i}\overline{X^{i}F^{j}}.
+\]
+
+\end_inset
+
+Entonces, para
+\begin_inset Formula $j=r-1$
+\end_inset
+
+,
+\begin_inset Formula $\overline{F^{r+1}}=0$
+\end_inset
+
+ y las primeras
+\begin_inset Formula $n$
+\end_inset
+
+ columnas de
+\begin_inset Formula $C$
+\end_inset
+
+ solo tienen entradas no nulas en las primeras
+\begin_inset Formula $n$
+\end_inset
+
+ filas y estas entradas son
+\begin_inset Formula
+\[
+\begin{pmatrix} & & & -F_{0}\\
+1 & & & -F_{1}\\
+ & \ddots & & \vdots\\
+ & & 1 & -F_{n-1}
+\end{pmatrix}=C(F),
+\]
+
+\end_inset
+
+mientras que para
+\begin_inset Formula $j<r-1$
+\end_inset
+
+,
+\begin_inset Formula $\overline{F^{j+1}}$
+\end_inset
+
+ es un elemento de la base y las columnas de
+\begin_inset Formula $C$
+\end_inset
+
+ correspondientes a
+\begin_inset Formula $\tilde{{\cal B}}_{j}$
+\end_inset
+
+ solo tienen entradas no nulas en las filas de
+\begin_inset Formula $\tilde{{\cal B}}_{j}$
+\end_inset
+
+, formando la submatriz
+\begin_inset Formula $C(F)$
+\end_inset
+
+, y en la columna de
+\begin_inset Formula $\overline{X^{n-1}F^{j}}$
+\end_inset
+
+ con la fila de
+\begin_inset Formula $\overline{F^{j+1}}$
+\end_inset
+
+, dando la submatriz
+\begin_inset Formula $U$
+\end_inset
+
+ de la definición de
+\begin_inset Formula $C_{r}(F)$
+\end_inset
+
+.
+ Finalmente, el
+\begin_inset Formula $K[X]$
+\end_inset
+
+-módulo generado por
+\begin_inset Formula $(\frac{K[X]}{(F^{r})},g)$
+\end_inset
+
+ es claramente
+\begin_inset Formula $\frac{K[X]}{(F^{r})}$
+\end_inset
+
+, y si
+\begin_inset Formula $\phi:M\to\frac{K[X]}{(F^{r})}$
+\end_inset
+
+ es el isomorfismo de la hipótesis, como
+\begin_inset Formula $\phi(f(v))=\phi(Xv)=X\phi(v)=g(\phi(v))$
+\end_inset
+
+, tomando la base
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ de
+\begin_inset Formula $V$
+\end_inset
+
+ inducida por
+\begin_inset Formula $\tilde{{\cal B}}$
+\end_inset
+
+ mediante
+\begin_inset Formula $\phi^{-1}$
+\end_inset
+
+ queda
+\begin_inset Formula $M_{{\cal B}}(f)=M_{\tilde{{\cal B}}}(g)=C_{r}(F)$
+\end_inset
+
+, y el polinomio característico de
+\begin_inset Formula $f$
+\end_inset
+
+ es el de
+\begin_inset Formula $C_{r}(F)$
+\end_inset
+
+ que es
+\begin_inset Formula $F^{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $3\implies1]$
+\end_inset
+
+ Tomando
+\begin_inset Formula $g$
+\end_inset
+
+ y
+\begin_inset Formula $\tilde{{\cal B}}$
+\end_inset
+
+ de la parte anterior de la prueba,
+\begin_inset Formula $M_{{\cal B}}(f)=C_{r}(f)=M_{\tilde{B}}(g)$
+\end_inset
+
+ y, como esto también significa que
+\begin_inset Formula $\dim V=\dim\frac{K[X]}{(F^{r})}$
+\end_inset
+
+, queda el isomorfismo
+\begin_inset Formula $M\to\frac{K[X]}{(F^{r})}$
+\end_inset
+
+ deseado, y como
+\begin_inset Formula $\text{ann}_{K[X]}(M)=\text{ann}_{K[X]}\frac{K[X]}{(F^{r})}=(F^{r})$
+\end_inset
+
+ y
+\begin_inset Formula $F^{r}$
+\end_inset
+
+ es mónico,
+\begin_inset Formula $F^{r}$
+\end_inset
+
+ es el polinomio mínimo de
+\begin_inset Formula $M$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Sea
+\begin_inset Formula $\phi:\frac{K[X]}{(F^{r})}\to M$
+\end_inset
+
+ un
+\begin_inset Formula $K[X]$
+\end_inset
+
+-isomorfismo, que induce un
+\begin_inset Formula $K$
+\end_inset
+
+-isomorfismo
+\begin_inset Formula $\phi:\frac{K[X]}{(F^{r})}\to V$
+\end_inset
+
+, como
+\begin_inset Formula $(\overline{1},\overline{X},\dots,\overline{X}^{rn-1})$
+\end_inset
+
+ es base de
+\begin_inset Formula $\frac{K[X]}{(F^{r})}$
+\end_inset
+
+, tomando
+\begin_inset Formula $v\coloneqq\phi(\overline{1})$
+\end_inset
+
+,
+\begin_inset Formula $(\overline{v},\overline{f(v)},\dots,\overline{f^{rn-1}(v)})$
+\end_inset
+
+ es base de
+\begin_inset Formula $V$
+\end_inset
+
+ y
+\begin_inset Formula $F(f)^{r}(v)=F^{r}(f)(v)=\overline{F^{r}}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\implies1]$
+\end_inset
+
+ Para
+\begin_inset Formula $w\in M=V$
+\end_inset
+
+, existen
+\begin_inset Formula $b_{s}\in K$
+\end_inset
+
+ con
+\begin_inset Formula $w=\sum_{s=0}^{rn-1}b_{s}f^{s}(v)=(\sum_{s=0}^{rn-1}b_{s}X^{s})v$
+\end_inset
+
+, luego
+\begin_inset Formula $M=(v)$
+\end_inset
+
+ y
+\begin_inset Formula $\pi:K[X]\twoheadrightarrow M$
+\end_inset
+
+ dada por
+\begin_inset Formula $\pi(G)\coloneqq Gv$
+\end_inset
+
+ es un epimorfismo, pero
+\begin_inset Formula $F^{r}\in\ker\pi$
+\end_inset
+
+, por lo que
+\begin_inset Formula $\pi$
+\end_inset
+
+ induce un epimorfismo
+\begin_inset Formula $\hat{\pi}:\frac{K[X]}{(F^{r})}\twoheadrightarrow M$
