From 4bc806acf2c4e893eaa09f13029332d26bd7e5f6 Mon Sep 17 00:00:00 2001 From: Hearot Date: Wed, 3 May 2023 17:22:17 +0200 Subject: [PATCH] feat(geometria): aggiunge un file unico per gli appunti sul prodotto scalare --- .../1. Introduzione al prodotto scalare.tex | 195 +++ ...eoremi principali sul prodotto scalare.tex | 430 +++++ ...ni, spazi euclidei e teorema spettrale.tex | 1550 +++++++++++++++++ .../logo.png | Bin 0 -> 95066 bytes .../main.pdf | Bin 0 -> 783726 bytes .../main.tex | 39 + .../main.pdf | Bin 241056 -> 241056 bytes .../main.tex | 2 +- tex/latex/style/recap.sty | 615 +++++++ 9 files changed, 2830 insertions(+), 1 deletion(-) create mode 100644 Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/1. Introduzione al prodotto scalare.tex create mode 100644 Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/2. Proprietà e teoremi principali sul prodotto scalare.tex create mode 100644 Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/3. Prodotti hermitiani, spazi euclidei e teorema spettrale.tex create mode 100644 Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/logo.png create mode 100644 Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/main.pdf create mode 100644 Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/main.tex create mode 100644 tex/latex/style/recap.sty diff --git a/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/1. Introduzione al prodotto scalare.tex b/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/1. Introduzione al prodotto scalare.tex new file mode 100644 index 0000000..62c9782 --- /dev/null +++ b/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/1. Introduzione al prodotto scalare.tex @@ -0,0 +1,195 @@ +\chapter{Introduzione al prodotto scalare} + +\begin{note} + Nel corso del documento, per $V$, qualora non specificato, si intenderà uno spazio vettoriale di dimensione + finita $n$. +\end{note} + +\begin{definition} + Un \textbf{prodotto scalare} su $V$ è una forma bilineare simmetrica $\varphi$ con argomenti in $V$. +\end{definition} + +\begin{example} + Sia $\varphi : M(n, \KK)^2 \to \KK$ tale che $\varphi(A, B) = \tr(AB)$. \\ + + \li $\varphi(A + A', B) = \tr((A + A')B) = \tr(AB + A'B) = \tr(AB) + \tr(A'B) = \varphi(A, B) + \varphi(A', B)$ (linearità + nel primo argomento), \\ + \li $\varphi(\alpha A, B) = \tr(\alpha A B) = \alpha \tr(AB) = \alpha \varphi(A, B)$ (omogeneità nel secondo argomento), \\ + \li $\varphi(A, B) = \tr(AB) = \tr(BA) = \varphi(B, A)$ (simmetria), \\ + \li poiché $\varphi$ è simmetrica, $\varphi$ è lineare e omogenea anche nel secondo argomento, e quindi è una + forma bilineare simmetrica, ossia un prodotto scalare su $M(n, \KK)$. +\end{example} + +\begin{definition} + Si definisce prodotto scalare \textit{canonico} di $\KK^n$ la forma bilineare simmetrica $\varphi$ con + argomenti in $\KK^n$ tale che: + + \[ \varphi((x_1, ..., x_n), (y_1, ..., y_n)) = \sum_{i=1}^n x_i y_i. \] +\end{definition} + +\begin{remark} + Si può facilmente osservare che il prodotto scalare canonico di $\KK^n$ è effettivamente un prodotto + scalare. \\ + + \li $\varphi((x_1, ..., x_n) + (x_1', ..., x_n'), (y_1, ..., y_n)) = \sum_{i=1}^n (x_i + x_i') y_i = + \sum_{i=1}^n \left[x_iy_i + x_i' y_i\right] = \sum_{i=1}^n x_i y_i + \sum_{i=1}^n x_i' y_i = + \varphi((x_1, ..., x_n), (y_1, ..., y_n)) + \varphi((x_1', ..., x_n'), (y_1, ..., y_n))$ (linearità nel + primo argomento), \\ + \li $\varphi(\alpha(x_1, ..., x_n), (y_1, ..., y_n)) = \sum_{i=1}^n \alpha x_i y_i = \alpha \sum_{i=1}^n x_i y_i = + \alpha \varphi((x_1, ..., x_n), (y_1, ..., y_n))$ (omogeneità nel primo argomento), \\ + \li $\varphi((x_1, ..., x_n), (y_1, ..., y_n)) = \sum_{i=1}^n x_i y_i = \sum_{i=1}^n y_i x_i = \varphi((y_1, ..., y_n), (x_1, ..., x_n))$ (simmetria), \\ + \li poiché $\varphi$ è simmetrica, $\varphi$ è lineare e omogenea anche nel secondo argomento, e quindi è una + forma bilineare simmetrica, ossia un prodotto scalare su $\KK^n$. +\end{remark} + +\begin{example} + Altri esempi di prodotto scalare sono i seguenti: \\ + + \li $\varphi(A, B) = \tr(A^\top B)$ per $M(n, \KK)$, \\ + \li $\varphi(p(x), q(x)) = p(a) q(a)$ per $\KK[x]$, con $a \in \KK$, \\ + \li $\varphi(p(x), q(x)) = \sum_{i=1}^n p(x_i) q(x_i)$ per $\KK[x]$, con $x_1$, ..., $x_n$ distinti, \\ + \li $\varphi(p(x), q(x)) = \int_a^b p(x)q(x) dx$ per lo spazio delle funzioni integrabili su $\RR$, con $a$, $b$ in $\RR$, \\ + \li $\varphi(\vec{x}, \vec{y}) = \vec{x}^\top A \vec{y}$ per $\KK^n$, con $A \in M(n, \KK)$ simmetrica. +\end{example} + +\begin{definition} + Sia\footnote{In realtà, la definizione è facilmente estendibile a qualsiasi campo, purché esso + sia ordinato.} $\KK = \RR$. Allora un prodotto scalare $\varphi$ si dice \textbf{definito positivo} se $\v \in V$, $\vec{v} \neq \vec{0} \implies + \varphi(\vec{v}, \vec{v}) > 0$. Analogamente $\varphi$ è \textbf{definito negativo} se $\vec{v} \neq \vec 0 \implies \varphi(\v, \v) < 0$. \\ + + Infine, $\varphi$ è \textbf{semidefinito positivo} se $\varphi(\v, \v) \geq 0$ $\forall \v \in V$ (o + \textbf{semidefinito negativo} se invece $\varphi(\v, \v) \leq 0$ $\forall \v \in V$). +\end{definition} + +\begin{example} + Il prodotto scalare canonico di $\RR^n$ è definito positivo: infatti $\varphi((x_1, ..., x_n), (x_1, ..., x_n)) = + \sum_{i=1}^n x_i^2 = 0 \iff x_i = 0$, $\forall 1 \leq i \leq n$ $\iff (x_1, ..., x_n) = \vec{0}$. \\ + + Al contrario, il prodotto scalare $\varphi : \RR^2 \to \RR$ tale che $\varphi((x_1, x_2), (y_1, y_2)) = x_1 y_1 - x_2 y_2$ non è definito positivo: $\varphi((x, y), (x, y)) = 0$, $\forall$ $(x, y) \mid x^2 = y^2$, ossia se + $y = x$ o $y = -x$. +\end{example} + +\begin{definition} + Dato un prodotto scalare $\varphi$ di $V$, ad ogni vettore $\vec{v} \in V$ si associa una \textbf{forma quadratica} + $q : V \to \KK$ tale che $q(\vec{v}) = \varphi(\vec{v}, \vec{v})$. +\end{definition} + +\begin{remark} + Si osserva che $q$ non è lineare in generale: infatti $q(\vec{v} + \vec{w}) \neq q(\vec{v}) + q(\vec{w})$ in + $\RR^n$. +\end{remark} + +\begin{definition} + Un vettore $\vec{v} \in V$ si dice \textbf{isotropo} rispetto al prodotto scalare $\varphi$ se $q(\vec{v}) = + \varphi(\vec{v}, \vec{v}) = 0$. +\end{definition} + +\begin{example} + Rispetto al prodotto scalare $\varphi : \RR^3 \to \RR$ tale che $\varphi((x_1, x_2, x_3), (y_1, y_2, y_3)) = + x_1 y_1 + x_2 y_2 - x_3 y_3$, i vettori isotropi $(x, y, z)$ sono quelli tali che $x^2 + y^2 = z^2$, ossia + i vettori stanti sul cono di eq.~$x^2 + y^2 = z^2$. +\end{example} + +\begin{remark} + Come già osservato in generale per le app.~multilineari, il prodotto scalare è univocamente determinato + dai valori che assume nelle coppie $\vv{i}, \vv{j}$ estraibili da una base $\basis$. Infatti, se + $\basis = (\vv1, ..., \vv{k})$, $\vec{v} = \sum_{i=1}^k \alpha_i \vv{i}$ e $\vec{w} = \sum_{i=1}^k \beta_i \vv{i}$, + allora: + + \[ \varphi(\vec{v}, \vec{w}) = \sum_{1 \leq i \leq j \leq k} \alpha_i \beta_j \, \varphi(\vv{i}, \vv{j}). \] +\end{remark} + +\begin{definition} + Sia $\varphi$ un prodotto scalare di $V$ e sia $\basis = (\vv1, ..., \vv{n})$ una base ordinata di $V$. Allora si denota con \textbf{matrice associata} + a $\varphi$ la matrice: + + \[ M_\basis(\varphi) = (\varphi(\vv{i}, \vv{j}))_{i,\,j = 1\text{---}n} \in M(n, \KK). \] +\end{definition} + +\begin{remark} + Si possono fare alcune osservazioni riguardo $M_\basis(\varphi)$. \\ + + \li $M_\basis(\varphi)$ è simmetrica, infatti $\varphi(\vv{i}, \vv{j}) = \varphi(\vv{j}, \vv{i})$ per + definizione di prodotto scalare, \\ + \li $\varphi(\vec{v}, \vec{w}) = [\vec{v}]_\basis^\top M_\basis(\varphi) [\vec{w}]_\basis$. +\end{remark} + +\begin{theorem} (di cambiamento di base per matrici di prodotti scalari) Siano $\basis$, $\basis'$ due + basi ordinate di $V$. Allora, se $\varphi$ è un prodotto scalare di $V$ e $P = M^{\basis'}_{\basis}(\Id_V)$, vale la seguente identità: + + \[ \underbrace{M_{\basis'}(\varphi)}_{A'} = P^\top \underbrace{M_{\basis}}_{A} P. \] +\end{theorem} + +\begin{proof} Siano $\basis = (\vv{1}, ..., \vv{n})$ e $\basis' = (\vec{w}_1, ..., \vec{w}_n)$. Allora + $A'_{ij} = \varphi(\vec{w}_i, \vec{w}_j) = [\vec{w}_i]_{\basis}^\top A [\vec{w}_j]_{\basis} = + (P^i)^\top A P^j = P_i^\top (AP)^j = (P^\top AP)_{ij}$, da cui la tesi. +\end{proof} + +\begin{definition} + Si definisce \textbf{congruenza} la relazione di equivalenza $\cong$ definita nel seguente + modo su $A, B \in M(n, \KK)$: + + \[ A \cong B \iff \exists P \in GL(n, \KK) \mid A = P^\top A P. \] +\end{definition} + +\begin{remark} + Si può facilmente osservare che la congruenza è in effetti una relazione di equivalenza. \\ + + \li $A = I^\top A I \implies A \cong A$ (riflessione), \\ + \li $A \cong B \implies A = P^\top B P \implies B = (P^\top)\inv A P\inv = (P\inv)^\top A P\inv \implies B \cong A$ (simmetria), \\ + \li $A \cong B \implies A = P^\top B P$, $B \cong C \implies B = Q^\top C Q$, quindi $A = P^\top Q^\top C Q P = + (QP)^\top C (QP) \implies A \cong C$ (transitività). +\end{remark} + +\begin{remark} + Si osservano alcune proprietà della congruenza. \\ + + \li Per il teorema di cambiamento di base del prodotto scalare, due matrici associate a uno stesso + prodotto scalare sono sempre congruenti (esattamente come due matrici associate a uno stesso + endomorfismo sono sempre simili). + \li Se $A$ e $B$ sono congruenti, $A = P^\top B P \implies \rg(A) = \rg(P^\top B P) = \rg(BP) = \rg(B)$, + dal momento che $P$ e $P^\top$ sono invertibili; quindi il rango è un invariante per congruenza. Allora + è ben definito il rango $\rg(\varphi)$ di un prodotto scalare come il rango di una sua qualsiasi matrice + associata. + \li Se $A$ e $B$ sono congruenti, $A = P^\top B P \implies \det(A) = \det(P^\top B P) = \det(P^\top) \det(B) \det(P)= + \det(P)^2 \det(B)$. Quindi, per $\KK = \RR$, il segno del determinante è invariante per congruenza. +\end{remark} + +\begin{definition} + Si dice \textbf{radicale} di un prodotto scalare $\varphi$ lo spazio: + + \[ V^\perp = \{ \vec{v} \in V \mid \varphi(\vec{v}, \vec{w}) = 0 \, \forall \vec{w} \in V \} \] + + \vskip 0.05in +\end{definition} + +\begin{remark} + Il radicale di $\RR^n$ con il prodotto scalare canonico ha dimensione nulla, dal momento che $\forall \vec{v} \in \RR^n \setminus \{\vec{0}\}$, $q(\vec{v}) = \varphi(\vec{v}, \vec{v}) > 0$. +\end{remark} + +\begin{definition} + Un prodotto scalare si dice \textbf{degenere} se il radicale dello spazio su tale prodotto scalare ha + dimensione non nulla. +\end{definition} + +%TODO: spiegare perché \alpha_\varphi è lineare e aggiungere esempi nella parte precedente. +%TODO: aggiungere osservazioni sul radicale (i.e. che è uno spazio, che ogni suo vettore è isotropo, ...). + +\begin{remark} + Si definisce l'applicazione lineare $\alpha_\varphi : V \to \dual{V}$ in modo tale che + $\alpha_\varphi(\vec{v}) = p$, dove $p(\vec{w}) = \varphi(\vec{v}, \vec{w})$. \\ + + Allora $V^\perp$ altro non è che $\Ker \alpha_\varphi$. Se $V$ ha dimensione finita, $\dim V = \dim \dual{V}$, + e si può allora concludere che $\dim V^\perp > 0 \iff \Ker \alpha_\varphi \neq \{\vec{0}\} \iff \alpha_\varphi$ non è + invertibile (infatti lo spazio di partenza e di arrivo di $\alpha_\varphi$ hanno la stessa dimensione). In + particolare, $\alpha_\varphi$ non è invertibile se e solo se $\det(\alpha_\varphi) = 0$. \\ + + Sia $\basis = (\vv{1}, ..., \vv{n})$ una base ordinata di $V$. Si consideri allora la base ordinata del + duale costruita su $\basis$, ossia $\dual{\basis} = (\vecdual{v_1}, ..., \vecdual{v_n})$. Allora + $M_{\basisdual}^\basis(\alpha_\varphi)^i = [\alpha_\varphi(\vv{i})]_{\basisdual} = \Matrix{\varphi(\vec{v_i}, \vec{v_1}) \\ \vdots \\ \varphi(\vec{v_i}, \vec{v_n})} \underbrace{=}_{\varphi \text{ è simmetrica}} + \Matrix{\varphi(\vec{v_1}, \vec{v_i}) \\ \vdots \\ \varphi(\vec{v_n}, \vec{v_i})} = M_\basis(\varphi)^i$. Quindi + $M_{\basisdual}^\basis(\alpha_\varphi) = M_\basis(\varphi)$. \\ + + Si conclude allora che $\varphi$ è degenere se e solo se $\det (M_\basis(\varphi)) = 0$ e che + $V^\perp \cong \Ker M_\basis(\varphi)$ con l'isomorfismo è il passaggio alle coordinate. +\end{remark} \ No newline at end of file diff --git a/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/2. Proprietà e teoremi principali sul prodotto scalare.tex b/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/2. Proprietà e teoremi principali sul prodotto scalare.tex new file mode 100644 index 0000000..f724bfd --- /dev/null +++ b/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/2. Proprietà e teoremi