+\end_inset
+
+, y como
+\begin_inset Formula $\dim_{K}\frac{K[X]}{(F^{r})}=rn=\dim_{K}M$
+\end_inset
+
+,
+\begin_inset Formula $\hat{\pi}$
+\end_inset
+
+ es un isomorfismo.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de clasificación de endomorfismos:
+\series default
+ Existen una base
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ de
+\begin_inset Formula $V$
+\end_inset
+
+,
+\begin_inset Formula $h_{1},\dots,h_{t}\in\mathbb{N}^{*}$
+\end_inset
+
+ y
+\begin_inset Formula $p_{1},\dots,p_{t}\in K[X]$
+\end_inset
+
+ irreducibles tales que
+\begin_inset Formula
+\[
+M_{{\cal B}}(f)=\begin{pmatrix}\boxed{C_{h_{1}}(p_{1})}\\
+ & \ddots\\
+ & & \boxed{C_{h_{t}}(p_{t})}
+\end{pmatrix},
+\]
+
+\end_inset
+
+siendo esta matriz, llamada
+\series bold
+forma canónica
+\series default
+ de
+\begin_inset Formula $f$
+\end_inset
+
+, unívocamente determinada por
+\begin_inset Formula $f$
+\end_inset
+
+ salvo reordenación de bloques y formada, exactamente, por
+\begin_inset Formula
+\[
+\frac{\text{rk}(p(f)^{h-1})+\text{rk}(p(f)^{h+1})-2\text{rk}(p(f)^{h})}{\text{rg}p}
+\]
+
+\end_inset
+
+bloques
+\begin_inset Formula $C_{h}(p)$
+\end_inset
+
+ para cada divisor irreducible mónico
+\begin_inset Formula $p$
+\end_inset
+
+ del polinomio característico de
+\begin_inset Formula $f$
+\end_inset
+
+ y cada
+\begin_inset Formula $h\leq\min\{s\in\mathbb{N}^{*}\mid\text{rk}(p(f)^{s})=\text{rk}(p(f)^{s+1})\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Sea
+\begin_inset Formula $M=\bigoplus_{i=1}^{k}\bigoplus_{j=1}^{r_{i}}N_{ij}$
+\end_inset
+
+ una descomposición canónica con cada
+\begin_inset Formula $N_{ij}\cong\frac{K[X]}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+, cada
+\begin_inset Formula $N_{ij}$
+\end_inset
+
+ es un subespacio
+\begin_inset Formula $f$
+\end_inset
+
+-invariante de
+\begin_inset Formula $V$
+\end_inset
+
+, por lo que existe una base
+\begin_inset Formula ${\cal B}_{ij}$
+\end_inset
+
+ de
+\begin_inset Formula $N_{ij}$
+\end_inset
+
+ como
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial con
+\begin_inset Formula $M_{{\cal B}_{ij}}(f|_{N_{ij}})=C_{n_{ij}}(p_{i})$
+\end_inset
+
+, y uniendo las bases se obtiene una base
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ con
+\begin_inset Formula $M_{{\cal B}}(f)$
+\end_inset
+
+ de la forma buscada.
+\end_layout
+
+\begin_layout Standard
+Si ahora
+\begin_inset Formula ${\cal B}'$
+\end_inset
+
+ es otra base tal que
+\begin_inset Formula $M_{{\cal B}}(f)$
+\end_inset
+
+ está formada por bloques diagonales
+\begin_inset Formula $(C_{h_{s}}(q_{s}))_{s=1}^{u}$
+\end_inset
+
+,
+\begin_inset Formula $V$
+\end_inset
+
+ se puede descomponer en suma directa interna de subespacios
+\begin_inset Formula $f$
+\end_inset
+
+-invariantes
+\begin_inset Formula $W_{s}$
+\end_inset
+
+ con bases
+\begin_inset Formula ${\cal B}_{s}$
+\end_inset
+
+ tales que, si
+\begin_inset Formula $\hat{f}_{s}\coloneqq f|_{W_{s}}:W_{s}\to W_{S}$
+\end_inset
+
+,
+\begin_inset Formula $M_{{\cal B}_{s}}(\hat{f}_{s})=C_{h_{s}}(q_{s})$
+\end_inset
+
+, con lo que el módulo generado por
+\begin_inset Formula $(W_{s},\hat{f}_{s})$
+\end_inset
+
+ es un submódulo no nulo de
+\begin_inset Formula $M$
+\end_inset
+
+ isomorfo a
+\begin_inset Formula $\frac{K[X]}{(q_{s}^{h_{s}})}$
+\end_inset
+
+, de modo que
+\begin_inset Formula $M=\bigoplus_{s=1}^{u}\frac{K[X]}{(q_{s}^{h_{s}})}$
+\end_inset
+
+ y, como las descomposiciones de esta forma son únicas, los bloques son
+ los mismos que en la descomposición que hemos encontrado y los irreducibles
+ que aparecen son los divisores irreducibles de
+\begin_inset Formula $f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Para la última parte, otra forma de obtener la forma canónica de cada
+\begin_inset Formula $M(p)$
+\end_inset
+
+ es usando los
+\begin_inset Formula $(F_{h})_{h=1}^{n}$
+\end_inset
+
+ con
+\begin_inset Formula $n\coloneqq\max_{i}r_{i}=\min\{s\in\mathbb{N}^{*}\mid\text{ann}_{M(p)}(p^{s})=\text{ann}_{M(p)}(p^{s+1})\}$
+\end_inset
+
+, cada
+\begin_inset Formula $F_{h}\subseteq\text{ann}_{M}(p^{h})$
+\end_inset
+
+ y tales que cada
+\begin_inset Formula $F_{h}\dot{\cup}pF_{h+1}\dot{\cup}\dots\dot{\cup}p^{n-h}F_{n}$
+\end_inset
+
+ induce una base de
+\begin_inset Formula $\frac{\text{ann}_{M}(p^{h})}{\text{ann}_{M}(p^{h-1})}=\frac{\ker(p(f)^{h})}{\ker(p(f)^{h-1})}$
+\end_inset
+
+ como
+\begin_inset Formula $\frac{K[X]}{(p)}$
+\end_inset
+
+-espacio vectorial.
+ Si
+\begin_inset Formula $\hat{f}\coloneqq f|_{M(p)}:M(p)\to M(p)$
+\end_inset
+
+,
+\begin_inset Formula $\text{ann}_{M(p)}(p^{s})=\ker(p(\hat{f})^{s})=\ker(p(f)^{s})$
+\end_inset
+
+ ya que
+\begin_inset Formula $p(f)^{s}(v)=0\implies p^{s}v=0\implies v\in\text{ann}_{M}(p^{s})\subseteq M(p)$
+\end_inset
+
+, de modo que
+\begin_inset Formula $n=\min\{s\in\mathbb{N}^{*}\mid\text{rk}(p(f)^{s})=\text{rk}(p(f)^{s+1})\}$
+\end_inset
+
+.
+ Además, el número de apariciones de
+\begin_inset Formula $p^{s}$
+\end_inset
+
+ como divisor elemental de
+\begin_inset Formula $M$
+\end_inset
+
+ es
+\begin_inset Formula $\mu_{h}=\delta_{h}-\delta_{h+1}\coloneqq\dim_{\frac{K[X]}{(p)}}\frac{\ker(p(f)^{h})}{\ker(p(f)^{h-1})}-\dim_{\frac{K[X]}{(p)}}\frac{\ker(p(f)^{h+1})}{\ker(p(f)^{h})}$
+\end_inset
+
+, pero es fácil ver que todo
+\begin_inset Formula $\frac{K[X]}{(p)}$
+\end_inset
+
+-espacio vectorial
+\begin_inset Formula $U$
+\end_inset
+
+ es un
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial y
+\begin_inset Formula $\dim_{\frac{K[X]}{(p)}}(U)=\frac{\dim_{K}(U)}{\text{gr}p}$
+\end_inset
+
+, luego
+\begin_inset Formula $\mu_{h}=\frac{1}{\text{gr}p}(\dim_{K}\ker(p(f)^{h})-\dim_{K}\ker(p(f)^{h-1})-\dim_{K}\ker(p(f)^{h+1})+\dim_{K}\ker(p(f)^{h}))$
+\end_inset
+
+ y el resultado sale de que
+\begin_inset Formula $\dim_{K}\ker(p(f)^{h})=\dim_{K}V-\text{rk}(p(f)^{h})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Como
+\series bold
+teorema
+\series default
+, toda
+\begin_inset Formula $C\in{\cal M}_{n}(K)$
+\end_inset
+
+ es semejante a una de la forma
+\begin_inset Formula
+\[
+\begin{pmatrix}\boxed{C_{h_{1}}(p_{1})}\\
+ & \ddots\\
+ & & \boxed{C_{h_{t}}(p_{t})}
+\end{pmatrix}
+\]
+
+\end_inset
+
+con los
+\begin_inset Formula $p_{i}\in K[X]$
+\end_inset
+
+ irreducibles, siendo esta matriz, llamada
+\series bold
+forma canónica
+\series default
+ de
+\begin_inset Formula $C$
+\end_inset
+
+, unívocamente determinada por
+\begin_inset Formula $C$
+\end_inset
+
+ salvo reordenación de bloques y formada, exactamente, por
+\begin_inset Formula
+\[
+\frac{\text{rk}(p(C)^{h-1})+\text{rk}(p(C)^{h+1})-2\text{rk}(p(C)^{h})}{\text{rg}p}
+\]
+
+\end_inset
+
+bloques
+\begin_inset Formula $C_{h}(p)$
+\end_inset
+
+ para cada divisor irreducible mónico
+\begin_inset Formula $p$
+\end_inset
+
+ del polinomio característico de
+\begin_inset Formula $p$
+\end_inset
+
+ y cada
+\begin_inset Formula $h\leq\min\{s\in\mathbb{N}^{*}\mid\text{rk}(p(f)^{s})=\text{rk}(p(f)^{s+1})\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{exinfo}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si
+\begin_inset Formula $F\in K[X]$
+\end_inset
+
+ es no constante con factorización irreducible
+\begin_inset Formula $F=p_{1}^{m_{1}}\cdots p_{k}^{m_{k}}$
+\end_inset
+
+ con los
+\begin_inset Formula $p_{i}$
+\end_inset
+
+ mónicos irreducibles distintos, la forma canónica de la matriz compañera
+
+\begin_inset Formula $C$
+\end_inset
+
+ de
+\begin_inset Formula $F$
+\end_inset
+
+ es
+\begin_inset Formula
+\[
+\begin{pmatrix}\boxed{C_{m_{1}}(p_{1})}\\
+ & \ddots\\
+ & & \boxed{C_{m_{k}}(p_{k})}
+\end{pmatrix},
+\]
+
+\end_inset
+
+y en particular
+\begin_inset Formula $C$
+\end_inset
+
+ tiene un único divisor elemental asociado a cada divisor mónico irreducible
+ de
+\begin_inset Formula $F$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{exinfo}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Formas de Jordan
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{exinfo}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Un
+\series bold
+valor propio
+\series default
+ de
+\begin_inset Formula $f$
+\end_inset
+
+ es un
+\begin_inset Formula $\lambda\in K$
+\end_inset
+
+ tal que
+\begin_inset Formula $X-\lambda$
+\end_inset
+
+ divide al polinomio característico de
+\begin_inset Formula $f$
+\end_inset
+
+, y su
+\series bold
+multiplicidad geométrica
+\series default
+ es
+\begin_inset Formula $\nu_{\text{g}}(\lambda)\coloneqq\dim_{K}\ker(f-\lambda1_{V})>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{exinfo}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Para
+\begin_inset Formula $\lambda\in K$
+\end_inset
+
+,
+\begin_inset Formula $C(X-\lambda)=(\lambda)\in{\cal M}_{1}(K)$
+\end_inset
+
+ y, para
+\begin_inset Formula $r>0$
+\end_inset
+
+, llamamos
+\series bold
+bloque de Jordan
+\series default
+ de tamaño
+\begin_inset Formula $r$
+\end_inset
+
+ asociado al valor propio
+\begin_inset Formula $\lambda$
+\end_inset
+
+ a
+\begin_inset Formula $J_{r}(\lambda)\coloneqq C_{r}(X-\lambda)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{samepage}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de Jordan:
+\end_layout
+
+\begin_layout Enumerate
+Si el polinomio característico de
+\begin_inset Formula $f$