principali sul prodotto scalare.tex @@ -0,0 +1,430 @@ +\chapter{Proprietà e teoremi principali sul prodotto scalare} + +\begin{note} + Nel corso del documento, per $V$ si intenderà uno spazio vettoriale di dimensione + finita $n$ e per $\varphi$ un suo prodotto scalare. Analogamente si intenderà lo stesso + per $V'$ e $\varphi'$. +\end{note} + +\begin{proposition} (formula delle dimensioni del prodotto scalare) + Sia $W \subseteq V$ un sottospazio di $V$. Allora vale la seguente identità: + + \[ \dim W + \dim W^\perp = \dim V + \dim (W \cap V^\perp). \] +\end{proposition} + +\begin{proof} + Si consideri l'applicazione lineare $f : V \to \dual W$ tale che $f(\vec v)$ è un funzionale di $\dual W$ tale che + $f(\vec v)(\vec w) = \varphi(\vec v, \vec w)$ $\forall \vec w \in W$. Si osserva che $W^\perp = \Ker f$, da cui, + per la formula delle dimensioni, $\dim V = \dim W^\perp + \rg f$. Inoltre, si osserva anche che + $f = i^\top \circ a_\varphi$, dove $i : W \to V$ è tale che $i(\vec w) = \vec w$, infatti $f(\vec v) = a_\varphi(\vec v) \circ i$ è un funzionale di $\dual W$ tale che $f(\vec v)(\vec w) = \varphi(\vec v, \vec w)$. Pertanto + $\rg f = \rg (i^\top \circ a_\varphi)$. \\ + + Si consideri ora l'applicazione $g = a_\varphi \circ i : W \to \dual W$. Sia ora $\basis_W$ una base di $W$ e + $\basis_V$ una base di $V$. Allora le matrice associate di $f$ e di $g$ sono le seguenti: + + \begin{enumerate}[(i)] + \item $M_{\dual \basis_W}^{\basis_V}(f) = M_{\dual \basis_W}^{\basis_V}(i^\top \circ a_\varphi) = + \underbrace{M_{\dual \basis_W}^{\dual \basis_V}(i^\top)}_A \underbrace{M_{\dual \basis_V}^{\basis_V}(a_\varphi)}_B = AB$, + \item $M_{\dual \basis_V}^{\basis_W}(g) = M_{\dual \basis_V}^{\basis_W}(a_\varphi \circ i) = + \underbrace{M_{\dual \basis_V}^{\basis_V}(a_\varphi)}_B \underbrace{M_{\basis_V}^{\basis_W}(i)}_{A^\top} = BA^\top \overbrace{=}^{B^\top = B} (AB)^\top$. + \end{enumerate} + + Poiché $\rg(A) = \rg(A^\top)$, si deduce che $\rg(f) = \rg(g) \implies \rg(i^\top \circ a_\varphi) = \rg(a_\varphi \circ i) = \rg(\restr{a_\varphi}{W}) = \dim W - \dim \Ker \restr{a_\varphi}{W} = \dim W - \dim (W \cap \underbrace{\Ker a_\varphi}_{V^\perp}) = \dim W - \dim (W \cap V^\perp)$. Si conclude allora, sostituendo quest'ultima + identità nell'identità ricavata a inizio dimostrazione che $\dim V = \dim W^\top + \dim W - \dim (W \cap V^\perp)$, + ossia la tesi. +\end{proof} + +\begin{remark} + Si possono fare alcune osservazioni sul radicale di un solo elemento $\vec w$ e su quello del suo sottospazio + generato $W = \Span(\vec w)$: \\ + + \li $\vec w ^\perp = W^\perp$, \\ + \li $\vec w \notin W^\perp \iff \Rad (\restr{\varphi}{W}) = W \cap W^\perp \iff \vec w \text{ non è isotropo } = \zerovecset \iff + V = W \oplus W^\perp$. +\end{remark} + +\begin{definition} + Si definisce \textbf{base ortogonale} di $V$ una base $\vv 1$, ..., $\vv n$ tale per cui $\varphi(\vv i, \vv j) = 0 + \impliedby i \neq j$, ossia per cui la matrice associata del prodotto scalare è diagonale. +\end{definition} + +\begin{proposition} (formula di polarizzazione) + Se $\Char \KK \neq 2$, un prodotto scalare è univocamente determinato dalla sua forma quadratica $q$. +\end{proposition} + +\begin{proof} + Si nota infatti che $q(\vec v + \vec w) - q(\vec v) - q(\vec w) = 2 \varphi(\vec v, \vec w)$, e quindi, + poiché $2$ è invertibile per ipotesi, che $\varphi(\vec v, \vec w) = 2\inv (q(\vec v + \vec w) - q(\vec v) - q(\vec w))$. +\end{proof} + +\begin{theorem}(di Lagrange) + Ogni spazio vettoriale $V$ su $\KK$ tale per cui $\Char \KK \neq 2$ ammette una base ortogonale. +\end{theorem} + +\begin{proof} + Sia dimostra il teorema per induzione su $n := \dim V$. Per $n \leq 1$, la dimostrazione è triviale. Sia + allora il teorema vero per $i \leq n$. Se $V$ ammette un vettore non isotropo $\vec w$, sia $W = \Span(\vec w)$ e si consideri la decomposizione $V = W \oplus W^\perp$. Poiché $W^\perp$ ha dimensione $n-1$, per ipotesi induttiva + ammette una base ortogonale. Inoltre, tale base è anche ortogonale a $W$, e quindi l'aggiunta di $\vec w$ a + questa base ne fa una base ortogonale di $V$. Se invece $V$ non ammette vettori non isotropi, ogni forma quadratica + è nulla, e quindi il prodotto scalare è nullo per la proposizione precedente. +\end{proof} + + +\begin{note} + D'ora in poi, nel corso del documento, si assumerà $\Char \KK \neq 2$. +\end{note} + +\begin{theorem} (di Sylvester, caso complesso) + Sia $\KK$ un campo i cui elementi sono tutti quadrati di un + altro elemento del campo (e.g.~$\CC$). Allora esiste una base + ortogonale $\basis$ tale per cui: + + \[ M_\basis(\varphi) = \Matrix{I_r & \rvline & 0 \\ \hline 0 & \rvline & 0\,}. \] +\end{theorem} + +\begin{proof} + Per il teorema di Lagrange, esiste una base ortogonale $\basis'$ di $V$. + Si riordini la base in modo tale che la forma quadratica valutata nei primi elementi sia sempre diversa da zero. Allora, poiché ogni + elemento di $\KK$ è per ipotesi quadrato di un altro elemento + di $\KK$, si sostituisca $\basis'$ con una base $\basis$ tale per + cui, se $q(\vv i) = 0$, $\vv i \mapsto \vv i$, e altrimenti + $\vv i \mapsto \frac{\vv i}{\sqrt{q(\vv i)}}$. Allora $\basis'$ + è una base tale per cui la matrice associata del prodotto scalare + in tale base è proprio come desiderata nella tesi, dove $r$ è + il numero di elementi tali per cui la forma quadratica valutata + in essi sia diversa da zero. +\end{proof} + +\begin{remark} + Si possono effettuare alcune considerazioni sul teorema di Sylvester + complesso. \\ + + \li Si può immediatamente concludere che il rango è un invariante + completo per la congruenza in un campo in cui tutti gli elementi + sono quadrati, ossia che $A \cong B \iff \rg(A) = \rg(B)$, se $A$ e + $B$ sono matrici simmetriche: infatti + ogni matrice simmetrica rappresenta una prodotto scalare, ed è + pertanto congruente ad una matrice della forma desiderata + nell'enunciato del teorema di Sylvester complesso. Poiché il rango + è un invariante della congruenza, si ricava che $r$ nella forma + della matrice di Sylvester, rappresentando il rango, è anche + il rango di ogni sua matrice congruente. In particolare, se due + matrici simmetriche hanno stesso rango, allora sono congruenti + alla stessa matrice di Sylvester, e quindi, essendo la congruenza + una relazione di congruenza, sono congruenti a loro volta. \\ + \li Due matrici simmetriche con stesso rango, allora, non solo + sono SD-equivalenti, ma sono anche congruenti. \\ + \li Ogni base ortogonale deve quindi avere lo stesso numero + di elementi nulli. +\end{remark} + +\begin{definition} (somma diretta ortogonale) + Siano i sottospazi $U$ e $W \subseteq V$ in somma diretta. Allora si dice che $U$ e $W$ sono in \textbf{somma + diretta ortogonale rispetto al prodotto scalare} $\varphi$ di $V$, ossia che $U \oplus W = U \oplus^\perp W$, se $\varphi(\vec u, \vec w) = 0$ $\forall \vec u \in U$, $\vec w \in W$. +\end{definition} + +\begin{definition} (cono isotropo) + Si definisce \textbf{cono isotropo} di $V$ rispetto al prodotto scalare $\varphi$ il seguente insieme: + + \[ \CI(\varphi) = \{ \v \in V \mid \varphi(\v, \v) = 0 \}, \] + + \vskip 0.05in + + ossia l'insieme dei vettori isotropi di $V$. +\end{definition} + +\begin{note} + La notazione $\varphi > 0$ indica che $\varphi$ è definito positivo (si scrive $\varphi \geq 0$ se invece è semidefinito + positivo). + Analogamente $\varphi < 0$ indica che $\varphi$ è definito negativo (e $\varphi \leq 0$ indica che è semidefinito negativo). +\end{note} + +\begin{exercise} Sia $\Char \KK \neq 2$. + Siano $\vv1$, ..., $\vv k \in V$ e sia $M = \left( \varphi(\vv i, \vv j) \right)_{i, j = 1\textrm{---}k} \in M(k, \KK)$, + dove $\varphi$ è un prodotto scalare di $V$. Sia inoltre $W = \Span(\vv 1, ..., \vv k)$. Si dimostrino + allora le seguenti affermazioni. + + \begin{enumerate}[(i)] + \item Se $M$ è invertibile, allora $\vv 1$, ..., $\vv k$ sono linearmente indipendenti. + + \item Siano $\vv 1$, ..., $\vv k$ linearmente indipendenti. Allora $M$ è invertibile $\iff$ $\restr{\varphi}{W}$ è non degenere $\iff$ $W \cap W^\perp = \zerovecset$. + + \item Siano $\vv1$, ..., $\vv k$ a due a due ortogonali tra loro. Allora $M$ è invertibile $\iff$ nessun + vettore $\vv i$ è isotropo. + + \item Siano $\vv1$, ..., $\vv k$ a due a due ortogonali tra loro e siano anche linearmente indipendenti. + Allora $M$ è invertibile $\implies$ si può estendere $\basis_W = \{\vv 1, \ldots, \vv k\}$ a una base ortogonale di $V$. + + \item Sia $\KK = \RR$. Sia inoltre $\varphi > 0$. Allora $\vv 1$, ..., $\vv k$ sono linearmente + indipendenti $\iff$ $M$ è invertibile. + + \item Sia $\KK = \RR$. Sia ancora $\varphi > 0$. Allora se $\vv 1$, ..., $\vv k$ sono a due a due + ortogonali e sono tutti non nulli, sono anche linearmente indipendenti. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate}[(i)] + \item Siano $a_1$, ..., $a_k \in \KK$ tali che $a_1 \vv 1 + \ldots + a_k \vv k = 0$. Vale in + particolare che $\vec 0 = \varphi(\vv i, \vec 0) = \varphi(\vv i, a_1 \vv 1 + \ldots + a_k \vv k) = + \sum_{j=1}^k a_j \varphi(\vv i, \vv j)$ $\forall 1 \leq i \leq k$. Allora $\sum_{j=1}^k a_j M^j = 0$. + Dal momento che $M$ è invertibile, $\rg(M) = k$, e quindi l'insieme delle colonne di $M$ è linearmente + indipendente, da cui si ricava che $a_j = 0$ $\forall 1 \leq j \leq k$, e quindi che $\vv 1$, ..., + $\vv k$ sono linearmente indipendenti. + + \item Poiché $\vv 1$, ..., $\vv k$ sono linearmente indipendenti, tali vettori formano una base di + $W$, detta $\basis$. In particolare, allora, vale che $M = M_\basis(\restr{\varphi}{W})$. Pertanto, + se $M$ è invertibile, $\Rad(\restr{\varphi}{W}) = \Ker M = \zerovecset$, e dunque $\restr{\varphi}{W}$ + è non degenere. Se invece $\restr{\varphi}{W}$ è non degenere, $\zerovecset = \Rad(\restr{\varphi}{W}) = W \cap W^\perp$. Infine, se $W \cap W^\perp = \zerovecset$, $\zerovecset = W \cap W^\perp = \Rad(\restr{\varphi}{W}) = \Ker M$, e quindi $M$ è iniettiva, e dunque invertibile. + + \item Dal momento che $\vv 1$, ..., $\vv k$ sono ortogonali tra loro, $M$ è una matrice diagonale. + Pertanto $M$ è invertibile se e solo se ogni suo elemento diagonale è diverso da $0$, ossia + se $\varphi(\vv i, \vv i) \neq 0$ $\forall 1 \leq i \leq k$, e dunque se e solo se nessun vettore + $\vv i$ è isotropo. + + \item Se $M$ è invertibile, da (ii) si deduce che $\Rad(\restr{\varphi}{W}) = W \cap W^\perp = \zerovecset$, + e quindi che $W$ e $W^\perp$ sono in somma diretta. Inoltre, per la formula delle dimensioni del prodotto + scalare, $\dim W + \dim W^\perp = \dim V + \underbrace{\dim (W \cap V^\perp)}_{\leq \dim (W \cap W^\perp) = 0} = \dim V$. Pertanto $V = W \oplus^\perp W^\perp$. \\ + + Allora, dacché $\Char \KK \neq 2$, per il teorema di Lagrange, $W^\perp$ ammette una base ortogonale $\basis_{W^\perp}$. Si conclude + dunque che $\basis = \basis_W \cup \basis_{W^\perp}$ è una base ortogonale di $V$. + + \item Se $M$ è invertibile, da (i) $\vv1$, ..., $\vv k$ sono linearmente indipendenti. Siano ora + invece $\vv 1$, ..., $\vv k$ linearmente indipendenti per ipotesi. Siano $a_1$, ..., $a_k \in \KK$ tali + che $a_1 M^1 + \ldots + a_k M^k = 0$, allora $a_1 \varphi(\vv i, \vv 1) + \ldots + a_k \varphi(\vv i, \vv k) = 0$ $\forall 1 \leq i \leq k$. Pertanto, detto $\v = a_1 \vv 1 + \ldots + a_k \vv k$, si ricava che: + + \[ \varphi(\v, \v) = \sum_{i=1}^k \sum_{j=1}^k a_j \, \varphi(\vv i, \vv j) = 0. \] + + Tuttavia questo è possibile solo se $\v = a_1 \vv 1 + \ldots + a_k \vv k = 0$. Dal momento che + $\vv 1$, ..., $\vv k$ sono linearmente indipendenti, si conclude che $a_1 = \cdots = a_k = 0$, ossia + che le colonne di $M$ sono tutte linearmente indipendenti e quindi che $\rg(M) = k \implies$ $M$ è invertibile. + + \item Poiché $\vv 1$, ..., $\vv k$ sono ortogonali a due a due tra loro, $M$ è una matrice diagonale. + Inoltre, dacché $\varphi > 0$ e $\vv i \neq \vec 0$ $\forall 1 \leq i \leq k$, gli elementi diagonali di $M$ sono sicuramente tutti diversi da zero, e quindi $\det (M) \neq 0$ $\implies$ $M$ è invertibile. Allora, + per il punto (v), $\vv 1$, ..., $\vv k$ sono linearmente indipendenti. + \end{enumerate} +\end{solution} + +\begin{definition} + Data una base ortogonale $\basis$ di $V$ rispetto al prodotto + scalare $\varphi$, + si definiscono i seguenti indici: + + \begin{align*} + \iota_+(\varphi) &= \max\{ \dim W \mid W \subseteq V \E \restr{\varphi}{W} > 0 \}, &\text{(}\textbf{indice di positività}\text{)} \\ + \iota_-(\varphi) &= \max\{ \dim W \mid W \subseteq V \E \restr{\varphi}{W} < 0 \}, &\text{(}\textbf{indice di negatività}\text{)}\\ + \iota_0(\varphi) &= \dim V^\perp &\text{(}\textbf{indice di nullità}\text{)} + \end{align*} + + Quando il prodotto scalare $\varphi$ è noto dal contesto, si omette + e si scrive solo $\iota_+$, $\iota_-$ e $\iota_0$. In particolare, + la terna $\sigma(\varphi) = \sigma = (i_+, i_-, i_0)$ è detta \textbf{segnatura} del + prodotto $\varphi$. +\end{definition} + +\begin{theorem} (di Sylvester, caso reale) Sia $\KK$ un campo ordinato + i cui elementi positivi sono tutti quadrati (e.g.