+\end_inset
+
+ se descompone completamente en
+\begin_inset Formula $K[X]$
+\end_inset
+
+, existe una base
+\begin_inset Formula ${\cal B}$
+\end_inset
+
+ de
+\begin_inset Formula $V$
+\end_inset
+
+ tal que
+\begin_inset Formula
+\[
+M_{{\cal B}}(f)=\begin{pmatrix}\boxed{J_{h_{1}}(\lambda_{1})}\\
+ & \ddots\\
+ & & \boxed{J_{h_{t}}(\lambda_{t})}
+\end{pmatrix}
+\]
+
+\end_inset
+
+para ciertos
+\begin_inset Formula $h_{i}>0$
+\end_inset
+
+ y
+\begin_inset Formula $\lambda_{i}\in K$
+\end_inset
+
+, siendo esta matriz unívocamente determinada por
+\begin_inset Formula $f$
+\end_inset
+
+ salvo reordenación de bloques y formada por
+\begin_inset Formula $\text{rk}((f-\lambda1_{V})^{h-1})+\text{rk}((f-\lambda1_{V})^{h+1})-2\text{rk}((f-\lambda1_{V})^{h})$
+\end_inset
+
+ bloques
+\begin_inset Formula $J_{h}(\lambda)$
+\end_inset
+
+ para cada valor propio
+\begin_inset Formula $\lambda$
+\end_inset
+
+ de
+\begin_inset Formula $f$
+\end_inset
+
+ y cada
+\begin_inset Formula $h\in\mathbb{N}^{*}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Por el teorema de clasificación de endomorfismos usando que los irreducibles
+ del polinomio característico son los
+\begin_inset Formula $X-\lambda$
+\end_inset
+
+ con
+\begin_inset Formula $\lambda$
+\end_inset
+
+ valor propio de
+\begin_inset Formula $f$
+\end_inset
+
+ y que el grado de estos es 1.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si
+\begin_inset Formula $C\in{\cal M}_{n}(K)$
+\end_inset
+
+ es una matriz cuadrada cuyo polinomio característico se descompone completament
+e en
+\begin_inset Formula $K[X]$
+\end_inset
+
+,
+\begin_inset Formula $C$
+\end_inset
+
+ es semejante a una matriz como la del apartado anterior, única salvo reordenaci
+ón de bloques y formada por
+\begin_inset Formula $\text{rk}((C-\lambda I)^{h-1})+\text{rk}((C-\lambda I)^{h+1})-2\text{rk}((C-\lambda I)^{h})$
+\end_inset
+
+ bloques
+\begin_inset Formula $J_{h}(\lambda)$
+\end_inset
+
+ para cada valor propio
+\begin_inset Formula $\lambda$
+\end_inset
+
+ de
+\begin_inset Formula $C$
+\end_inset
+
+ y cada
+\begin_inset Formula $h\in\mathbb{N}^{*}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{samepage}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sean
+\begin_inset Formula $\varphi$
+\end_inset
+
+ el polinomio característico de
+\begin_inset Formula $f$
+\end_inset
+
+ y
+\begin_inset Formula $p$
+\end_inset
+
+ un divisor mónico irreducible de grado
+\begin_inset Formula $d$
+\end_inset
+
+ y multiplicidad 1:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $M(p)=\ker(p(f))\cong\frac{K[X]}{(p)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Claramente
+\begin_inset Formula $\ker(p(f))\subseteq M(p)$
+\end_inset
+
+, y si
+\begin_inset Formula $x\in M(p)$
+\end_inset
+
+, existe
+\begin_inset Formula $s>0$
+\end_inset
+
+ con
+\begin_inset Formula $p^{s}x=0$
+\end_inset
+
+ y
+\begin_inset Formula $x\in\ker(p(f)^{s})$
+\end_inset
+
+, pero como la multiplicidad de
+\begin_inset Formula $p$
+\end_inset
+
+ en
+\begin_inset Formula $\varphi$
+\end_inset
+
+ es 1,
+\begin_inset Formula $\ker(p(f))=\ker(p(f)^{s})$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Para todo
+\begin_inset Formula $v\in M(p)\setminus\{0\}$
+\end_inset
+
+,
+\begin_inset Formula ${\cal B}\coloneqq\{f^{s}(v)\}_{s\in\mathbb{N}_{d}}$
+\end_inset
+
+ es una base de
+\begin_inset Formula $\ker(p(f))$
+\end_inset
+
+ y
+\begin_inset Formula $M_{{\cal B}}(f|_{M(p)}:M(p)\to M(p))=C_{1}(p)$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean
+\begin_inset Formula $\phi_{0}:\frac{K[X]}{(p)}\to M(p)$
+\end_inset
+
+ un isomorfismo,
+\begin_inset Formula $\overline{q}\coloneqq(\phi_{0})^{-1}(v)\neq0$
+\end_inset
+
+ y
+\begin_inset Formula $\pi:\frac{K[X]}{(p)}\twoheadrightarrow\frac{K[X]}{(p)}$
+\end_inset
+
+ el epimorfismo
+\begin_inset Formula $\pi(\overline{F})\coloneqq\overline{qF}$
+\end_inset
+
+, como
+\begin_inset Formula $\gcd\{p,q\}=1$
+\end_inset
+
+, existe una identidad de Bézout
+\begin_inset Formula $1=pR+qS$
+\end_inset
+
+, luego
+\begin_inset Formula $\overline{1}=\overline{qS}\in\text{Im}\pi$
+\end_inset
+
+ y
+\begin_inset Formula $\pi$
+\end_inset
+
+ es un isomorfismo.
+ Por tanto
+\begin_inset Formula $\phi\coloneqq\phi_{0}\circ\pi L\frac{K[X]}{(p)}\to M(p)$
+\end_inset
+
+ es un isomorfismo con
+\begin_inset Formula $\phi(\overline{1})=v$
+\end_inset
+
+ y, como
+\begin_inset Formula $(X^{s})_{s\in\mathbb{N}_{d}}$
+\end_inset
+
+ es base de
+\begin_inset Formula $\frac{K[X]}{(p)}$
+\end_inset
+
+ como
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial,
+\begin_inset Formula ${\cal B}\coloneqq(f^{s}(v))_{s\in\mathbb{N}_{d}}$
+\end_inset
+
+ es base de
+\begin_inset Formula $M(p)$
+\end_inset
+
+ como
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial.
+ Ahora bien, si
+\begin_inset Formula $b_{i}\coloneqq f^{i}(v)$
+\end_inset
+
+, para
+\begin_inset Formula $i\in\{0,\dots,d-2\}$
+\end_inset
+
+,
+\begin_inset Formula $f(b_{i})=f(f^{i}(v))=f^{i+1}(v)=b_{i+1}$
+\end_inset
+
+, y para
+\begin_inset Formula $d-1$
+\end_inset
+
+,
+\begin_inset Formula
+\[
+f(b_{d-1})=f^{d}(v)=\phi(X^{d})=\phi(X^{d}-p)=\phi\left(-\sum_{i=0}^{d-1}p_{i}X^{i}\right)=\sum_{i=0}^{d-1}-p_{i}b_{i},
+\]
+
+\end_inset
+
+lo que nos da
+\begin_inset Formula $M_{{\cal B}}(f)=C(p)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Análogamente, si
+\begin_inset Formula $C\in{\cal M}_{n}(K)$
+\end_inset
+
+ y
+\begin_inset Formula $p\in K[X]$
+\end_inset
+
+ es un irreducible con multiplicidad 1 en el polinomio característico de
+
+\begin_inset Formula $C$
+\end_inset
+
+, la forma canónica de
+\begin_inset Formula $C$
+\end_inset
+
+ tiene exactamente un bloque de la forma
+\begin_inset Formula $C_{h}(p)$
+\end_inset
+
+ que es precisamente
+\begin_inset Formula $C(p)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{samepage}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Un
+\begin_inset Formula $\lambda\in\mathbb{R}$
+\end_inset
+
+ es un
+\series bold
+valor propio simple
+\series default
+ de
+\begin_inset Formula $f$
+\end_inset
+
+ o de
+\begin_inset Formula $C\in{\cal M}_{n}(K)$
+\end_inset
+
+ si
+\begin_inset Formula $X-\lambda$
+\end_inset
+
+ es divisor de su polinomio característico con multiplicidad 1, en cuyo
+ caso:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $M(X-\lambda)=\ker((X-\lambda)(f))=\{v\in V\mid f(v)=\lambda v\}\cong\frac{K[X]}{(X-\lambda)}$
+\end_inset
+
+ es el subespacio propio de
+\begin_inset Formula $V$
+\end_inset
+
+ asociado al valor propio
+\begin_inset Formula $\lambda$
+\end_inset
+
+ de
+\begin_inset Formula $f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Para todo
+\begin_inset Formula $v\in M(X-\lambda)\setminus\{0\}$
+\end_inset
+
+,
+\begin_inset Formula $M(X-\lambda)=(v)$
+\end_inset
+
+ y
+\begin_inset Formula $f|_{(v)}$
+\end_inset
+
+ es el producto por
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+La forma canónica de
+\begin_inset Formula $C$
+\end_inset
+
+ tiene un único bloque de la forma
+\begin_inset Formula $J_{h}(\lambda)$
+\end_inset
+
+, que es
+\begin_inset Formula $J(\lambda)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{samepage}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Anillos de polinomios y matrices
+\end_layout
+
+\begin_layout Standard
+Si
+\begin_inset Formula $B\in\text{GL}_{s}(K)$
+\end_inset
+
+ y
+\begin_inset Formula
+\[
+C\coloneqq\begin{pmatrix} & \boxed{B} & \boxed{I_{s}}\\
+ & & \ddots & \ddots\\
+ & & & \ddots & \boxed{I_{s}}\\
+ & & & & \boxed{B}\\
+\\
+\end{pmatrix}\in{\cal M}_{rs}(K),
+\]
+
+\end_inset
+
+para
+\begin_inset Formula $k\in\{1,\dots,r-1\}$
+\end_inset
+
+, viendo
+\begin_inset Formula $C^{k}$
+\end_inset
+
+ por bloques como elemento de
+\begin_inset Formula ${\cal M}_{r}({\cal M}_{s}(K))$
+\end_inset
+
+, su
+\begin_inset Formula $k$
+\end_inset
+
+-ésima diagonal por encima de la principal está formada por copias de
+\begin_inset Formula $B^{k}$
+\end_inset
+
+ y las de debajo de dicha diagonal son nulas, y
+\begin_inset Formula $C^{r}=0\neq C^{r-1}$
+\end_inset
+
+.
+
+\series bold
+Demostración:
+\series default
+
+\begin_inset Formula $\phi:{\cal M}_{rs}(K)\to{\cal M}_{r}({\cal M}_{s}(K))$
+\end_inset
+
+ que agrupa las matrices en bloques es un isomorfismo de anillos, pues clarament
+e conserva la suma y la identidad y, para el producto, haciendo los índices
+ de matrices empezar por 0 por simplicidad,
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Como debería ser siempre.
+\end_layout
+
+\end_inset
+
+ si
+\begin_inset Formula $A,B\in{\cal M}_{rs}(K)$
+\end_inset
+
+, para
+\begin_inset Formula $i,j\in\{0,\dots,r-1\}$
+\end_inset
+
+ y
+\begin_inset Formula $k,l\in\{1,\dots,s\}$
+\end_inset
+
+,
+\begin_inset Formula
+\begin{align*}
+(\phi(A)\phi(B))_{ijkl} & =\left(\sum_{p\in\mathbb{N}_{r}}\phi(A)_{ir}\phi(B)_{rj}\right)_{kl}=\sum_{p\in\mathbb{N}_{r}}\left(\phi(A)_{ip}\phi(B)_{pj}\right)_{kl}=\\
+ & =\sum_{p\in\mathbb{N}_{r}}\sum_{q\in\mathbb{N}_{s}}\phi(A)_{ipkq}\phi(B)_{pjql}=\sum_{p\in\mathbb{N}_{r}}\sum_{q\in\mathbb{N}_{s}}A_{is+k,ps+q}B_{ps+q,js+l}=\\
+ & =\sum_{z\in\mathbb{N}_{rs}}A_{is+k,z}B_{z,js+l}=(AB)_{is+k,js+l}=\phi(AB)_{ijkl}.
+\end{align*}
+
+\end_inset
+
+Entonces, si
+\begin_inset Formula $C\in{\cal M}_{r}({\cal M}_{s}(K))$
+\end_inset
+
+, queremos ver que cada
+\begin_inset Formula $(C^{k})_{ij}=\binom{k}{2k+i-j}B^{2k+i-j}$
+\end_inset
+
+, con lo que
+\begin_inset Formula $(C^{k})_{i,i+k}=\binom{k}{k}B^{k}=B^{k}$
+\end_inset
+
+ y, para
+\begin_inset Formula $j<i+k$
+\end_inset
+
+,
+\begin_inset Formula $2k+i-j>k$
+\end_inset
+
+ y
+\begin_inset Formula $\binom{k}{2k+i-j}=0$
+\end_inset
+
+.