~$\RR$). Allora + esiste una base ortogonale $\basis$ tale per cui: + + \[ M_\basis(\varphi) = \Matrix{I_{\iota_+} & \rvline & 0 & \rvline & 0 \\ \hline 0 & \rvline & -I_{\iota_-} & \rvline & 0 \\ \hline 0 & \rvline & 0 & \rvline & 0\cdot I_{\iota_0} }. \] + + \vskip 0.05in + + Inoltre, per ogni base ortogonale, esistono esattamente + $\iota_+$ vettori della base con forma quadratica positiva, + $\iota_-$ con forma negativa e $\iota_0$ con + forma nulla. +\end{theorem} + +\begin{proof} + Per il teorema di Lagrange, esiste una base ortogonale $\basis'$ di $V$. + Si riordini la base in modo tale che la forma quadratica valutata nei primi elementi sia strettamente positiva, che nei secondi elementi sia strettamente negativa e che negli ultimi sia nulla. Si sostituisca + $\basis'$ con una base $\basis$ tale per cui, se $q(\vv i) > 0$, + allora $\vv i \mapsto \frac{\vv i}{\sqrt{q(\vv i)}}$; se + $q(\vv i) < 0$, allora $\vv i \mapsto \frac{\vv i}{\sqrt{-q(\vv i)}}$; + altrimenti $\vv i \mapsto \vv i$. Si è allora trovata una base + la cui matrice associata del prodotto scalare è come desiderata nella + tesi. \\ + + Sia ora $a$ il numero di vettori della base con forma quadratica + positiva, $b$ il numero di vettori con forma negativa e $c$ quello + dei vettori con forma nulla. Si consideri $W_+ = \Span(\vv 1, ..., \vv a)$, $W_- = \Span(\vv{a+1}, ..., \vv b)$, $W_0 = \Span(\vv{b+1}, ..., \vv c)$. \\ + + Sia $M = M_\basis(\varphi)$. Si osserva che $c = n - \rg(M) = \dim \Ker(M) = \dim V^\perp = \iota_0$. Inoltre $\forall \v \in W_+$, dacché + $\basis$ è ortogonale, + $q(\v) = q(\sum_{i=1}^a \alpha_i \vv i) = \sum_{i=1}^a \alpha_i^2 q(\vv i) > 0$, e quindi $\restr{\varphi}{W_+} > 0$, da cui $\iota_+ \geq a$. + Analogamente $\iota_- \geq b$. \\ + + Si mostra ora che è impossibile che $\iota_+ > a$. Se così infatti + fosse, sia $W$ tale che $\dim W = \iota_+$ e che $\restr{\varphi}{W} > 0$. $\iota_+ + b + c$ sarebbe maggiore di $a + b + c = n := \dim V$. Quindi, per la formula di Grassman, $\dim(W + W_- + W_0) = \dim W + + \dim(W_- + W_0) - \dim (W \cap (W_- + W_0)) \implies \dim (W \cap (W_- + W_0)) = \dim W + + \dim(W_- + W_0) - \dim(W + W_- + W_0) > 0$, ossia esisterebbe + $\v \neq \{\vec 0\} \mid \v \in W \cap (W_- + W_0)$. Tuttavia + questo è assurdo, dacché dovrebbe valere sia $q(\v) > 0$ che + $q(\v) < 0$, \Lightning. Quindi $\iota_+ = a$, e analogamente + $\iota_- = b$. +\end{proof} + +\begin{definition} + Si dice \textbf{base di Sylvester} una base di $V$ tale per cui la + matrice associata di $\varphi$ sia esattamente nella forma + vista nella dimostrazione del teorema di Sylvester. Analogamente + si definisce tale matrice come \textbf{matrice di Sylvester}. +\end{definition} + +\begin{remark} \nl + \li Si può dunque definire la segnatura di una matrice simmetrica + come la segnatura di una qualsiasi sua base ortogonale, dal + momento che tale segnatura è invariante per cambiamento di base. \\ + \li La segnatura è un invariante completo per la congruenza nel caso reale. Se infatti due matrici hanno la stessa segnatura, sono + entrambe congruenti alla matrice come vista nella dimostrazione + della forma reale del teorema di Sylvester, e quindi, essendo + la congruenza una relazione di equivalenza, sono congruenti + tra loro. Analogamente vale il viceversa, dal momento che ogni + base ortogonale di due matrici congruenti devono contenere gli + stessi numeri $\iota_+$, $\iota_-$ e $\iota_0$ di vettori + di base con forma quadratica positiva, negativa e nulla. \\ + \li Se $\ww 1$, ..., $\ww k$ sono tutti i vettori di una base + ortogonale $\basis$ con forma quadratica nulla, si osserva che $W = \Span(\ww 1, ..., \ww k)$ altro non è che $V^\perp$ stesso. Infatti, come + visto anche nella dimostrazione del teorema di Sylvester reale, vale + che $\dim W = \dim \Ker (M_\basis(\varphi)) = \dim V^\perp$. Inoltre, + se $\w \in W$ e $\v \in V$, $\varphi(\w, \v) = \varphi(\sum_{i=1}^k + \alpha_i \ww i, \sum_{i=1}^k \beta_i \ww i + \sum_{i=k+1}^n \beta_i \vv i) + = \sum_{i=1}^k \alpha_i \beta_i q(\ww i) = 0$, e quindi + $W \subseteq V^\perp$, da cui si conclude che $W = V^\perp$. \\ + \li Vale in particolare che $\rg(\varphi) = \iota_+ + \iota_-$, mentre + $\dim \Ker(\varphi) = \iota_0$, e quindi $n = \iota_+ + \iota_- + \iota_0$. \\ + \li Se $V = U \oplusperp W$, allora $\iota_+(\varphi) = \iota_+(\restr{\varphi}{V}) + \iota_+(\restr{\varphi}{W})$, e + analogamente per gli altri indici. +\end{remark} + +\begin{definition} (isometria) + Dati due spazi vettoriali $(V, \varphi)$ e + $(V', \varphi')$ dotati di prodotto scalare sullo stesso campo $\KK$, si dice che + $V$ e $V'$ sono \textbf{isometrici} se esiste un isomorfismo + $f$, detto \textit{isometria}, che preserva tali che prodotti, ossia tale che: + + \[ \varphi(\vec v, \vec w) = \varphi'(f(\vec v), f(\vec w)). \] +\end{definition} + +\begin{exercise} Sia $f : V \to V'$ un isomorfismo. Allora $f$ è un'isometria $\iff$ $\forall$ base $\basis = \{ \vv 1, \ldots, \vv n \}$ di $V$, $\basis' = \{ f(\vv 1), \ldots, f(\vv n) \}$ è una base di $V'$ e $\varphi(\vv i, \vv j) = \varphi'(f(\vv i), f(\vv j))$ $\forall 1 \leq i, j \leq n$ $\iff$ $\exists$ base $\basis = \{ \vv 1, \ldots, \vv n \}$ di $V$ tale che $\basis' = \{ f(\vv 1), \ldots, f(\vv n) \}$ è una base di $V'$ e $\varphi(\vv i, \vv j) = \varphi'(f(\vv i), f(\vv j))$ $\forall 1 \leq i, j \leq n$. +\end{exercise} + +\begin{solution} Se $f$ è un'isometria, detta $\basis$ una base di $V$, $\basis' = f(\basis)$ è una base di $V'$ + dal momento che $f$ è anche un isomorfismo. Inoltre, dacché $f$ è un'isometria, vale sicuramente che + $\varphi(\vv i, \vv j) = \varphi'(f(\vv i), f(\vv j))$ $\forall 1 \leq i, j \leq n$. \\ + + Sia ora assunto per ipotesi che $\forall$ base $\basis = \{ \vv 1, \ldots, \vv n \}$ di $V$, $\basis' = \{ f(\vv 1), \ldots, f(\vv n) \}$ è una base di $V'$ e $\varphi(\vv i, \vv j) = \varphi'(f(\vv i), f(\vv j))$ $\forall 1 \leq i, j \leq n$. Allora, analogamente a prima, detta $\basis = \{ \vv 1, \ldots, \vv n \}$ una base di $V$, $\basis' = f(\basis)$ è una base di $V'$, e in quanto tale, + per ipotesi, è tale che $\varphi(\vv i, \vv j) = \varphi'(f(\vv i), f(\vv j))$ $\forall 1 \leq i, j \leq n$. \\ + + Sia infine assunto per ipotesi che $\exists$ base $\basis = \{ \vv 1, \ldots, \vv n \}$ di $V$ tale che $\basis' = \{ f(\vv 1), \ldots, f(\vv n) \}$ è una base di $V'$ e $\varphi(\vv i, \vv j) = \varphi'(f(\vv i), f(\vv j))$ $\forall 1 \leq i, j \leq n$. Siano $\v$, $\w \in V$. Allora $\exists a_1$, ..., $a_n$, $b_1$, ..., $b_n \in \KK$ + tali che $\v = a_1 \vv 1 + \ldots + a_n \vv n$ e $\w = b_1 \vv 1 + \ldots + b_n \vv n$. Si ricava pertanto + che: + + \[ \varphi'(f(\v), f(\w)) = \sum_{i=1}^n \sum_{j=1}^n a_i b_j \, \varphi'(f(\vv i), f(\vv j)) = + \sum_{i=1}^n \sum_{j=1}^n a_i b_j \, \varphi(\vv i, \vv j) = \varphi(\v, \w), \] + + da cui la tesi. +\end{solution} + +\begin{proposition} Sono equivalenti le seguenti affermazioni: + + \begin{enumerate}[(i)] + \item $V$ e $V'$ sono isometrici; + \item $\forall$ base $\basis$ di $V$, base $\basis'$ di $V'$, + $M_\basis(\varphi)$ e $M_{\basis'}(\varphi')$ sono congruenti; + \item $\exists$ base $\basis$ di $V$, base $\basis'$ di $V'$ tale che + $M_\basis(\varphi)$ e $M_{\basis'}(\varphi')$ sono congruenti. + \end{enumerate} +\end{proposition} + +\begin{proof} Se $V$ e $V'$ sono isometrici, sia $f : V \to V'$ un'isometria. Sia $\basisC = \{ \vv 1, \ldots, \vv n \}$ una base di $V$. Allora, poiché $f$ è anche un isomorfismo, $\basisC' = f(\basisC)$ è una base di $V$ tale che + $\varphi(\vv i, \vv j) = \varphi'(f(\vv i), f(\vv j))$ $\forall 1 \leq i, j \leq n$. Pertanto $M_\basisC(\varphi) = M_{\basisC'}(\varphi')$. Si conclude allora che, cambiando base in $V$ (o in $V'$), la matrice associata + al prodotto scalare varia per congruenza dalla formula di cambiamento di base per il prodotto scalare, da cui si ricava che per ogni scelta di $\basis$ base di $V$ e di $\basis'$ base di $V'$, $M_\basis(\varphi) \cong M_{\basis'}(\varphi')$. Inoltre, se tale risultato è vero per ogni $\basis$ base di $V$ e di $\basis'$ base di $V'$, dal momento che sicuramente esistono due basi $\basis$, $\basis'$ di $V$ e $V'$, vale anche (ii) $\implies$ (iii). \\ + + Si dimostra ora (iii) $\implies$ (i). Per ipotesi $M_\basis(\varphi) \cong M_{\basis'}(\varphi')$, quindi + $\exists P \in \GL(n, \KK) \mid M_{\basis'}(\varphi') = P^\top M_\basis(\varphi) P$. Allora $\exists$ $\basis''$ + base di $V'$ tale che $P = M_{\basis''}^{\basis'}(\Idv)$, da cui $P\inv = M_{\basis'}^{\basis''}(\varphi)$. Per la formula di cambiamento di base del prodotto + scalare, $M_{\basis''}(\varphi) = (P\inv)^\top M_{\basis'} P\inv = M_\basis(\varphi)$. Detta + $\basis'' = \{ \ww 1, \ldots, \ww n \}$, si costruisce allora l'isomorfismo $f : V \to V'$ tale + che $f(\vv i) = \ww i$ $\forall 1 \leq i \leq n$.. Dal momento che per costruzione $M_\basis(\varphi) = M_{\basis''}(\varphi')$, + $\varphi(\vv i, \vv j) = \varphi'(\ww i, \ww j) = \varphi'(f(\vv i), f(\vv j))$ $\forall 1 \leq i, j \leq n$. + Si conclude dunque che $\varphi(\v, \w) = \varphi'(f(\v), f(\w))$ $\forall \v, \w \in V$, e dunque + che $f$ è un'isometria, come desiderato dalla tesi. +\end{proof} + +\begin{proposition} $(V, \varphi)$ e $(V', \varphi')$ spazi vettoriali + su $\RR$ sono + isometrici $\iff$ $\varphi$ e $\varphi'$ hanno la stessa segnatura. +\end{proposition} + +\begin{proof}\nl\nl + \rightproof Per la precedente proposizione, esistono due basi $\basis$ e $\basis'$, una di $V$ e una di $V'$, + tali che $M_\basis(\varphi) \cong M_{\basis'}(\varphi)$. Allora queste due matrici condividono la stessa + segnatura, e così quindi anche $\varphi$ e $\varphi'$. \\ + + \leftproof Se $\varphi$ e $\varphi'$ hanno la stessa segnatura, esistono due basi $\basis = \{ \vv 1, \ldots, \vv n \}$ e $\basis' = \{ \ww 1, \ldots, \ww n \}$, una + di $V$ e una di $V'$, tali che $M = M_\basis(\varphi) = M_{\basis'}(\varphi')$ e che $M$ è una matrice di + Sylvester. Allora si costruisce $f : V \to V'$ tale che $f(\vv i) = \ww i$. Esso è un isomorfismo, e per + costruzione $\varphi(\vv i, \vv j) = \varphi'(\ww i, \ww j) = \varphi'(f(\vv i), f(\vv j))$ $\forall 1 \leq i, j \leq n$, da cui + si conclude che $\varphi(\v, \w) = \varphi'(f(\v), f(\w))$ $\forall \v$, $\w \in V$, e quindi che $V$ e + $V'$ sono isometrici. +\end{proof} + +% \begin{example} + % Per $\varphi = x_1 y_1 + x_2 y_2 - x_3 y_3$. %TODO: guarda dalle slide. + % \end{example} + +\begin{definition} (sottospazio isotropo) + Sia $W$ un sottospazio di $V$. Allora $W$ si dice \textbf{sottospazio isotropo} di $V$ + se $\restr{\varphi}{W} = 0$. +\end{definition} + +\begin{remark}\nl + \li $V^\perp$ è un sottospazio isotropo di $V$. \\ + \li $\vec{v}$ è un vettore isotropo $\iff$ $W = \Span(\vec v)$ è un sottospazio isotropo di $V$. \\ + \li $W \subseteq V$ è isotropo $\iff$ $W \subseteq W^\perp$. +\end{remark} + +\begin{proposition} + Sia $\varphi$ non degenere. Se $W$ è un sottospazio isotropo di $V$, allora + $\dim W \leq \frac{1}{2} \dim V$. +\end{proposition} + +\begin{proof} + Poiché $W$ è un sottospazio isotropo di $V$, $W \subseteq W^\perp \implies \dim W \leq \dim W^\perp$. + Allora, poiché $\varphi$ è non degenere, $\dim W + \dim W^\perp = \dim V$, $\dim W \leq \dim V - \dim W$, + da cui $\dim W \leq \frac{1}{2} \dim V$. +\end{proof} + +\begin{definition} + Si definisce \textbf{indice di Witt} $W(\varphi)$ di $(V, \varphi)$ + come la massima dimensione di un sottospazio isotropo. +\end{definition} + +\begin{remark}\nl + \li Se $\varphi > 0$ o $\varphi < 0$, $W(\varphi) = 0$. +\end{remark} + +\begin{proposition} + Sia $\KK = \RR$. Sia $\varphi$ non degenere e sia $\sigma(\varphi) = (\iota_+(\varphi), \iota_-(\varphi), 0)$. Allora + $W(\varphi) = \min\{\iota_+(\varphi), \iota_-(\varphi)\}$. +\end{proposition} + +\begin{proof} + Senza perdità di generalità si assuma $\iota_-(\varphi) \leq \iota_+(\varphi)$ (il caso $\iota_-(\varphi) > \iota_+(\varphi)$ è analogo). Sia $W$ un sottospazio con $\dim W > \iota_-(\varphi)$. Sia $W^+$ + un sottospazio con $\dim W^+ = \iota_+(\varphi)$ e $\restr{\varphi}{W^+} > 0$. Allora, per la formula + di Grassmann, $\dim W + \dim W^+ > n \implies \dim W + \dim W^+ > \dim W + \dim W^+ - \dim (W \cap W^+) \implies \dim (W \cap W^+) > 0$. Quindi $\exists \w \in W$, $\w \neq \vec 0$ tale che $\varphi(\w, \w) > 0$, da cui + si ricava che $W$ non è isotropo. Pertanto $W(\varphi) \leq \iota_-(\varphi)$. \\ + + Sia $a := \iota_+(\varphi)$ e sia $b := \iota_-(\varphi)$. + Sia ora $\basis = \{ \vv 1, \ldots, \vv a, \ww 1, \ldots, \ww b \}$ una base tale per cui $M_\basis(\varphi)$ è la matrice di Sylvester per $\varphi$. Siano $\vv 1$, ..., $\vv a$ tali che $\varphi(\vv i, \vv i) = 1$ + con $1 \leq i \leq a$. Analogamente siano $\ww 1$, ..., $\ww b$ tali che $\varphi(\ww i, \ww i) = -1$ con + $1 \leq i \leq b$. Detta allora $\basis' = \{ \vv 1 ' := \vv 1 + \ww 1, \ldots, \vv b ' := \vv b + \ww b \}$, sia $W = \Span(\basis')$. \\ + + Si osserva che $\basis'$ è linearmente indipendente, e dunque che $\dim W = \iota_-$. Inoltre + $\varphi(\vv i ', \vv j ') = \varphi(\vv i + \ww i, \vv j + \ww j)$. Se $i \neq j$, allora + $\varphi(\vv i ', \vv j ') = 0$, dal momento che i vettori di $\basis$ sono a due a due ortogonali + tra loro. Se invece $i = j$, allora $\varphi(\vv i ', \vv j ') = \varphi(\vv i, \vv i) + \varphi(\ww i, \ww i) = 1-1=0$. Quindi $M_{\basis'}(\restr{\varphi}{W}) = 0$, da cui si conclude che $\restr{\varphi}{W} = 0$. + Pertanto $W(\varphi) \geq i_-(\varphi)$, e quindi $W(\varphi) = i_-(\varphi)$, da cui la tesi. +\end{proof} \ No newline at end of file diff --git a/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/3. Prodotti hermitiani, spazi euclidei e teorema spettrale.tex b/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/3. Prodotti hermitiani, spazi euclidei e teorema spettrale.tex new file mode 100644 index 0000000..4ea65a5 --- /dev/null +++ b/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/3. Prodotti hermitiani, spazi euclidei e teorema spettrale.tex @@ -0,0 +1,1550 @@ +\chapter{Prodotti hermitiani, spazi euclidei e teorema spettrale} + +\begin{note} + Nel corso del documento, per $V$ si intenderà uno spazio vettoriale di dimensione + finita $n$ e per $\varphi$ un suo prodotto, hermitiano o scalare + dipendentemente dal contesto. +\end{note} + +\begin{definition} (prodotto hermitiano) Sia $\KK = \CC$. Una mappa $\varphi : V \times V \to \CC$ si dice \textbf{prodotto hermitiano} se: + + \begin{enumerate}[(i)] + \item $\varphi$ è $\CC$-lineare nel secondo argomento, ossia se $\varphi(\v, \U + \w) = \varphi(\v, \U) + \varphi(\v, \w)$ e + $\varphi(\v, a \w) = a \, \varphi(\v, \w)$, + \item $\varphi(\U, \w) = \conj{\varphi(\w, \U)}$. + \end{enumerate} +\end{definition} + +\begin{definition} (prodotto hermitiano canonico in $\CC^n$) Si definisce + \textbf{prodotto hermitiano canonico} di $\CC^n$ il prodotto $\varphi : \CC^n \times \CC^n \to \CC$ tale per cui, detti $\v = (z_1 \cdots z_n)^\top$ e $\w = (w_1 \cdots w_n)^\top$, $\varphi(\v, \w) = \sum_{i=1}^n \conj{z_i} w_i$. +\end{definition} + +\begin{remark}\nl + \li $\varphi(\U + \w, \v) = \conj{\varphi(\v, \U + \w)} = + \conj{\varphi(\v, \U) + \varphi(\v, \w)} = \conj{\varphi(\v, \U)} + \conj{\varphi(\v, \U)} = \varphi(\w, \v) + \varphi(\U, \v)$, ossia + $\varphi$ è additiva anche nel primo argomento. \\ + \li $\varphi(a \v, \w) = \conj{\varphi(\w, a \v)} = \conj{a} \conj{\varphi(\w, \v)} = \conj{a} \, \varphi(\v, \w)$. \\ + \li $\varphi(\v, \v) = \conj{\varphi(\v, \v)}$, e quindi $\varphi(\v, \v) \in \RR$. \\ + \li Sia $\v = \sum_{i=1}^n x_i \vv i$ e sia $\w = \sum_{i=1}^n y_i \vv i$, allora $\varphi(\v, \w) = \sum_{i =1}^n \sum_{j=1}^n \conj{x_i} y_i \varphi(\vv i, \vv j)$. \\ + \li $\varphi(\v, \w) = 0 \iff \varphi(\w, \v) = 0$. +\end{remark} + +\begin{proposition} + Data la forma quadratica $q : V \to \RR$ del prodotto hermitiano $\varphi$ tale che $q(\v) = \varphi(\v, \v) \in \RR$, tale + forma quadratica individua univocamente il prodotto hermitiano $\varphi$. +\end{proposition} + +\begin{proof} + Innanzitutto si osserva che: + + \[ \varphi(\v, \w) = \frac{\varphi(\v, \w) + \conj{\varphi(\v, \w)}}{2} + \frac{\varphi(\v, \w) . \conj{\varphi(\v, \w)}}{2}. \] + + \vskip 0.05in + + Si considerano allora le due identità: + + \[ q(\v + \w) - q(\v) - q(\w) = + \varphi(\v, \w) + \conj{\varphi(\w, \v)} = 2 \, \Re(\varphi(\v, \w)), \] + + \[ q(i\v + \w) - q(\v) - q(\w) = -i(\varphi(\v, \w) - \conj{\varphi(\v, \w)}) = 2 \, \imm(\varphi(\v, \w)), \] + + \vskip 0.05in + + da cui si conclude che il prodotto $\varphi$ è univocamente + determinato dalla sua forma quadratica. +\end{proof} + +\begin{definition} + Si definisce \textbf{matrice aggiunta} di $A \in M(n, \KK)$ la matrice coniugata della trasposta di $A$, ossia: + + \[ A^* = \conj{A^\top} = \conj{A}^\top. \] +\end{definition} + +\begin{remark} + Per quanto riguarda la matrice aggiunta valgono le principali proprietà della matrice trasposta: + + \begin{itemize} + \item $(A + B)^* = A^* + B^*$, + \item $(AB)^* = B^* A^*$, + \item $(A\inv)^* = (A^*)\inv$, se $A$ è invertibile. + \end{itemize} +\end{remark} + +%TODO: aggiungere tr(conj(A^t) B) + +\begin{definition} (matrice associata del prodotto hermitiano) Analogamente + al caso del prodotto scalare, data una base $\basis = \{\vv 1, \ldots, \vv n\}$ si definisce + come \textbf{matrice associata del prodotto hermitiano} $\varphi$ + la matrice $M_\basis(\varphi) = (\varphi(\vv i, \vv j))_{i,j = 1 \textrm{---} n}$. +\end{definition} + +\begin{remark} + Si osserva che, analogamente al caso del prodotto scalare, vale + la seguente identità: + + \[ \varphi(\v, \w) = [\v]_\basis^* M_\basis(\varphi) [\w]_\basis. \] +\end{remark} + +\begin{proposition} + (formula del cambiamento di base per i prodotto hermitiani) Siano + $\basis$, $\basis'$ due basi di $V$. Allora vale la seguente + identità: + + \[ M_{\basis'} = M_{\basis}^{\basis'}(\Idv)^* M_\basis(\varphi) M_{\basis}^{\basis'}(\Idv). \] +\end{proposition} + +\begin{proof} + Siano $\basis = \{ \vv 1, \ldots, \vv n \}$ e $\basis' = \{ \ww 1, \ldots, \ww n \}$. Allora $\varphi(\ww i, \ww j) = [\ww i]_\basis^* M_\basis(\varphi) [\ww j]_\basis = \left( M_\basis^{\basis'}(\Idv)^i \right)^* M_\basis(\varphi) M_\basis^{\basis'}(\Idv)^j = + \left(M_\basis^{\basis'}(\Idv)\right)^*_i M_\basis(\varphi) M_\basis^{\basis'}(\Idv)^j$, da cui si ricava l'identità + desiderata. +\end{proof} + +\begin{definition} (radicale di un prodotto hermitiano) + Analogamente al caso del prodotto scalare, si definisce il \textbf{radicale} del prodotto $\varphi$ come il seguente sottospazio: + + \[ V^\perp = \{ \v \in V \mid \varphi(\v, \w) = 0 \, \forall \w \in V \}. \] +\end{definition} + +\begin{proposition} + Sia $\basis$ una base di $V$ e $\varphi$ un prodotto hermitiano. Allora $V^\perp = [\cdot]_\basis \inv (\Ker M_\basis(\varphi))$\footnote{Stavolta non è sufficiente considerare la mappa $f : V \to V^*$ tale che $f(\v) = \left[ \w \mapsto \varphi(\v, \w) \right]$, dal momento che $f$ non è lineare, bensì antilineare, ossia $f(a \v) = \conj a f(\v)$.}. +\end{proposition} + +\begin{proof} + Sia $\basis = \{ \vv 1, \ldots, \vv n \}$ e sia $\v \in V^\perp$. + Siano $a_1$, ..., $a_n \in \KK$ tali che $\v = a_1 \vv 1 + \ldots + a_n \vv n$. Allora, poiché $\v \in V$, $0 = \varphi(\vv i, \v) + = a_1 \varphi(\vv i, \vv 1) + \ldots + a_n \varphi(\vv i, \vv n) = M_i [\v]_\basis$, da cui si ricava che $[\v]_\basis \in \Ker M_\basis(\varphi)$, e quindi che $V^\perp \subseteq [\cdot]_\basis \inv (\Ker M_\basis(\varphi))$. \\ + + Sia ora $\v \in V$ tale che $[\v]_\basis \in \Ker M_\basis(\varphi)$. + Allora, per ogni $\w \in V$, $\varphi(\w, \v) = [\w]_\basis^* M_\basis(\varphi) [\v]_\basis = [\w]_\basis^* 0 = 0$, da cui si + conclude che $\v \in V^\perp$, e quindi che $V^\perp \supseteq [\cdot]_\basis \inv (\Ker M_\basis(\varphi))$, da cui + $V^\perp = [\cdot]_\basis \inv (\Ker M_\basis(\varphi))$, ossia + la tesi. +\end{proof} + +\begin{remark} + Come conseguenza della proposizione appena dimostrata, valgono + le principali proprietà già viste per il prodotto scalare. \\ + + \li $\det(M_\basis(\varphi)) = 0 \iff V^\perp \neq \zerovecset \iff \varphi$ è degenere, \\ + \li Vale il teorema di Lagrange, e quindi quello di Sylvester, benché con alcune accortezze: si + introduce, come nel caso di $\RR$, il concetto di segnatura, che diventa l'invariante completo + della nuova congruenza hermitiana, che ancora una volta si dimostra essere una relazione + di equivalenza. \\ + \li Come mostrato nei momenti finali del documento (vd.~\textit{Esercizio 3}), vale + la formula delle dimensioni anche nel caso del prodotto hermitiano. +\end{remark} + +\hr + +\begin{definition} (restrizione ai reali di uno spazio) Sia $V$ + uno spazio vettoriale su $\CC$ con base $\basis$. Si definisce allora lo spazio $V_\RR$, detto + \textbf{spazio di restrizione su $\RR$} di $V$, come uno spazio su $\RR$ generato da + $\basis_\RR = \basis \cup i \basis$. +\end{definition} + +\begin{example} + Si consideri $V = \CC^3$. Una base di $\CC^3$ è chiaramente $\{ \e1, \e2, \e3 \}$. Allora + $V_\RR$ sarà uno spazio vettoriale su $\RR$ generato dai vettori $\{ \e1, \e2, \e3, i\e1, i\e2, i\e3 \}$. +\end{example} + +\begin{remark} + Si osserva che lo spazio di restrizione su $\RR$ e lo spazio di partenza condividono lo stesso insieme + di vettori. Infatti, $\Span_\CC(\basis) = \Span_\RR(\basis \cup i\basis)$. Ciononostante, $\dim V_\RR = 2 \dim V$\footnote{Si sarebbe potuto ottenere lo stesso risultato utilizzando il teorema delle torri algebriche: $[V_\RR : \RR] = [V: \CC] [\CC: \RR] = 2 [V : \CC]$.}, se $\dim V \in \NN$. +\end{remark} + +\begin{definition} (complessificazione di uno spazio) Sia $V$ uno spazio vettoriale su $\RR$. + Si definisce allora lo \textbf{spazio complessificato} $V_\CC = V \times V$ su $\CC$ con le seguenti operazioni: + + \begin{itemize} + \item $(\v, \w) + (\v', \w') = (\v + \v', \w + \w')$, + \item $(a+bi)(\v, \w) = (a\v - b\w, a\w + b\v)$. + \end{itemize} +\end{definition} + +\begin{remark} + La costruzione dello spazio complessificato emula in realtà la costruzione di $\CC$ come spazio + $\RR \times \RR$. Infatti se $z = (c, d)$, vale che $(a + bi)(c, d) = (ac - bd, ad + bc)$, mentre + si mantiene l'usuale operazione di addizione. In particolare si può identificare l'insieme + $V \times \zerovecset$ come $V$, mentre $\zerovecset \times V$ viene identificato come l'insieme + degli immaginari $iV$ di $V_\CC$. Infine, moltiplicare per uno scalare reale un elemento di + $V \times \zerovecset$ equivale a moltiplicare la sola prima componente con l'usuale operazione + di moltiplicazione di $V$. Allora, come accade per $\CC$, si può sostituire la notazione + $(\v, \w)$ con la più comoda notazione $\v + i \w$. +\end{remark} + +\begin{remark} + Sia $\basis = \{ \vv 1, \ldots, \vv n \}$ una base di $V$. Innanzitutto si osserva che + $(a+bi)(\v, \vec 0) = (a\v, b\v)$. Pertanto si può concludere che $\basis \times \zerovecset$ è + una base dello spazio complessificato $V_\CC$ su $\CC$. \\ + + Infatti, se $(a_1 + b_1 i)(\vv 1, \vec 0) + \ldots + (a_n + b_n i)(\vv n, \vec 0) = (\vec 0, \vec 0)$, + allora $(a_1 \vv 1 + \ldots + a_n \vv n, b_1 \vv 1 + \ldots + b_n \vv n) = (\vec 0, \vec 0)$. + Poiché però $\basis$ è linearmente indipendente per ipotesi, l'ultima identità implica che + $a_1 = \cdots = a_n = b_1 = \cdots = b_n = 0$, e quindi che $\basis \times \zerovecset$ è linearmente + indipendente. \\ + + Inoltre $\basis \times \zerovecset$ genera $V_\CC$. Se infatti $\v = (\U, \w)$, e vale che: + + \[ \U = a_1 \vv 1 + \ldots + a_n \vv n, \quad \w = b_1 \vv 1 + \ldots + b_n \vv n, \] + + \vskip 0.1in + + allora $\v = (a_1 + b_1 i) (\vv 1, \vec 0) + \ldots + (a_n + b_n i) (\vv n, \vec 0)$. Quindi + $\dim V_\CC = \dim V$. +\end{remark} + +\begin{definition} + Sia $f$ un'applicazione $\CC$-lineare di $V$ spazio vettoriale su $\CC$. Allora + si definisce la \textbf{restrizione su} $\RR$ di $f$, detta $f_\RR : V_\RR \to V_\RR$, + in modo tale che $f_\RR(\v) = f(\v)$. +\end{definition} + +\begin{remark} + Sia $\basis = \{\vv 1, \ldots, \vv n\}$ una base di $V$ su $\CC$. Sia $A = M_\basis(f)$. Si + osserva allora che, se $\basis' = \basis \cup i \basis$ e $A = A' + i A''$ con $A'$, $A'' \in M(n, \RR)$, + vale la seguente identità: + + \[ M_{\basis'}(f_\RR) = \Matrix{ A' & \rvline & -A'' \\ \hline A'' & \rvline & A' }. \] + + Infatti, se $f(\vv i) = (a_1 + b_1 i) \vv 1 + \ldots + (a_n + b_n i) \vv n$, vale che + $f_\RR(\vv i) = a_1 \vv 1 + \ldots + a_n \vv n + b_1 (i \vv 1) + \ldots + b_n (i \vv n)$, + mentre $f_\RR(i \vv i) = i f(\vv i) = - b_1 \vv 1 + \ldots - b_n \vv n + a_1 (i \vv 1) + \ldots + a_n (i \vv n)$. +\end{remark} + +\begin{definition} + Sia $f$ un'applicazione $\RR$-lineare di $V$ spazio vettoriale su $\RR$. Allora + si definisce la \textbf{complessificazione} di $f$, detta $f_\CC : V_\CC \to V_\CC$, + in modo tale che $f_\CC(\v + i \w) = f(\v) + i f(\w)$. +\end{definition} + +\begin{remark} + Si verifica infatti che $f_\CC$ è $\CC$-lineare. + \begin{itemize} + \item $f_\CC((\vv1 + i \ww1) + (\vv2 + i \ww2)) = f_\CC((\vv1 + \vv2) + i (\ww1 + \ww2)) = + f(\vv1 + \vv2) + i f(\ww1 + \ww2) = (f(\vv1) + i f(\ww1)) + (f(\vv2) + i f(\ww2)) = + f_\CC(\vv1 + i\ww1) + f_\CC(\vv2 + i\ww2)$. + + \item $f_\CC((a+bi)(\v + i\w)) = f_\CC(a\v-b\w + i(a\w+b\v)) = f(a\v - b\w) + i f(a\w + b\v) = + af(\v) - bf(\w) + i(af(\w) + bf(\v)) = (a+bi)(f(\v) + if(\w)) = (a+bi) f_\CC(\v + i\w)$. + \end{itemize} +\end{remark} + +\begin{proposition} + Sia $f_\CC$ la complessificazione di $f \in \End(V)$, dove $V$ è uno spazio vettoriale su $\RR$. + Sia inoltre $\basis = \{ \vv 1, \ldots, \vv n \}$ una base di $V$. Valgono allora i seguenti risultati: + + \begin{enumerate}[(i)] + \item $\restr{(f_\CC)_\RR}{V}$ assume gli stessi valori di $f$, + \item $M_\basis(f_\CC) = M_\basis(f) \in M(n, \RR)$, + \item $M_{\basis \cup i \basis}((f_\CC)_\RR) = \Matrix{M_\basis(f) & \rvline & 0 \\ \hline 0 & \rvline & M_\basis(f)}$. + \end{enumerate} +\end{proposition} + +\begin{proof}Si dimostrano i risultati separatamente. + \begin{enumerate}[(i)] + \item Si osserva che $(f_\CC)_\RR(\vv i) = f_\CC(\vv i) = f(\vv i)$. Dal momento che + $(f_\CC)_\RR$ è $\RR$-lineare, si conclude che $(f_\CC)_\RR$ assume gli stessi valori + di $f$. + + \item Dal momento che $\basis$, nell'identificazione di $(\v, \vec 0)$ come $\v$, è + sempre una base di $V_\CC$, e $f_\CC(\vv i) = f(\vv i)$, chiaramente + $[f_\CC(\vv i)]_\basis = [f(\vv i)]_\basis$, e quindi $M_\basis(f_\CC) = M_\basis(f)$, + dove si osserva anche che $M_\basis(f) \in M(n, \RR)$, essendo $V$ uno spazio vettoriale + su $\RR$. + + \item Sia $f(\vv i) = a_1 \vv 1 + \ldots + a_n \vv n$ con $a_1$, ..., $a_n \in \RR$. Come + osservato in (i), $\restr{(f_\CC)_\RR}{\basis} = \restr{(f_\CC)_\RR}{\basis}$, e quindi + la prima metà di $M_{\basis \cup i \basis}((f_\CC)_\RR)$ è formata da due blocchi: uno + verticale coincidente con $M_\basis(f)$ e un altro completamente nullo, dal momento che + non compare alcun termine di $i \basis$ nella scrittura di $(f_\CC)_\RR(\vv i)$. Al + contrario, per $i \basis$, $(f_\CC)_\RR(i \vv i) = f_\CC(i \vv i) = i f(\vv i) = a_1 (i \vv 1) + + \ldots + a_n (i \vv n)$; pertanto la seconda metà della matrice avrà i due blocchi della prima metà, + benché scambiati. + \end{enumerate} +\end{proof} + +\begin{remark} + Dal momento che $M_\basis(f_\CC) = M_\basis(f)$, $f_\CC$ e $f$ condividono lo stesso polinomio caratteristico + e vale che $\Sp(f) \subseteq \Sp(f_\CC)$, dove vale l'uguaglianza se e solo se tale polinomio caratteristico + è completamente riducibile in $\RR$. Inoltre, se $V_\lambda$ è l'autospazio su $V$ dell'autovalore $\lambda$, l'autospazio + su $V_\CC$, rispetto a $f_\CC$, è invece ${V_\CC}_\lambda = V_\lambda + i V_\lambda$, la cui + dimensione rimane invariata rispetto a $V_\lambda$, ossia $\dim V_\lambda = \dim {V_\CC}_\lambda$ + (infatti, analogamente a prima, una base di $V_\lambda$ può essere identificata come base + anche per ${V_\CC}_\lambda$). +\end{remark} + +\begin{proposition} + Sia $f_\CC$ la complessificazione di $f \in \End(V)$, dove $V$ è uno spazio vettoriale su $\RR$. + Sia inoltre $\basis = \{ \vv 1, \ldots, \vv n \}$ una base di $V$. Allora un endomorfismo + $\tilde g : V_\CC \to V_\CC$ complessifica un endomorfismo $g \in \End(V)$ $\iff$ $M_\basis(\tilde g) \in M(n, \RR)$. +\end{proposition} + +\begin{proof} + Se $\tilde g$ complessifica $g \in \End(V)$, allora, per la proposizione precedente, + $M_\basis(\tilde g) = M_\basis(g) \in M(n, \RR)$. Se invece $A = M_\basis(\tilde g) \in M(n, \RR)$, + si considera $g = M_\basis\inv(A) \in \End(V)$. Si verifica facilemente che $\tilde g$ non è altro che + il complessificato di tale $g$: + + \begin{itemize} + \item $\tilde g (\vv i) = g(\vv i)$, dove l'uguaglianza è data dal confronto delle matrici associate, + e quindi $\restr{\tilde g}{V} = g$; + \item $\tilde g(\v + i\w) = \tilde g(\v) + i \tilde g(\w) = g(\v) + i g(\w)$, da cui la tesi. + \end{itemize} +\end{proof} + +\begin{proposition} + Sia $\varphi$ un prodotto scalare di $V$ spazio vettoriale su $\RR$. Allora esiste un + unico prodotto hermitiano $\varphi_\CC : V_\CC \times V_\CC \to \CC$ che estende $\varphi$ (ossia tale che + $\restr{\varphi_\CC}{V \times V} = \varphi$), il quale assume la stessa segnatura + di $\varphi$. +\end{proposition} + +\begin{proof} + Sia $\basis$ una base di Sylvester per $\varphi$. Si consideri allora il prodotto + $\varphi_\CC$ tale che: + + \[ \varphi_\CC(\vv1 + i\ww1, \vv2 + i\ww2) = \varphi(\vv1, \vv2) + \varphi(\ww1, \ww2) + i(\varphi(\vv1, \ww1) - \varphi(\ww1, \vv2)). \] + + Chiaramente $\restr{\varphi_\CC}{V \times V} = \varphi$. Si verifica allora che $\varphi_\CC$ è hermitiano: + + \begin{itemize} + \item $\varphi_\CC(\v + i\w, (\vv1 + i\ww1) + (\vv2 + i\ww2))$ $= \varphi(\v, \vv1 + \vv2) + \varphi(\w, \ww1 + \ww2)$ $+ i(\varphi(\v, \ww1 + \ww2)$ $- \varphi(\w, \vv1 + \vv2))$ $= [\varphi(\v, \vv1) + \varphi(\w, \ww1) + i(\varphi(\v, \ww1) - \varphi(\w, \vv1))]$ $+ [\varphi(\v, \vv2) + \varphi(\w, \ww2) + i(\varphi(\v, \ww2) - \varphi(\w, \vv2))] = \varphi_\CC(\v + i\w, \vv1 + i\ww1) + + \varphi_\CC(\v + i\w, \vv2 + i\ww2)$ (additività nel secondo argomento), + + \item $\varphi_\CC(\v + i\w, (a+bi)(\vv1 + i\ww1)) = \varphi_\CC(\v + i\w, a\vv1-b\ww1 + i(b\vv1+a\ww1)) = + \varphi(\v, a\vv1-b\ww1) + \varphi(\w, b\vv1+a\ww1) + i(\varphi(\v, b\vv1+a\ww1) - \varphi(\w, a\vv1-b\ww1))= + a\varphi(\v, \vv1) - b\varphi(\v, \ww1) + b\varphi(\w, \vv1) + a\varphi(\w, \ww1) + i(b\varphi(\v, \vv1) + a\varphi(\v, \ww1) - a\varphi(\w, \vv1) + b\varphi(\w, \ww1)) = a(\varphi(\v, \vv1) + \varphi(\w, \ww1)) - + b(\varphi(\v, \ww1) - \varphi(\w, \vv1)) + i(a(\varphi(\v, \ww1) - \varphi(\w, \vv1)) + b(\varphi(\v, \vv1) + \varphi(\w, \ww1))) = (a+bi)(\varphi(\v, \vv1) + \varphi(\w, \ww1) + i(\varphi(\v, \ww1) - \varphi(\w, \vv1))) = (a+bi) \varphi_\CC(\v + \w, \vv1 + i\ww1)$ (omogeneità nel secondo argomento), + + \item $\varphi_\CC(\vv1 + i\ww1, \vv2 + i\ww2) = \varphi(\vv1, \vv2) + \varphi(\ww1, \ww2) + i(\varphi(\vv1, \ww2) - \varphi(\ww1, \vv2)) = \conj{\varphi(\vv1, \vv2) + \varphi(\ww1, \ww2) + i(\varphi(\ww1, \vv2) - \varphi(\vv1, \ww2))} = \conj{\varphi(\vv2, \vv1) + \varphi(\ww2, \ww1) + i(\varphi(\vv2, \ww1) - \varphi(\ww2, \vv1))} = \conj{\varphi_\CC(\vv2 + \ww2, \vv1 + \ww1)}$ (coniugio nello scambio degli argomenti). + \end{itemize} + + Ogni prodotto hermitiano $\tau$ che estende il prodotto scalare $\varphi$ ha la stessa matrice associata nella + base $\basis$, essendo $\tau(\vv i, \vv i) = \varphi(\vv i, \vv i)$ vero per ipotesi. Pertanto $\tau$ è + unico, e vale che $\tau = \varphi_\CC$. Dal momento che $M_\basis(\varphi_\CC) = M_\basis(\varphi)$ è + una matrice di Sylvester, $\varphi_\CC$ mantiene anche la stessa segnatura di $\varphi$. +\end{proof} + +\hr + +\begin{theorem} (di rappresentazione di Riesz per il prodotto scalare) + Sia $V$ uno spazio vettoriale e sia $\varphi$ un suo prodotto scalare + non degenere. Allora per ogni $f \in V^*$ esiste un unico $\v \in V$ tale che + $f(\w) = \varphi(\v, \w)$ $\forall \w \in V$. +\end{theorem} + +\begin{proof} + Si consideri l'applicazione $a_\varphi$. Poiché $\varphi$ non è degenere, $\Ker a_\varphi = V^\perp = \zerovecset$, da cui si deduce che $a_\varphi$ è un isomorfismo. Quindi $\forall f \in V^*$ esiste + un unico $\v \in V$ tale per cui $a_\varphi(\v) = f$, e dunque tale per cui $\varphi(\v, \w) = a_\varphi(\v)(\w) = f(\w)$ $\forall \w \in V$. +\end{proof} + +\begin{proof}[Dimostrazione costruttiva] + Sia $\basis = \{ \vv 1, \ldots, \vv n \}$ una base ortogonale di $V$ per $\varphi$. Allora $\basis^*$ è una base di $V^*$. In + particolare $f = f(\vv 1) \vec{v_1^*} + \ldots + f(\vv n) \vec{v_n^*}$. Sia $\v = \frac{f(\vv 1)}{\varphi(\vv 1, \vv 1)} \vv 1 + \ldots + \frac{f(\vv n)}{\varphi(\vv n, \vv n)}$. Detto $\w = a_1 \vv 1 + \ldots + a_n \vv n$, + si deduce che $\varphi(\v, \w) = a_1 f(\vv 1) + \ldots + a_n f(\vv n) = f(\w)$. Se esistesse $\v' \in V$ con + la stessa proprietà di $\v$, $\varphi(\v, \w) = \varphi(\v', \w) \implies \varphi(\v - \v', \w)$ $\forall \w \in V$. Si deduce dunque che $\v - \v' \in V^\perp$, contenente solo $\vec 0$ dacché $\varphi$ è non degenere; + e quindi si conclude che $\v = \v'$, ossia che esiste solo un vettore con la stessa proprietà di $\v$. +\end{proof} + +\begin{theorem} (di rappresentazione di Riesz per il prodotto hermitiano) + Sia $V$ uno spazio vettoriale su $\CC$ e sia $\varphi$ un suo prodotto hermitiano non + degenere. Allora per ogni $f \in V^*$ esiste un unico $\v \in V$ tale che + $f(\w) = \varphi(\v, \w)$ $\forall \w \in V$. +\end{theorem} + +\begin{proof} + Sia $\basis = \{ \vv 1, \ldots, \vv n \}$ una base ortogonale di $V$ per $\varphi$. Allora $\basis^*$ è una base di $V^*$. In + particolare $f = f(\vv 1) \vec{v_1^*} + \ldots + f(\vv n) \vec{v_n^*}$. Sia $\v = \frac{\conj{f(\vv 1)}}{\varphi(\vv 1, \vv 1)} \vv 1 + \ldots + \frac{\conj{f(\vv n)}}{\varphi(\vv n, \vv n)}$. Detto $\w = a_1 \vv 1 + \ldots + a_n \vv n$, + si deduce che $\varphi(\v, \w) = a_1 f(\vv 1) + \ldots + a_n f(\vv n) = f(\w)$. Se esistesse $\v' \in V$ con + la stessa proprietà di $\v$, $\varphi(\v, \w) = \varphi(\v', \w) \implies \varphi(\v - \v', \w)$ $\forall \w \in V$. Si deduce dunque che $\v - \v' \in V^\perp$, contenente solo $\vec 0$ dacché $\varphi$ è non degenere; + e quindi si conclude che $\v = \v'$, ossia che esiste solo un vettore con la stessa proprietà di $\v$. +\end{proof} + +\begin{proposition} + Sia $V$ uno spazio vettoriale con prodotto scalare $\varphi$ non degenere. + Sia $f \in \End(V)$. Allora esiste un unico endomorfismo + $f_\varphi^\top : V \to V$, detto il \textbf{trasposto di} $f$ e indicato con $f^\top$ in assenza + di ambiguità\footnote{Si tenga infatti in conto della differenza tra $f_\varphi^\top : V \to V$, di cui si discute + nell'enunciato, e $f^\top : V^* \to V^*$ che invece è tale che $f^top(g) = g \circ f$.}, tale che: + + \[ a_\varphi \circ g = f^\top \circ a_\varphi, \] + + \vskip 0.05in + + ossia che: + + \[ \varphi(\v, f(\w)) = \varphi(g(\v), \w) \, \forall \v, \w \in V. \] +\end{proposition} + +\begin{proof} + Si consideri $(f^\top \circ a_\varphi)(\v) \in V^*$. Per il teorema di rappresentazione di Riesz per + il prodotto scalare, esiste un unico $\v'$ tale che $(f^\top \circ a_\varphi)(\v)(\w) = \varphi(\v', \w) \implies \varphi(\v, f(\w)) = \varphi(\v', \w)$ $\forall \w \in V$. Si costruisce allora una mappa + $f_\varphi^\top : V \to V$ che associa a $\v$ tale $\v'$. Si dimostra che $f_\varphi^\top$ è un'applicazione lineare, e che + dunque è un endomorfismo: + + \begin{enumerate}[(i)] + \item Siano $\vv 1$, $\vv 2 \in V$. Si deve dimostrare innanzitutto che $f_\varphi^\top(\vv 1 + \vv 2) = f_\varphi^\top(\vv 1) + f_\varphi^\top(\vv 2)$, ossia che $\varphi(f_\varphi^\top(\vv 1) + f_\varphi^\top(\vv 2), \w) = \varphi(\vv 1 + \vv 2, f(\w))$ $\forall \w \in V$. \\ + + Si osservano le seguenti identità: + \begin{align*} + &\varphi(\vv 1 + \vv 2, f(\w)) = \varphi(\vv 1, f(\w)) + \varphi(\vv 2, f(\w)) = (*), \\ + &\varphi(f_\varphi^\top(\vv 1) + f_\varphi^\top(\vv 2), \w) = \varphi(f_\varphi^\top(\vv 1), \w) + \varphi(f_\varphi^\top(\vv 2), \w) = (*), + \end{align*} + + da cui si deduce l'uguaglianza desiderata, essendo $f_\varphi^\top(\vv 1 + \vv 2)$ l'unico vettore di $V$ + con la proprietà enunciata dal teorema di rappresentazione di Riesz. + + \item Sia $\v \in V$. Si deve dimostrare che $f_\varphi^\top(a \v) = a f_\varphi^\top(\v)$, ossia che $\varphi(a f_\varphi^\top(\v), \w) = + \varphi(a\v, f(\w))$ $\forall a \in \KK$, $\w \in V$. È + sufficiente moltiplicare per $a$ l'identità $\varphi(f_\varphi^\top(\v), \w) = \varphi(\v, f(\w))$. Analogamente + a prima, si deduce che $f_\varphi^\top(a \v) = a f_\varphi^\top(\v)$, essendo $f_\varphi^\top(a \v)$ l'unico vettore di $V$ con la + proprietà enunciata dal teorema di rappresentazione di Riesz. + \end{enumerate} + + Infine si dimostra che $f_\varphi^\top$ è unico. Sia infatti $g$ un endomorfismo di $V$ che condivide la stessa + proprietà di $f_\varphi^\top$. Allora $\varphi(f_\varphi^\top(\v), \w) = \varphi(\v, f(\w)) = \varphi(g(\v), \w)$ $\forall \v$, $\w \in V$, da cui si deduce che $\varphi(f_\varphi^\top(\v) - '(\v), \w) = 0$ $\forall \v$, $\w \in V$, ossia che + $f_\varphi^\top(\v) - g(\v) \in V^\perp$ $\forall \v \in V$. Tuttavia $\varphi$ è non degenere, e quindi $V^\perp = \zerovecset$, da cui si deduce che deve valere l'identità $f_\varphi^\top(\v) = g(\v)$ $\forall \v \in V$, ossia + $g = f_\varphi^\top$. +\end{proof} + +\begin{proposition} + Sia $V$ uno spazio vettoriale su $\CC$ e sia $\varphi$ un suo prodotto hermitiano. Allora esiste un'unica + mappa\footnote{Si osservi che $f^*$ non è un'applicazione lineare, benché sia invece \textit{antilineare}.