+ Por inducción, para
+\begin_inset Formula $k=1$
+\end_inset
+
+,
+\begin_inset Formula $C_{i,i+1}=B=\binom{1}{1}B^{1}$
+\end_inset
+
+,
+\begin_inset Formula $C_{i,i+2}=I_{s}=\binom{1}{0}B^{0}$
+\end_inset
+
+ y el resto de entradas son nulas, y para
+\begin_inset Formula $k>1$
+\end_inset
+
+,
+\begin_inset Formula
+\begin{align*}
+(C^{k})_{ij} & =\sum_{l=1}^{r}(C^{k-1})_{il}C_{lj}=\sum_{l=1}^{r}\binom{k-1}{2k-2+i-l}\binom{1}{2+l-j}B^{2k-2+i-l+2-j+l}=\\
+ & =\sum_{l}\binom{k-1}{(1-k-i)+l}\binom{1}{(2-j)+l}B^{2k+i-j}=\binom{k}{2k+i-j}B^{2k+i-j},
+\end{align*}
+
+\end_inset
+
+donde en la última igualdad hemos usado que
+\begin_inset Formula $\sum_{k}\binom{r}{m+k}\binom{s}{n+k}=\binom{r+s}{r-m+n}$
+\end_inset
+
+ y en la penúltima hemos usado que
+\begin_inset Formula $(k-1)-(2k-2+i-l)=1-k-i+l$
+\end_inset
+
+ y que podemos expandir el rango del sumatorio ya que, si el producto de
+ los dos coeficientes no se anula, entonces
+\begin_inset Formula $2+l-j\in\{0,1\}\implies l\leq j-1<r$
+\end_inset
+
+ y
+\begin_inset Formula $0\leq1-k-i+l\leq k-1\implies k-1\leq l-i\leq2(k-1)\implies l\geq k+i-1>1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sean
+\begin_inset Formula $C\in{\cal M}_{n}(K)$
+\end_inset
+
+,
+\begin_inset Formula $P\in\text{GL}_{n}(K)$
+\end_inset
+
+ y
+\begin_inset Formula $C'\coloneqq PCP^{-1}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+Para
+\begin_inset Formula $F\in K[X]$
+\end_inset
+
+,
+\begin_inset Formula $F(C')=PF(C)P^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para
+\begin_inset Formula $k\in\mathbb{N}$
+\end_inset
+
+,
+\begin_inset Formula $(PCP^{-1})^{k}=PC^{k}P^{-1}$
+\end_inset
+
+, con lo que
+\begin_inset Formula $F(PCP^{-1})=\sum_{k}F_{k}PC^{k}P^{-1}\overset{F_{k}\in K}{=}P(\sum_{k}F_{k}C^{k})P^{-1}=PF(C)P^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $C$
+\end_inset
+
+ y
+\begin_inset Formula $C'$
+\end_inset
+
+ tienen el mismo polinomio mínimo.
+
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Por lo anterior, usando que el polinomio mínimo de una matriz
+\begin_inset Formula $C$
+\end_inset
+
+ es el menor
+\begin_inset Formula $d_{t}$
+\end_inset
+
+ con
+\begin_inset Formula $d_{t}(C)=0$
+\end_inset
+
+ y que
+\begin_inset Formula $F(C')=PF(C)P^{-1}=0\iff F(C)=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Formas canónicas reales
+\end_layout
+
+\begin_layout Standard
+Si
+\begin_inset Formula $(a,b)\in\mathbb{R}\times\mathbb{R}^{*}$
+\end_inset
+
+ y
+\begin_inset Formula $r>0$
+\end_inset
+
+, llamamos
+\begin_inset Formula
+\begin{align*}
+J(a,b) & \coloneqq\begin{pmatrix}a & -b\\
+b & a
+\end{pmatrix},
+\end{align*}
+
+\end_inset
+
+con polinomio característico irreducible
+\begin_inset Formula $p\coloneqq(X-a)^{2}+b^{2}$
+\end_inset
+
+, pues
+\begin_inset Formula $p=X^{2}-2aX+a^{2}+b^{2}$
+\end_inset
+
+ y
+\begin_inset Formula $(-2a)^{2}-4(a^{2}+b^{2})=-b^{2}<0$
+\end_inset
+
+.
+ Entonces, para
+\begin_inset Formula $r\in\mathbb{N}^{*}$
+\end_inset
+
+, llamamos
+\series bold
+bloque de Jordan real
+\series default
+ de tamaño
+\begin_inset Formula $r$
+\end_inset
+
+ asociado a
+\begin_inset Formula $(a,b)$
+\end_inset
+
+ o a
+\begin_inset Formula $p$
+\end_inset
+
+ a
+\begin_inset Formula
+\[
+J_{r}(a,b)\coloneqq\begin{pmatrix}\boxed{J(a,b)} & \boxed{I_{2}}\\
+ & \ddots & \ddots\\
+ & & \ddots & \boxed{I_{2}}\\
+ & & & \boxed{J(a,b)}
+\end{pmatrix}\in{\cal M}_{2r}(\mathbb{R}).
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Toda
+\begin_inset Formula $C\in{\cal M}_{n}(\mathbb{R})$
+\end_inset
+
+ es semejante a una matriz de la forma
+\begin_inset Formula
+\[
+\begin{pmatrix}\boxed{J_{r_{1}}(a_{1},b_{1})}\\
+ & \ddots\\
+ & & \boxed{J_{r_{t}}(a_{t},b_{t})}\\
+ & & & \boxed{J_{h_{1}}(\lambda_{1})}\\
+ & & & & \ddots\\
+ & & & & & \boxed{J_{h_{s}}(\lambda_{s})}
+\end{pmatrix},
+\]
+
+\end_inset
+
+única salvo reordenación de bloques, formada por
+\begin_inset Formula
+\[
+\text{rk}((C-\lambda I)^{h-1})+\text{rk}((C-\lambda I)^{h+1})-2\text{rk}((C-\lambda I)^{h})
+\]
+
+\end_inset
+
+bloques
+\begin_inset Formula $J_{h}(\lambda)$
+\end_inset
+
+ para cada
+\begin_inset Formula $h\in\mathbb{N}^{*}$
+\end_inset
+
+ y
+\begin_inset Formula $\lambda$
+\end_inset
+
+ valor propio real de
+\begin_inset Formula $C$
+\end_inset
+
+ y
+\begin_inset Formula
+\[
+\frac{1}{2}(\text{rk}(p(C)^{r-1})+\text{rk}(p(C)^{r+1})-2\text{rk}(p(C)^{r})
+\]
+
+\end_inset
+
+bloques
+\begin_inset Formula $J_{r}(a,b)$
+\end_inset
+
+ para cada
+\begin_inset Formula $r\in\mathbb{N}^{*}$
+\end_inset
+
+ y
+\begin_inset Formula $p=(X-a)^{2}+b^{2}$
+\end_inset
+
+ divisor irreducible cuadrático del polinomio característico de
+\begin_inset Formula $C$
+\end_inset
+
+.
+
+\series bold
+Demostración:
+\series default
+ Por el teorema de clasificación de matrices cuadradas y el hecho de que
+ todos los irreducibles en
+\begin_inset Formula $\mathbb{R}[X]$
+\end_inset
+
+ son de grado 1 o 2, solo hay que ver que
+\begin_inset Formula $J_{r}(a,b)$
+\end_inset
+
+ es semejante a
+\begin_inset Formula $C_{r}(p)$
+\end_inset
+
+, ambas con polinomio característico
+\begin_inset Formula $p^{r}$
+\end_inset
+
+.
+ Pero si
+\begin_inset Formula $J\coloneqq J_{r}(a,b)$
+\end_inset
+
+,
+\begin_inset Formula $(J-aI)=J_{r}(0,b)$
+\end_inset
+
+ y, viendo
+\begin_inset Formula $J_{r}(0,b)\in{\cal M}_{r}({\cal M}_{2}(K))$
+\end_inset
+
+,
+\begin_inset Formula
+\[
+J_{r}(0,b)_{ij}=\begin{cases}
+J(0,b), & j=i;\\
+I_{2}, & j=i+1;\\
+0, & \text{en otro caso},
+\end{cases}
+\]
+
+\end_inset
+
+y como además
+\begin_inset Formula $J(0,b)^{2}=-b^{2}I_{2}\in\text{GL}_{2}(\mathbb{R})$
+\end_inset
+
+,
+\begin_inset Formula
+\[
+(J_{r}(0,b)^{2})_{ij}=\begin{cases}
+J(0,b)^{2}=-b^{2}I_{2}, & j=i;\\
+2J(0,b), & j=i+1;\\
+I_{2}, & j=i+2;\\
+0, & \text{en otro caso},
+\end{cases}
+\]
+
+\end_inset
+
+con lo que
+\begin_inset Formula $p(J)=(J-aI)^{2}+b^{2}$
+\end_inset
+
+ tiene la forma de la matriz del resultado anterior y
+\begin_inset Formula $p(J)^{r}=0\neq p(J)^{r-1}$
+\end_inset
+
+.
+ Entonces el
+\begin_inset Formula $\mathbb{R}[X]$
+\end_inset
+
+-módulo
+\begin_inset Formula $M$
+\end_inset
+
+ asociado a
+\begin_inset Formula $(\mathbb{R}^{2r},v\mapsto Jv)$
+\end_inset
+
+ tiene un sumando directo isomorfo a
+\begin_inset Formula $\frac{\mathbb{R}[X]}{(p^{r})}$
+\end_inset
+
+, y como
+\begin_inset Formula $\dim_{\mathbb{R}}\frac{\mathbb{R}[X]}{(p^{r})}=2h=\dim_{\mathbb{R}}M$
+\end_inset
+
+,
+\begin_inset Formula $M\cong\frac{\mathbb{R}[X]}{(p^{r})}$
+\end_inset
+
+.
+ Pero por el teorema de clasificación de endomorfismos,
+\begin_inset Formula $v\mapsto Jv$
+\end_inset
+
+ se expresa como
+\begin_inset Formula $C_{r}(p)$
+\end_inset
+
+ en alguna base de
+\begin_inset Formula $\mathbb{R}^{2r}$
+\end_inset
+
+ y por tanto en alguna de
+\begin_inset Formula $M$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Series de Taylor pero en álgebra y son un porro
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+En realidad el porro es todo lo de antes.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sean
+\begin_inset Formula $\lambda\in K$
+\end_inset
+
+,
+\begin_inset Formula $r,k\in\mathbb{N}^{*}$
+\end_inset
+
+ y
+\begin_inset Formula $J\coloneqq J_{r}(\lambda)$
+\end_inset
+
+, si
+\begin_inset Formula $k<r$
+\end_inset
+
+,
+\begin_inset Formula $(J-\lambda I_{r})^{k}$
+\end_inset
+
+ tiene a 1 las celdas de la diagonal
+\begin_inset Formula $k$
+\end_inset
+
+-ésima por encima de la diagonal principal y a 0 el resto, y si
+\begin_inset Formula $k\geq r$
+\end_inset
+
+,
+\begin_inset Formula $(J-\lambda I_{r})^{k}=0$
+\end_inset
+
+.
+
+\series bold
+Demostración:
+\series default
+ Esto equivale a que, en cualquier caso,
+\begin_inset Formula $((J-\lambda I_{r})^{k})_{ij}\equiv\delta_{i-j,k}$
+\end_inset
+
+.
+ Para
+\begin_inset Formula $k=1$
+\end_inset
+
+ esto es claro, y para
+\begin_inset Formula $k>1$
+\end_inset
+
+,
+\begin_inset Formula $((J-\lambda I_{r})^{k})_{ij}=\sum_{l=1}^{r}\delta_{i-l,k-1}\delta_{l-j,1}=\delta_{i-j,k}$
+\end_inset
+
+, pues lo de dentro del sumatorio vale 1 si y sólo si
+\begin_inset Formula $i-l=k-1$
+\end_inset
+
+ y
+\begin_inset Formula $l-j=1$
+\end_inset
+
+, si y sólo si
+\begin_inset Formula $l=j+1$
+\end_inset
+
+ e
+\begin_inset Formula $i=j+k$
+\end_inset
+
+, pero si
+\begin_inset Formula $j+k\leq r$
+\end_inset
+
+,
+\begin_inset Formula $l\leq r$
+\end_inset
+
+ está dentro de rango y hay exactamente un sumando en que se da esto, y
+ si
+\begin_inset Formula $j+k>r$
+\end_inset
+
+, esto no se da en ningún sumando pero tampoco se da
+\begin_inset Formula $i-j=k$
+\end_inset
+
+ porque entonces sería
+\begin_inset Formula $i>r$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sean
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+ igual a
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ o
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+,
+\begin_inset Formula $D\subseteq\mathbb{K}$
+\end_inset
+
+ abierto,
+\begin_inset Formula $\psi:D\to\mathbb{K}$
+\end_inset
+
+ infinitamente derivable,
+\begin_inset Formula $\lambda\in D$
+\end_inset
+
+ y
+\begin_inset Formula $J\coloneqq J_{r}(\lambda)$
+\end_inset
+
+, llamamos
+\series bold
+valor
+\series default
+ o
+\series bold
+evaluación
+\series default
+ de
+\begin_inset Formula $\psi$
+\end_inset
+
+ en
+\begin_inset Formula $J$
+\end_inset
+
+ a
+\begin_inset Formula $\psi(J)$
+\end_inset
+
+, que es un polinomio en
+\begin_inset Formula $J$
+\end_inset
+
+.
+ En efecto,
+\begin_inset Formula $\psi$
+\end_inset
+
+ tiene una serie de Taylor
+\begin_inset Formula $\psi(x)=\sum_{n\geq0}\frac{\psi^{(n)}(\lambda)}{n!}(x-\lambda)^{n}$
+\end_inset
+
+ y entonces
+\begin_inset Formula $\psi(J)=\sum_{n\geq0}\frac{\psi^{(n)}(\lambda)}{n!}(J-\lambda I)^{n}$
+\end_inset
+
+, pero para
+\begin_inset Formula $n\geq r$
+\end_inset
+
+ es
+\begin_inset Formula $(J-\lambda I)^{n}=0$
+\end_inset
+
+, por lo que queda una suma finita que es un polinomio en
+\begin_inset Formula $J$
+\end_inset
+
+.
+ Además:
+\end_layout
+
+\begin_layout Enumerate
+Para
+\begin_inset Formula $k\in\{1,\dots,r-1\}$
+\end_inset
+
+,
+\begin_inset Formula
+\[
+(J^{k})_{ij}=\binom{k}{j-i}\lambda^{k-j+i},
+\]
+
+\end_inset
+
+tomando el criterio
+\begin_inset Formula $0\cdot\infty=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para
+\begin_inset Formula $k=1$
+\end_inset
+
+ es claro, pues para
+\begin_inset Formula $j=i$
+\end_inset
+
+ es
+\begin_inset Formula $J_{ij}=\lambda=\binom{1}{0}\lambda^{1}$
+\end_inset
+
+, para
+\begin_inset Formula $j=i+1$
+\end_inset
+
+ es
+\begin_inset Formula $J_{ij}=1=\binom{1}{1}\lambda^{0}$
+\end_inset
+
+ y en otro caso la fórmula da 0, usando el criterio si fuese necesario.
+ Para
+\begin_inset Formula $k>1$
+\end_inset
+
+, por inducción,
+\begin_inset Formula
+\begin{align*}
+(J^{k})_{ij} & =\sum_{l=1}^{r}(J^{k-1})_{il}J_{lj}=\sum_{l=1}^{r}\binom{k-1}{l-i}\binom{1}{j-l}\lambda^{(k-1-l+i)+(1-j+l)}=\\
+ & =\sum_{l}\binom{k-1}{l-i}\binom{1}{(j-i)-(l-i)}\lambda^{k+i-j}=\binom{k}{j-i}\lambda^{k+i-j},
+\end{align*}
+
+\end_inset
+
+donde justificamos expandir el rango del sumatorio viendo que, si
+\begin_inset Formula $0\leq l-i\leq k-1$
+\end_inset
+
+ y
+\begin_inset Formula $0\leq j-l\leq1$
+\end_inset
+
+, entonces por lo primero
+\begin_inset Formula $i\leq l$
+\end_inset
+
+ y por lo segundo
+\begin_inset Formula $l\leq j$
+\end_inset
+
+, luego
+\begin_inset Formula $l\in\{1,\dots,r\}$
+\end_inset
+
+.
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula
+\[
+(\psi(J))_{ij}=\begin{cases}
+\frac{\psi^{(j-i)}(\lambda)}{(j-i)!}, & j\geq i;\\
+0, & \text{en otro caso}.
+\end{cases}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\psi(J)=\sum_{n\geq0}\frac{\psi^{(n)}(\lambda)}{n!}(J-\lambda I)^{n}$
+\end_inset
+
+, con lo que
+\begin_inset Formula
+\[
+(\psi(J))_{ij}=\sum_{n\geq0}\frac{\psi^{(n)}(\lambda)}{n!}\delta_{j-i,n}=\begin{cases}
+\frac{\psi^{(n)}(\lambda)}{n!}, & n\coloneqq j-i\geq0;\\
+0, & \text{en otro caso}.
+\end{cases}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Sean
+\begin_inset Formula $C\in{\cal M}_{n}(\mathbb{K})$
+\end_inset
+
+ y
+\begin_inset Formula $P\in\text{GL}_{n}(\mathbb{K})$
+\end_inset
+
+ son tales que
+\begin_inset Formula $P^{-1}CP\eqqcolon\text{diag}(J_{1},\dots,J_{t})$
+\end_inset
+
+ con los
+\begin_inset Formula $J_{i}$
+\end_inset
+
+ bloques de Jordan,
+\begin_inset Formula $D\subseteq\mathbb{K}$
+\end_inset
+
+ es un abierto que contiene a todos los valores propios de
+\begin_inset Formula $C$
+\end_inset
+
+ y
+\begin_inset Formula $\psi:D\to\mathbb{K}$
+\end_inset
+
+ es infinitamente derivable, llamamos
+\series bold
+valor
+\series default
+ o
+\series bold
+evaluación
+\series default
+ de
+\begin_inset Formula $\psi$
+\end_inset
+
+ en
+\begin_inset Formula $C$
+\end_inset
+
+ a
+\begin_inset Formula $\psi(C)\coloneqq P(\psi(J_{1})\oplus\dots\oplus\psi(J_{t}))P^{-1}$
+\end_inset
+
+, que no depende de la
+\begin_inset Formula $P$
+\end_inset
+
+ elegida.
+\end_layout
+
+\end_body
+\end_document
generated by cgit v1.2.3 (git 2.46.0) at 2025-10-26 23:35:50 +0000