} $f^* : V \to V$, detta \textbf{aggiunto di} $f$, tale che $\varphi(\v, f(\w)) = \varphi(f^*(\v), \w)$ $\forall \v$, $\w \in V$. +\end{proposition} + +\begin{proof} + Sia $\v \in V$. Si consideri il funzionale $\sigma$ tale che $\sigma(\w) = \varphi(\v, f(\w))$. Per il + teorema di rappresentazione di Riesz per il prodotto scalare esiste un unico $\v' \in V$ tale per cui + $\varphi(\v, f(\w)) = \sigma(\w) = \varphi(\v', \w)$. Si costruisce allora una mappa $f^*$ che associa + $\v$ a tale $\v'$. \\ + + Si dimostra infine che la mappa $f^*$ è unica. Sia infatti $\mu : V \to V$ che condivide la stessa + proprietà di $f^*$. Allora $\varphi(f^*(\v), \w) = \varphi(\v, f(\w)) = \varphi(\mu(\v), \w)$ $\forall \v$, $\w \in V$, da cui si deduce che $\varphi(f^*(\v) - \mu(\v), \w) = 0$ $\forall \v$, $\w \in V$, ossia che + $f^*(\v) - \mu(\v) \in V^\perp$ $\forall \v \in V$. Tuttavia $\varphi$ è non degenere, e quindi $V^\perp = \zerovecset$, da cui si deduce che deve valere l'identità $f^*(\v) = \mu(\v)$ $\forall \v \in V$, ossia + $\mu = f^*$. +\end{proof} + +\begin{remark} + L'operazione di trasposizione di un endomorfismo sul prodotto scalare non degenere $\varphi$ è un'involuzione. Infatti valgono + le seguenti identità $\forall \v$, $\w \in V$: + + \[ \system{\varphi(\w, f^\top(\v)) = \varphi(f^\top(\v), \w) = \varphi(\v, f(\w)), \\ \varphi(\w, f^\top(\v)) = \varphi((f^\top)^\top(\w), \v) = + \varphi(\v, (f^\top)^\top(\w)).} \] + + \vskip 0.05in + + Si conclude allora, poiché $\varphi$ è non degenere, che + $f(\w) = (f^\top)^\top(\w)$ $\forall \w \in V$, ossia che $f = (f^\top)^\top$. +\end{remark} + +\begin{remark} + Analogamente si può dire per l'operazione di aggiunta per un prodotto hermitiano $\varphi$ non degenere. + Valgono infatti le seguenti identità $\forall \v$, $\w \in V$: + + \[ \system{\conj{\varphi(\w, f^*(\v))} = \varphi(f^*(\v), \w) = \varphi(\v, f(\w)), \\ \conj{\varphi(\w, f^*(\v))} = \conj{\varphi((f^*)^*(\w), \v)} = + \varphi(\v, (f^*)^*(\w)),} \] + + \vskip 0.05in + + da cui si deduce, come prima, che $f = (f^*)^*$. +\end{remark} + +\begin{definition} (base ortonormale) + Si definisce \textbf{base ortonormale} di uno spazio vettoriale $V$ su un suo prodotto $\varphi$ + una base ortogonale $\basis = \{ \vv 1, \ldots, \vv n \}$ tale che $\varphi(\vv i, \vv j) = \delta_{ij}$. +\end{definition} + +\begin{proposition} + Sia $\varphi$ un prodotto scalare non degenere di $V$. Sia $f \in \End(V)$. Allora + vale la seguente identità: + + \[ M_\basis(f_\varphi^\top) = M_\basis(\varphi)\inv M_\basis(f)^\top M_\basis(\varphi), \] + + dove $\basis$ è una base di $V$. +\end{proposition} + +\begin{proof} + Sia $\basis^*$ la base relativa a $\basis$ in $V^*$. Per la proposizione precedente vale la seguente identità: + + \[ a_\varphi \circ f_\varphi^\top = f^\top \circ a_\varphi. \] + + Pertanto, passando alle matrici associate, si ricava che: + + \[ M_{\basis^*}^\basis(a_\varphi) M_\basis(f_\varphi^\top) = M_{\basis^*}(f^\top) M_{\basis^*}^\basis(a_\varphi). \] + + Dal momento che valgono le seguenti due identità: + + \[ M_{\basis^*}^\basis(a_\varphi) = M_\basis(\varphi), \qquad M_{\basis^*}(f^\top) = M_\basis(f)^\top, \] + + e $a_\varphi$ è invertibile (per cui anche $M_\basis(\varphi)$ lo è), si conclude che: + + \[ M_\basis(\varphi) M_\basis(f_\varphi^\top) = M_\basis(f)^\top M_\basis(\varphi) \implies M_\basis(f_\varphi^\top) = M_\basis(\varphi)\inv M_\basis(f)^\top M_\basis(\varphi), \] + + da cui la tesi. +\end{proof} + +\begin{corollary} Sia $\varphi$ un prodotto scalare di $V$. + Se $\basis$ è una base ortonormale, $\varphi$ è non degenere e $M_\basis(f_\varphi^\top) = M_\basis(f)^\top$. +\end{corollary} + +\begin{proof} + Se $\basis$ è una base ortonormale, $M_\basis(\varphi) = I_n$. Pertanto $\varphi$ è + non degenere. Allora, per la proposizione precedente: + + \[ M_\basis(f_\varphi^\top) = M_\basis(\varphi)\inv M_\basis(f)^\top M_\basis(\varphi) = M_\basis(f)^\top. \] +\end{proof} + +\begin{proposition} + Sia $\varphi$ un prodotto hermitiano non degenere di $V$. Sia $f \in \End(V)$. Allora + vale la seguente identità: + + \[ M_\basis(f_\varphi^*) = M_\basis(\varphi)\inv M_\basis(f)^* M_\basis(\varphi), \] + + dove $\basis$ è una base di $V$. +\end{proposition} + +\begin{proof} Sia $\basis = \{ \vv 1, \ldots, \vv n\}$. + Dal momento che $\varphi$ è non degenere, $\Ker M_\basis(\varphi) = V^\perp = \zerovecset$, e quindi + $M_\basis(\varphi)$ è invertibile. \\ + + Dacché allora $\varphi(f^*(\v), \w) = \varphi(\v, f(\w))$ $\forall \v$, $\w \in V$, + vale la seguente identità: + + \[ [f^*(\v)]_\basis^* M_\basis(\varphi) [\w]_\basis = [\v]_\basis^* M_\basis(\varphi) [f(\w)]_\basis, \] + + ossia si deduce che: + + \[ [\v]_\basis^* M_\basis(f^*)^* M_\basis(\varphi) [\w]_\basis = [\v]_\basis^* M_\basis(\varphi) M_\basis(f) [\w]_\basis. \] + + Sostituendo allora a $\v$ e $\w$ i vettori della base $\basis$, si ottiene che: + + \begin{gather*} + (M_\basis(f^*)^* M_\basis(\varphi))_{ij} = [\vv i]_\basis^* M_\basis(f^*)^* M_\basis(\varphi) [\vv j]_\basis = \\ = [\vv i]_\basis^* M_\basis(\varphi) M_\basis(f) [\vv j]_\basis = (M_\basis(\varphi) M_\basis(f))_{ij}, + \end{gather*} + + e quindi che $M_\basis(f^*)^* M_\basis(\varphi) = M_\basis(\varphi) M_\basis(f)$. Moltiplicando + a destra per l'inversa di $M_\basis(\varphi)$ e prendendo l'aggiunta di ambo i membri (ricordando + che $M_\basis(\varphi)^* = M_\basis(\varphi)$, essendo $\varphi$ un prodotto hermitiano), si ricava + l'identità desiderata. + +\end{proof} + +\begin{corollary} Sia $\varphi$ un prodotto hermitiano di $V$ spazio vettoriale su $\CC$. + Se $\basis$ è una base ortonormale, $\varphi$ è non degenere e $M_\basis(f_\varphi^*) = M_\basis(f)^*$. +\end{corollary} + +\begin{proof} + Se $\basis$ è una base ortonormale, $M_\basis(\varphi) = I_n$. Pertanto $\varphi$ è + non degenere. Allora, per la proposizione precedente: + + \[ M_\basis(f_\varphi^*) = M_\basis(\varphi)\inv M_\basis(f)^* M_\basis(\varphi) = M_\basis(f)^*. \] +\end{proof} + +\begin{note} + D'ora in poi, nel corso del documento, s'intenderà per $\varphi$ un prodotto scalare (o eventualmente hermitiano) non degenere di $V$. +\end{note} + +\begin{definition} (operatori simmetrici) + Sia $f \in \End(V)$. Si dice allora che $f$ è \textbf{simmetrico} (o \textit{autoaggiunto}) se $f = f^\top$. +\end{definition} + +\begin{definition} (applicazioni e matrici ortogonali) + Sia $f \in \End(V)$. Si dice allora che $f$ è \textbf{ortogonale} se $\varphi(\v, \w) = \varphi(f(\v), f(\w))$, + ossia se è un'isometria in $V$. + Sia $A \in M(n, \KK)$. Si dice dunque che $A$ è \textbf{ortogonale} se $A^\top A = A A^\top = I_n$. +\end{definition} + +\begin{definition} + Le matrici ortogonali di $M(n, \KK)$ formano un sottogruppo moltiplicativo di $\GL(n, \KK)$, detto \textbf{gruppo ortogonale}, + e indicato con $O_n$. Il sottogruppo di $O_n$ contenente solo le matrici con determinante pari a $1$ è + detto \textbf{gruppo ortogonale speciale}, e si denota con $SO_n$. +\end{definition} + +\begin{remark} + Si possono classificare in modo semplice alcuni di questi gruppi ortogonali per $\KK = \RR$. \\ + + \li $A \in O_n \implies 1 = \det(I_n) = \det(A A^\top) = \det(A)^2 \implies \det(A) = \pm 1$. + \li $A = (a) \in O_1 \iff A^\top A = I_1 \iff a^2 = 1 \iff a = \pm 1$, da cui si ricava che l'unica matrice + di $SO_1$ è $(1)$. Si osserva inoltre che $O_1$ è abeliano di ordine $2$, e quindi che $O_1 \cong \ZZ/2\ZZ$. \\ + \li $A = \Matrix{a & b \\ c & d} \in O_2 \iff \Matrix{a^2 + b^2 & ab + cd \\ ab + cd & c^2 + d^2} = A^\top A = I_2.$ \\ + + Pertanto deve essere soddisfatto il seguente sistema di equazioni: + + \[ \system{a^2 + b^2 = c^2 + d^2 = 1, \\ ac + bd = 0.} \] + + Si ricava dunque che si può identificare + $A$ con le funzioni trigonometriche $\cos(\theta)$ e $\sin(\theta)$ con $\theta \in [0, 2\pi)$ nelle due forme: + \begin{align*} + &A = \Matrix{\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)} \quad &\text{(}\!\det(A) = 1, A \in SO_2\text{)}, \\ + &A = \Matrix{\cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta)} \quad &\text{(}\!\det(A) = -1\text{)}. + \end{align*} +\end{remark} + +\begin{definition} (applicazioni e matrici hermitiane) + Sia $f \in \End(V)$ e si consideri il prodotto hermitiano $\varphi$. Si dice allora che + $f$ è \textbf{hermitiano} se $f = f^*$. Sia $A \in M(n, \CC)$. Si dice dunque che $A$ + è \textbf{hermitiana} se $A = A^*$. +\end{definition} + +\begin{definition} (applicazioni e matrici unitarie) + Sia $f \in \End(V)$ e si consideri il prodotto hermitiano $\varphi$. Si dice allora che + $f$ è \textbf{unitario} se $\varphi(\v, \w) = \varphi(f(\v), f(\w))$. Sia $A \in M(n, \CC)$. + Si dice dunque che $A$ è \textbf{unitaria} se $A^* A = A A^* = I_n$. +\end{definition} + +\begin{definition} + Le matrici unitarie di $M(n, \CC)$ formano un sottogruppo moltiplicativo di $\GL(n, \CC)$, detto \textbf{gruppo unitario}, + e indicato con $U_n$. Il sottogruppo di $U_n$ contenente solo le matrici con determinante pari a $1$ è + detto \textbf{gruppo unitario speciale}, e si denota con $SU_n$. +\end{definition} + +\begin{remark}\nl + Si possono classificare in modo semplice alcuni di questi gruppi unitari. + + \li $A \in U_n \implies 1 = \det(I_n) = \det(A A^*) = \det(A) \conj{\det(A)} = \abs{\det(A)}^2 = 1$. + \li $A = (a) \in U_1 \iff A^* A = I_1 \iff \abs{a}^2 = 1 \iff a = e^{i\theta}$, $\theta \in [0, 2\pi)$, ossia il numero complesso $a$ appartiene alla circonferenza di raggio unitario. + \li $A = \Matrix{a & b \\ c & d} \in SU_2 \iff A A^* = \Matrix{\abs{a}^2 + \abs{b}^2 & a\conj c + b \conj d \\ \conj a c + \conj b d & \abs{c}^2 + \abs{d}^2} = I_2$, $\det(A) = 1$, ossia se il seguente + sistema di equazioni è soddisfatto: + + \[ \system{\abs{a}^2 + \abs{b}^2 = \abs{c}^2 + \abs{d}^2 = 1, \\ a\conj c + b \conj d = 0, \\ ad-bc=1,} \] + + le cui soluzioni riassumono il gruppo $SU_2$ nel seguente modo: + + \[ SU_2 = \left\{ \Matrix{x & -y \\ \conj y & \conj x} \in M(2, \CC) \;\middle\vert\; \abs{x}^2 + \abs{y}^2 = 1 \right\}. \] + +\end{remark} + +\begin{definition} (spazio euclideo reale) + Si definisce \textbf{spazio euclideo reale} uno spazio vettoriale $V$ su $\RR$ dotato + del prodotto scalare standard $\varphi = \innprod{\cdot, \cdot}$. +\end{definition} + +\begin{definition} (spazio euclideo complesso) + Si definisce \textbf{spazio euclideo complesso} uno spazio vettoriale $V$ su $\CC$ dotato + del prodotto hermitiano standard $\varphi = \innprod{\cdot, \cdot}$. +\end{definition} + +\begin{proposition} + Sia $(V, \varphi)$ uno spazio euclideo reale e sia $\basis$ una base ortonormale di $V$. Allora $f \in \End(V)$ è simmetrico $\iff$ $M_\basis(f) = M_\basis(f)^\top$ $\iff$ $M_\basis(f)$ è simmetrica. +\end{proposition} + +\begin{proof} + Per il corollario precedente, $f$ è simmetrico $\iff f = f^\top \iff M_\basis(f) = M_\basis(f^\top) = + M_\basis(f)^\top$. +\end{proof} + +\begin{proposition} + Sia $(V, \varphi)$ uno spazio euclideo reale e sia $\basis$ una base ortonormale di $V$. Allora + $f \in \End(V)$ è ortogonale $\iff$ $M_\basis(f) M_\basis(f)^\top = M_\basis(f)^\top M_\basis(f) = I_n$ $\defiff$ $M_\basis(f)$ è ortogonale. +\end{proposition} + +\begin{proof} + Si osserva che $M_\basis(\varphi) = I_n$. Sia $\basis = \{ \vv 1, \ldots, \vv n\}$. Se $f$ è ortogonale, allora + $[\v]_\basis^\top \, [\w]_\basis = [\v]_\basis^\top \, M_\basis(\varphi) [\w]_\basis = \varphi(\v, \w) = + \varphi(f(\v), f(\w)) = (M_\basis(f) [\v]_\basis)^\top \, M_\basis(\varphi) (M_\basis(f) [\w]_\basis) = + [\v]_\basis^\top M_\basis(f)^\top M_\basis(\varphi) M_\basis(f) [\w]_\basis = [\v]_\basis^\top M_\basis(f)^\top M_\basis(f) [\w]_\basis$. Allora, come visto nel corollario precedente, si ricava che $M_\basis(f)^\top M_\basis(f) = I_n$. Dal momento che gli inversi sinistri sono anche inversi destri, $M_\basis(f)^\top M_\basis(f) = M_\basis(f) M_\basis(f)^\top = I_n$. \\ + + Se invece $M_\basis(f)^\top M_\basis(f) = M_\basis(f) M_\basis(f)^\top = I_n$, $\varphi(\v, \w) = [\v]_\basis^\top [\w]_\basis = [\v]_\basis^\top M_\basis(f)^\top M_\basis(f) [\w]_\basis = + (M_\basis(f) [\v]_\basis)^\top (M_\basis(f) [\w]_\basis) =$ $(M_\basis(f) [\v]_\basis)^\top M_\basis(\varphi) (M_\basis(f) [\w]_\basis) = \varphi(f(\v), f(\w))$, e quindi + $f$ è ortogonale. +\end{proof} + +\begin{proposition} + Sia $(V, \varphi)$ uno spazio euclideo complesso e sia $\basis$ una base ortonormale di $V$. Allora $f \in \End(V)$ è hermitiano $\iff$ $M_\basis(f) = M_\basis(f)^*$ $\defiff$ $M_\basis(f)$ è hermitiana. +\end{proposition} + +\begin{proof} + Per il corollario precedente, $f$ è hermitiana $\iff$ $f = f^*$ $\iff M_\basis(f) = M_\basis(f^*) = M_\basis(f)^*$. +\end{proof} + +\begin{proposition} + Sia $(V, \varphi)$ uno spazio euclideo complesso e sia $\basis$ una base ortonormale di $V$. Allora $f \in \End(V)$ è unitario $\iff$ $M_\basis(f) M_\basis(f)^* = M_\basis(f)^* M_\basis(f) = I_n$ $\defiff$ $M_\basis(f)$ è unitaria. +\end{proposition} + +\begin{proof} + Si osserva che $M_\basis(\varphi) = I_n$. Sia $\basis = \{ \vv 1, \ldots, \vv n\}$. Se $f$ è unitario, allora + $[\v]_\basis^* \, [\w]_\basis = [\v]_\basis^* \, M_\basis(\varphi) [\w]_\basis = \varphi(\v, \w) = + \varphi(f(\v), f(\w)) = (M_\basis(f) [\v]_\basis)^* \, M_\basis(\varphi) (M_\basis(f) [\w]_\basis) = + [\v]_\basis^* M_\basis(f)^* M_\basis(\varphi) M_\basis(f) [\w]_\basis = [\v]_\basis^* M_\basis(f)^* M_\basis(f) [\w]_\basis$. Allora, come visto nel corollario precedente, si ricava che $M_\basis(f)^* M_\basis(f) = I_n$. Dal momento che gli inversi sinistri sono anche inversi destri, $M_\basis(f)^* M_\basis(f) = M_\basis(f) M_\basis(f)^* = I_n$. \\ + + Se invece $M_\basis(f)^* M_\basis(f) = M_\basis(f) M_\basis(f)^* = I_n$, $\varphi(\v, \w) = [\v]_\basis^* [\w]_\basis = [\v]_\basis^* M_\basis(f)^* M_\basis(f) [\w]_\basis$ $= + (M_\basis(f) [\v]_\basis)^* (M_\basis(f) [\w]_\basis)$ $= (M_\basis(f) [\v]_\basis)^* M_\basis(\varphi) (M_\basis(f) [\w]_\basis) = \varphi(f(\v), f(\w))$, e quindi + $f$ è unitario. +\end{proof} + +\begin{remark} + Se $\basis$ è una base ortonormale di $(V, \varphi)$, ricordando che $M_\basis(f^\top) = M_\basis(f)^\top$ e che $M_\basis(f^*) = M_\basis(f)^*$, sono equivalenti allora i seguenti fatti: \\ + + \li $f \circ f^\top = f^\top \circ f = \Idv$ $\iff$ $M_\basis(f)$ è ortogonale $\iff$ $f$ è ortogonale, \\ + \li $f \circ f^* = f^* \circ f = \Idv$ $\iff$ $M_\basis(f)$ è unitaria $\iff$ $f$ è unitario (se $V$ è uno spazio vettoriale su $\CC$). +\end{remark} + +\begin{proposition} + Sia $V = \RR^n$ uno spazio vettoriale col prodotto scalare standard $\varphi$. Allora sono equivalenti i seguenti fatti: + + \begin{enumerate}[(i)] + \item $A \in O_n$, + \item $f_A$ è un operatore ortogonale, + \item le colonne e le righe di $A$ formano una base ortonormale di $V$. + \end{enumerate} +\end{proposition} + +\begin{proof} + Sia $\basis$ la base canonica di $V$. Allora $M_\basis(f_A) = A$, e quindi, per una proposizione + precedente, $f_A$ è un operatore ortogonale. Viceversa si deduce che se $f_A$ è un operatore ortogonale, + $A \in O_n$. Dunque è sufficiente dimostrare che $A \in O_n \iff$ le colonne e le righe di $A$ formano una + base ortonormale di $V$. \\ + + \rightproof Se $A \in O_n$, in particolare $A \in \GL(n, \RR)$, e quindi $A$ è invertibile. Allora le + sue colonne e le sue righe formano già una base di $V$, essendo $n$ vettori di $V$ linearmente indipendenti. + Inoltre, poiché $A \in O_n$, $\varphi(\e i, \e j) = \varphi(A \e i, A \e j)$, e quindi le colonne di $A$ si mantengono a due a due ortogonali tra di loro, mentre $\varphi(A \e i, A \e i) = \varphi(\e i, \e i) = 1$. + Pertanto le colonne di $A$ formano una base ortonormale di $V$. \\ + + Si osserva che anche $A^\top \in O_n$. Allora le righe di $A$, che non sono altro che + le colonne di $A^\top$, formano anch'esse una base ortonormale di $V$. \\ + + \leftproof Nel moltiplicare $A^\top$ con $A$ altro non si sta facendo che calcolare il prodotto + scalare $\varphi$ tra ogni riga di $A^\top$ e ogni colonna di $A$ , ossia $(A^* A)_{ij} = \varphi((A^\top)_i, A^j) = \varphi(A^i, A^j) = \delta_{ij}$. + Quindi $A^\top A = A A^\top = I_n$, da cui si deduce che $A \in O_n$. +\end{proof} + +\begin{proposition} + Sia $V = \CC^n$ uno spazio vettoriale col prodotto hermitiano standard $\varphi$. Allora sono equivalenti i seguenti fatti: + + \begin{enumerate}[(i)] + \item $A \in U_n$, + \item $f_A$ è un operatore unitario, + \item le colonne e le righe di $A$ formano una base ortonormale di $V$. + \end{enumerate} +\end{proposition} + +\begin{proof} + Sia $\basis$ la base canonica di $V$. Allora $M_\basis(f_A) = A$, e quindi, per una proposizione + precedente, $f_A$ è un operatore unitario. Viceversa si deduce che se $f_A$ è un operatore unitario, + $A \in U_n$. Dunque è sufficiente dimostrare che $A \in U_n \iff$ le colonne e le righe di $A$ formano una + base ortonormale di $V$. \\ + + \rightproof Se $A \in U_n$, in particolare $A \in \GL(n, \RR)$, e quindi $A$ è invertibile. Allora le + sue colonne e le sue righe formano già una base di $V$, essendo $n$ vettori di $V$ linearmente indipendenti. + Inoltre, poiché $A \in U_n$, $\varphi(\e i, \e j) = \varphi(A \e i, A \e j)$, e quindi le colonne di $A$ si mantengono a due a due ortogonali tra di loro, mentre $\varphi(A \e i, A \e i) = \varphi(\e i, \e i) = 1$. + Pertanto le colonne di $A$ formano una base ortonormale di $V$. \\ + + Si osserva che anche $A^\top \in U_n$. Allora le righe di $A$, che non sono altro che + le colonne di $A^\top$, formano anch'esse una base ortonormale di $V$. \\ + + \leftproof Nel moltiplicare $A^*$ con $A$ altro non si sta facendo che calcolare il prodotto + hermitiano $\varphi$ tra ogni riga coniugata di $A^*$ e ogni colonna di $A$, ossia $(A^* A)_{ij} = \varphi((A^\top)_i, A^j) = \varphi(A^i, A^j) = \delta_{ij}$. + Quindi $A^* A = A A^* = I_n$, da cui si deduce che $A \in U_n$. +\end{proof} + +\begin{proposition} + Sia $(V, \varphi)$ uno spazio euclideo reale. Allora valgono i seguenti tre risultati: + + \begin{enumerate}[(i)] + \item $(V_\CC, \varphi_\CC)$ è uno spazio euclideo complesso. + + \item Se $f \in \End(V)$ è simmetrico, allora $f_\CC \in \End(V)$ è hermitiano. + + \item Se $f \in \End(V)$ è ortogonale, allora $f_\CC \in \End(V)$ è unitario. + \end{enumerate} +\end{proposition} + +\begin{proof} + Dacché $\varphi$ è il prodotto scalare standard dello spazio euclideo reale $V$, esiste una base ortnormale di $V$. Sia allora $\basis$ una base ortonormale di $V$. Si dimostrano i tre risultati separatamente. + + \begin{itemize} + \item È sufficiente dimostrare che $\varphi_\CC$ altro non è che il prodotto hermitiano standard. + Come si è già osservato precedentemente, $M_\basis(\varphi_\CC) = M_\basis(\varphi)$, e quindi, + dacché $M_\basis(\varphi) = I_n$, essendo $\basis$ ortonormale, vale anche che $M_\basis(\varphi_\CC) = I_n$, + ossia $\varphi_\CC$ è proprio il prodotto hermitiano standard. + + \item Poiché $f$ è simmetrico, $M_\basis(f) = M_\basis(f)^\top$, e quindi anche + $M_\basis(f_\CC) = M_\basis(f_\CC)^\top$. Dal momento che $M_\basis(f) \in M(n, \RR)$, + $M_\basis(f) = \conj{M_\basis(f)} \implies M_\basis(f_\CC)^\top = M_\basis(f_\CC)^*$. + Quindi $M_\basis(f_\CC) = M_\basis(f_\CC)^*$, ossia $M_\basis(f_\CC)$ è hermitiana, + e pertanto anche $f_\CC$ è hermitiano. + + \item Poiché $f$ è ortogonale, $M_\basis(f) M_\basis(f)^\top = I_n$, e quindi + anche $M_\basis(f_\CC) M_\basis(f_\CC)^\top = I_n$. Allora, come prima, si deduce + che $M_\basis(f_\CC)^\top = M_\basis(f_\CC)^*$, essendo $M_\basis(f_\CC) = M_\basis(f) \in M(n, \RR)$, + da cui + si ricava che $M_\basis(f_\CC) M_\basis(f_\CC)^* = M_\basis(f_\CC) M_\basis(f_\CC)^\top = I_n$, ossia che $f_\CC$ è unitario. \\ \qedhere + \end{itemize} +\end{proof} + +\begin{exercise} + Sia $(V, \varphi)$ uno spazio euclideo reale. Allora valgono i seguenti risultati: + + \begin{itemize} + \item Se $f$, $g \in \End(V)$ commutano, allora anche $f_\CC$, $g_\CC \in \End(V_\CC)$ commutano. + \item Se $f \in \End(V)$, $(f^\top)_\CC = (f_\CC)^*$. + \item Se $f \in \End(V)$, $f$ diagonalizzabile $\iff$ $f^\top$ diagonalizzabile. + \end{itemize} +\end{exercise} + +\begin{solution} + Dacché $\varphi$ è il prodotto scalare standard dello spazio euclideo reale $V$, esiste una base ortonormale $\basis = \{ \vv 1, \ldots, \vv n\}$ di $V$. Si dimostrano allora separatamente i tre risultati. + + \begin{itemize} + \item Si osserva che $M_\basis(f_\CC) M_\basis(g_\CC) = M_\basis(f) M_\basis(g) = + M_\basis(g) M_\basis(f) = M_\basis(g_\CC) M_\basis(f_\CC)$, e quindi + che $f_\CC \circ g_\CC = g_\CC \circ f_\CC$. + + \item Si osserva che $M_\basis(f) \in M(n, \RR) \implies M_\basis(f)^\top = M_\basis(f)^*$, e quindi che $M_\basis((f^\top)_\CC) = M_\basis(f^\top) = M_\basis(f)^\top = M_\basis(f)^* = M_\basis(f_\CC)^* = M_\basis((f_\CC)^*)$. Allora + $(f^\top)_\CC= (f_\CC)^*$. + + \item Poiché $\basis$ è ortonormale, $M_\basis(f^\top) = M_\basis(f)^\top$. Allora, se + $f$ è diagonalizzabile, anche $M_\basis(f)$ lo è, e quindi $\exists P \in \GL(n, \KK)$, + $D \in M(n, \KK)$ diagonale tale che $M_\basis(f) = P D P\inv$. Allora $M_\basis(f^\top) = + M_\basis(f)^\top = (P^\top)\inv D^\top P^\top$ è simile ad una matrice diagonale, e + pertanto $M_\basis(f^\top)$ è diagonalizzabile. Allora anche $f^\top$ è diagonalizzabile. + Vale anche il viceversa considerando l'identità $f = (f^\top)^\top$ e l'implicazione + appena dimostrata. + \end{itemize} +\end{solution} + +\hr + +\begin{note} + D'ora in poi, qualora non specificato diversamente, si assumerà che $V$ sia uno spazio + euclideo, reale o complesso. +\end{note} + +\begin{definition} (norma euclidea) + Sia $(V, \varphi)$ un qualunque spazio euclideo. Si definisce \textbf{norma} la mappa + $\norm{\cdot} : V \to \RR^+$ tale che $\norm{\v} = \sqrt{\varphi(\v, \v)}$. +\end{definition} + +\begin{definition} (distanza euclidea tra due vettori) + Sia $(V, \varphi)$ un qualunque spazio euclideo. Si definisce \textbf{distanza} la mappa + $d : V \times V \to \RR^+$ tale che $d(\v, \w) = \norm{\v - \w}$. +\end{definition} + +\begin{remark}\nl + \li Si osserva che in effetti $\varphi(\v, \v) \in \RR^+$ $\forall \v \in V$. Infatti, sia + per il caso reale che per il caso complesso, $\varphi$ è definito positivo. \\ + \li Vale che $\norm{\v} = 0 \iff \v = \vec 0$. Infatti, se $\v = \vec 0$, chiaramente + $\varphi(\v, \v) = 0 \implies \norm{\v} = 0$; se invece $\norm{\v} = 0$, + $\varphi(\v, \v) = 0$, e quindi $\v = \vec 0$, dacché $V^\perp = \zerovecset$, essendo + $\varphi$ definito positivo. \\ + \li Inoltre, vale chiaramente che $\norm{\alpha \v} = \abs{\alpha} \norm{\v}$. \\ + \li Se $f$ è un operatore ortogonale (o unitario), allora $f$ mantiene sia le + norme che le distanze tra vettori. Infatti $\norm{\v - \w}^2 = \varphi(\v - \w, \v - \w) = + \varphi(f(\v - \w), f(\v - \w)) = \varphi(f(\v) - f(\w), f(\v) - f(\w)) = \norm{f(\v) - f(\w)}^2$, + da cui segue che $\norm{\v - \w} = \norm{f(\v) - f(\w)}$. +\end{remark} + +\begin{proposition} [disuguaglianza di Cauchy-Schwarz] + Vale che $\norm{\v} \norm{\w} \geq \abs{\varphi(\v, \w)}$, $\forall \v$, $\w \in V$, dove + l'uguaglianza è raggiunta soltanto se $\v$ e $\w$ sono linearmente dipendenti. +\end{proposition} + +\begin{proof} + Si consideri innanzitutto il caso $\KK = \RR$, e quindi il caso in cui $\varphi$ è + il prodotto scalare standard. Siano $\v$, $\w \in V$. + Si consideri la disuguaglianza $\norm{\v + t\w}^2 \geq 0$, valida + per ogni elemento di $V$. Allora $\norm{\v + t \w}^2 = \norm{\v}^2 + 2 \varphi(\v, \w) t + \norm{\w}^2 t^2 \geq 0$. L'ultima disuguaglianza è possibile se e solo se $\frac{\Delta}{4} \leq 0$, e quindi se e solo + se $\varphi(\v, \w)^2 - \norm{\v}^2 \norm{\w}^2 \leq 0 \iff \norm{\v} \norm{\w} \geq \varphi(\v, \w)$. + Vale in particolare l'equivalenza se e solo se $\norm{\v + t\w} = 0$, ossia se $\v + t\w = \vec 0$, da cui + la tesi. \\ + + Si consideri ora il caso $\KK = \CC$, e dunque il caso in cui $\varphi$ è il prodotto hermitiano + standard. Siano $\v$, $\w \in V$, e siano $\alpha$, $\beta \in \CC$. Si consideri allora + la disuguaglianza $\norm{\alpha \v + \beta \w}^2 \geq 0$, valida per ogni elemento di $V$. Allora + $\norm{\alpha \v + \beta \w}^2 = \norm{\alpha \v}^2 + \varphi(\alpha \v, \beta \w) + \varphi(\beta \w, \alpha \v) + \norm{\beta \w}^2 = \abs{\alpha}^2 \norm{\v}^2 + \conj{\alpha} \beta \, \varphi(\v, \w) + + \alpha \conj{\beta} \, \varphi(\w, \v) + \abs{\beta}^2 \norm{\w}^2 \geq 0$. Ponendo allora + $\alpha = \norm{\w}^2$ e $\beta = -\varphi(\w, \v) = \conj{-\varphi(\v, \w)}$, si deduce che: + + \[ \norm{\v}^2 \norm{\w}^4 - \norm{\w}^2 \abs{\varphi(\v, \w)} \geq 0. \] + + \vskip 0.05in + + Se $\w = \vec 0$, la disuguaglianza di Cauchy-Schwarz è già dimostrata. Altrimenti, è sufficiente + dividere per $\norm{\w}^2$ (dal momento che $\w \neq \vec 0 \iff \norm{\w} \neq 0$) per ottenere + la tesi. Come prima, is osserva che l'uguaglianza si ottiene se e solo se $\v$ e $\w$ sono + linearmente dipendenti. +\end{proof} + +\begin{proposition} [disuguaglianza triangolare] + $\norm{\v + \w} \leq \norm{\v} + \norm{\w}$. +\end{proposition} + +\begin{proof} + Si osserva che $\norm{\v + \w}^2 = \norm{\v}^2 + \varphi(\v, \w) + \varphi(\w, \v) + \norm{\w}^2$. + Se $\varphi$ è il prodotto scalare standard, si ricava che: + \[ \norm{\v + \w}^2 = \norm{\v}^2 + 2 \varphi(\v, \w) + \norm{\w}^2 + \leq \norm{\v}^2 + 2 \norm{\v} \norm{\w} + \norm{\w}^2 = + (\norm{\v} + \norm{\w})^2,\] + + dove si è utilizzata la disuguaglianza di Cauchy-Schwarz. Da quest'ultima disuguaglianza si ricava, prendendo la radice quadrata, la disuguaglianza + desiderata. \\ + + Se invece $\varphi$ è il prodotto hermitiano standard, $\norm{\v + \w}^2 = \norm{\v}^2 + 2 \, \Re(\varphi(\v, \w)) + \norm{\w}^2 \leq \norm{\v}^2 + 2 \abs{\varphi(\v, \w)} + \norm{\w}^2$. Allora, riapplicando + la disuguaglianza di Cauchy-Schwarz, si ottiene che: + + \[ \norm{\v + \w}^2 \leq (\norm{\v} + \norm{\w})^2, \] + + da cui, come prima, si ottiene la disuguaglianza desiderata. +\end{proof} + +\begin{remark} + Utilizzando il concetto di norma euclidea, si possono ricavare due teoremi fondamentali della geometria, + e già noti dalla geometria euclidea. \\ + + \li Se $\v \perp \w$, allora $\norm{\v + \w}^2 = \norm{\v}^2 + \overbrace{(\varphi(\v, \w) + \varphi(\w, \v))}^{=\,0} + \norm{\w}^2 = \norm{\v}^2 + \norm{\w}^2$ (teorema di Pitagora), \\ + \li Se $\norm{\v} = \norm{\w}$ e $\varphi$ è un prodotto scalare, allora $\varphi(\v + \w, \v - \w) = \norm{\v}^2 - \varphi(\v, \w) + \varphi(\w, \v) - \norm{\w}^2 = \norm{\v}^2 - \norm{\w}^2 = 0$, e quindi + $\v + \w \perp \v - \w$ (le diagonali di un rombo sono ortogonali tra loro). +\end{remark} + +\begin{remark} + Sia $\basis = \{ \vv 1, \ldots, \vv n \}$ è una base ortogonale di $V$ per $\varphi$. \\ + + \li Se $\v = a_1 \vv 1 + \ldots + a_n \vv n$, con $a_1$, ..., $a_n \in \KK$, si osserva + che $\varphi(\v, \vv i) = a_i \varphi(\vv i, \vv i)$. Quindi $\v = \sum_{i=1}^n \frac{\varphi(\v, \vv i)}{\varphi(\vv i, \vv i)} \, \vv i$. In particolare, $\frac{\varphi(\v, \vv i)}{\varphi(\vv i, \vv i)}$ è + detto \textbf{coefficiente di Fourier} di $\v$ rispetto a $\vv i$, e si indica con $C(\v, \vv i)$. Se $\basis$ è ortonormale, + $\v = \sum_{i=1}^n \varphi(\v, \vv i) \, \vv i$. \\ + \li Quindi $\norm{\v}^2 = \varphi(\v, \v) = \sum_{i=1}^n \frac{\varphi(\v, \vv i)^2}{\varphi(\vv i, \vv i)}$. In + particolare, se $\basis$ è ortonormale, $\norm{\v}^2 = \sum_{i=1}^n \varphi(\v, \vv i)^2$. In tal caso, + si può esprimere la disuguaglianza di Bessel: $\norm{\v}^2 \geq \sum_{i=1}^k \varphi(\v, \vv i)^2$ per $k \leq n$. +\end{remark} + +\begin{remark} (algoritmo di ortogonalizzazione di Gram-Schmidt) + Se $\varphi$ è non degenere (o in generale, se $\CI(\varphi) = \zerovecset$) ed è + data una base $\basis = \{ \vv 1, \ldots, \vv n \}$ per $V$ (dove si ricorda che deve valere + $\Char \KK \neq 2$), è possibile + applicare l'\textbf{algoritmo di ortogonalizzazione di Gram-Schmidt} per ottenere + da $\basis$ una nuova base $\basis' = \{ \vv 1', \ldots, \vv n' \}$ con le seguenti proprietà: + + \begin{enumerate}[(i)] + \item $\basis'$ è una base ortogonale, + \item $\basis'$ mantiene la stessa bandiera di $\basis$ (ossia $\Span(\vv 1, \ldots, \vv i) = \Span(\vv 1', \ldots, \vv i')$ per ogni $1 \leq i \leq n$). + \end{enumerate} + + L'algoritmo si applica nel seguente modo: si prenda in considerazione $\vv 1$ e sottragga ad ogni altro vettore + della base il vettore $C(\vv 1, \vv i) \vv 1 = \frac{\varphi(\vv 1, \vv i)}{\varphi(\vv 1, \vv 1)} \vv 1$, + rendendo ortogonale ogni altro vettore della base con $\vv 1$. Pertanto si applica la mappa + $\vv i \mapsto \vv i - \frac{\varphi(\vv 1, \vv i)}{\varphi(\vv 1, \vv 1)} \vv i = \vv i ^{(1)}$. + Si verifica infatti che $\vv 1$ e $\vv i ^{(1)}$ sono ortogonali per $2 \leq i \leq n$: + + \[ \varphi(\vv 1, \vv i^{(1)}) = \varphi(\vv 1, \vv i) - \varphi\left(\vv 1, \frac{\varphi(\vv 1, \vv i)}{\varphi(\vv 1, \vv 1)} \vv i\right) = \varphi(\vv 1, \vv i) - \varphi(\vv 1, \vv i) = 0. \] + + Poiché $\vv 1$ non è isotropo, si deduce la decomposizione $V = \Span(\vv 1) \oplus \Span(\vv 1)^\perp$. + In particolare $\dim \Span(\vv 1)^\perp = n-1$: essendo allora i vettori $\vv 2 ^{(1)}, \ldots, \vv n ^{(1)}$ + linearmente indipendenti e appartenenti a $\Span(\vv 1)^\perp$, ne sono una base. Si conclude quindi + che vale la seguente decomposizione: + + \[ V = \Span(\vv 1) \oplus^\perp \Span(\vv 2 ^{(1)}, \ldots, \vv n ^{(1)}). \] + + \vskip 0.05in + + Si riapplica dunque l'algoritmo di Gram-Schmidt prendendo come spazio vettoriale lo spazio generato dai + vettori a cui si è applicato precedentemente l'algoritmo, ossia $V' = \Span(\vv 2 ^{(1)}, \ldots, \vv n ^{(1)})$, + fino a che non si ottiene $V' = \zerovecset$. \\ + + Si può addirittura ottenere una base ortonormale a partire da $\basis'$ normalizzando ogni vettore (ossia + dividendo per la propria norma), se si sta considerando uno spazio euclideo. +\end{remark} + +\begin{remark} + Poiché la base ottenuta tramite Gram-Schmidt mantiene la stessa bandiera della base di partenza, + ogni matrice triangolabile è anche triangolabile mediante una base ortogonale. +\end{remark} + +\begin{example} + Si consideri $V = (\RR^3, \innprod{\cdot, \cdot})$, ossia $\RR^3$ dotato del prodotto scalare standard. + Si applica l'algoritmo di ortogonalizzazione di Gram-Schmidt sulla seguente base: + + \[ \basis = \Biggl\{ \underbrace{\Vector{1 \\ 0 \\ 0}}_{\vv 1 \, = \, \e1}, \underbrace{\Vector{1 \\ 1 \\ 0}}_{\vv 2}, \underbrace{\Vector{1 \\ 1 \\ 1}}_{\vv 3} \Biggl\}. \] + + \vskip 0.05in + + Alla prima iterazione dell'algoritmo si ottengono i seguenti vettori: + + \begin{itemize} + \item $\vv 2 ^{(1)} = \vv 2 - \frac{\varphi(\vv 1, \vv 2)}{\varphi(\vv 1, \vv 1)} \vv 1 = \vv 2 - \vv 1 = \Vector{0 \\ 1 \\ 0} = \e 2$, + \item $\vv 3 ^{(1)} = \vv 3 - \frac{\varphi(\vv 1, \vv 3)}{\varphi(\vv 1, \vv 1)} \vv 1 = \vv 3 - \vv 1 = \Vector{0 \\ 1 \\ 1}$. + \end{itemize} + + Si considera ora $V' = \Span(\vv 2 ^{(1)}, \vv 3 ^{(1)})$. Alla seconda iterazione dell'algoritmo si + ottiene allora il seguente vettore: + + \begin{itemize} + \item $\vv 3 ^{(2)} = \vv 3 ^{(1)} - \frac{\varphi(\vv 2 ^{(1)}, \vv 3 ^{(1)})}{\varphi(\vv 2 ^{(1)}, \vv 2 ^{(1)})} \vv 2 ^{(1)} = \vv 3 ^{(1)} - \vv 2 ^{(1)} = \Vector{0 \\ 0 \\ 1} = \e 3$. + \end{itemize} + + Quindi la base ottenuta è $\basis' = \{\e1, \e2, \e3\}$, ossia la base canonica di $\RR^3$, già + ortonormale. +\end{example} + +\begin{remark} + Si osserva adesso che se $(V, \varphi)$ è uno spazio euclideo (e quindi $\varphi > 0$), e $W$ è + un sottospazio di $V$, vale la seguente decomposizione: + + \[ V = W \oplus^\perp W^\perp. \] + + Pertanto ogni vettore $\v \in V$ può scriversi come $\w + \w'$ dove $\w \in W$ e $\w' \in W^\perp$, + dove $\varphi(\w, \w') = 0$. +\end{remark} + +\begin{definition} (proiezione ortogonale) + Si definisce l'applicazione $\pr_W : V \to V$, detta \textbf{proiezione ortogonale} su $W$, + in modo tale che $\pr_W(\v) = \w$, dove $\v = \w + \w'$, con $\w \in W$ e $\w' \in W^\perp$. +\end{definition} + +\begin{remark}\nl + \li Dacché la proiezione ortogonale è un caso particolare della classica applicazione lineare + di proiezione su un sottospazio di una somma diretta, $\pr_W$ è un'applicazione lineare. \\ + \li Vale chiaramente che $\pr_W^2 = \pr_W$, da cui si ricava, se $W^\perp \neq \zerovecset$, che + $\varphi_{\pr_W}(\lambda) = \lambda (\lambda -1)$, ossia che $\Sp(\pr_W) = \{0, 1\}$. Infatti + $\pr_W(\v)$ appartiene già a $W$, ed essendo la scrittura in somma di due elementi, uno di $W$ e + uno di $W'$, unica, $\pr_W(\pr_W(\v)) = \pr_W(\v)$, da cui l'identità $\pr_W^2 = \pr_W$. \\ + \li Seguendo il ragionamento di prima, vale anche che $\restr{\pr_W}{W} = \Idw$ e che + $\restr{\pr_W}{W^\perp} = 0$. \\ + \li Inoltre, vale la seguente riscrittura di $\v \in V$: $\v = \pr_W(\v) + \pr_{W^\perp}(\v)$. \\ + \li Se $\basis = \{ \vv1, \ldots, \vv n \}$ è una base ortogonale di $W$, allora + $\pr_W(\v) = \sum_{i=1}^n \frac{\varphi(\v, \vv i)}{\varphi(\vv i, \vv i)} \vv i = \sum_{i=1}^n C(\v, \vv i) \vv i$. Infatti $\v -\sum_{i=1}^n C(\v, \vv i) \vv i \in W^\perp$. \\ + \li $\pr_W$ è un operatore simmetrico (o hermitiano se lo spazio è complesso). Infatti $\varphi(\pr_W(\v), \w) = + \varphi(\pr_W(\v), \pr_W(\w) + \pr_{W^\perp}(\w)) = \varphi(\pr_W(\v), \pr_W(\w)) = \varphi(\pr_W(\v) + \pr_{W^\perp}(\v), \pr_W(\w)) = \varphi(\v, \pr_W(\w))$. +\end{remark} + +\begin{proposition} + Sia $(V, \varphi)$ uno spazio euclideo. Allora valgono i seguenti risultati: + + \begin{enumerate}[(i)] + \item Siano $U$, $W \subseteq V$ sono sottospazi di $V$, allora $U \perp W$, ossia\footnote{È sufficiente che valga $U \subseteq W^\perp$ affinché valga anche $W \subseteq U^\perp$. Infatti $U \subseteq W^\perp \implies W = W^\dperp \subseteq U^\perp$. Si osserva che in generale vale che $W \subseteq W^\dperp$, dove vale l'uguaglianza nel caso di un prodotto $\varphi$ non degenere, com'è nel caso di uno spazio euclideo, + essendo $\varphi > 0$ per ipotesi.} $U \subseteq W^\perp$, $\iff \pr_U \circ \pr_W = \pr_W \circ \pr_U = 0$. + + \item Sia $V = W_1 \oplus \cdots \oplus W_n$. Allora $\v = \sum_{i=1}^n \pr_{W_i}(\v)$ $\iff$ $W_i \perp W_j$ $\forall i \neq j$, $1 \leq i, j \leq n$. + \end{enumerate} +\end{proposition} + +\begin{proof} + Si dimostrano i due risultati separatamente. + + \begin{enumerate}[(i)] + \item Sia $\v \in V$. Allora $\pr_W(\v) \in W = W^\dperp \subseteq U^\perp$. Pertanto + $\pr_U(\pr_W(\v)) = \vec 0$. Analogamente $\pr_W(\pr_U(\v)) = \vec 0$, da cui la tesi. + + \item Sia vero che $\v = \sum_{i=1}^n \pr_{W_i}(\v)$ $\forall \v \in V$. Sia $\w \in W_j$. Allora $\w = \sum_{i=1}^n \pr_{W_i}(\w) = \w + \sum_{\substack{i=1 \\ i \neq j}} \pr_{W_i}(\w) \implies \pr_{W_i}(\w) = \vec 0$ $\forall i \neq j$. Quindi $\w \in W_i^\perp$ $\forall i \neq j$, e si conclude che $W_i \subseteq W_j^\perp + \implies W_i \perp W_j$. Se invece $W_i \perp W_j$ $\forall i \neq j$, sia $\basis_i = \left\{ \w_i^{(1)}, \ldots, \w_i^{(k_i)} \right\}$ una base ortogonale di $W_i$. Allora $\basis = \cup_{i=1}^n \basis_i$ è anch'essa + una base ortogonale di $V$, essendo $\varphi\left(\w_i^{(t_i)}, \w_j^{(t_j)}\right) = 0$ per ipotesi. + Pertanto $\v = \sum_{i=1}^n \sum_{j=1}^{k_i} C\left(\v, \w_i^{(j)}\right) \w_i^{(j)} = \sum_{i=1}^n \pr_{W_i}(\v)$, + da cui la tesi. \qedhere + \end{enumerate} +\end{proof} + +\begin{definition} (inversione ortogonale) + Si definisce l'applicazione $\rho_W : V \to V$, detta \textbf{inversione ortogonale}, in modo tale che, detto $\v = \w + \w' \in V$ con $\w \in W$, $\w \in W^\perp$, $\rho_W(\v) = \w - \w'$. Se $\dim W = \dim V - 1$, + si dice che $\rho_W$ è una \textbf{riflessione}. +\end{definition} + +\begin{remark}\nl + \li Si osserva che $\rho_W$ è un'applicazione lineare. \\ + \li Vale l'identità $\rho_W^2 = \Idv$, da cui si ricava che $\varphi_{\rho_W}(\lambda) \mid (\lambda-1)(\lambda+1)$. In particolare, se $W^\perp \neq \zerovecset$, vale proprio + che $\Sp(\rho_W) = \{\pm1\}$, dove $V_1 = W$ e $V_{-1} = W^\perp$. \\ + \li $\rho_W$ è ortogonale (o unitaria, se $V$ è uno spazio euclideo complesso). Infatti se $\vv 1 = \ww 1 + \ww 1'$ e $\vv 2 = \ww 2 + \ww 2 '$, con $\ww 1$, $\ww 2 \in W$ e $\ww 1'$, $\ww 2' \in W$, $\varphi(\rho_W(\vv 1), \rho_W(\vv 2)) = \varphi(\ww 1 - \ww 1', \ww 2 - \ww 2') = \varphi(\ww 1, \ww 2) \underbrace{- \varphi(\ww 1', \ww 2) - \varphi(\ww 1, \ww 2')}_{=\,0} + \varphi(\ww 1', \ww 2') = \varphi(\ww 1 - \ww 1', \ww 2 - \ww 2')$. \\ + + Quindi $\varphi(\rho_W(\vv 1), \rho_W(\vv 2)) = \varphi(\ww 1, \ww 2) + \varphi(\ww 1', \ww 2) + \varphi(\ww 1, \ww 2') + \varphi(\ww 1', \ww 2') = \varphi(\vv 1, \vv 2)$. +\end{remark} + +\begin{lemma} Sia $(V, \varphi)$ uno spazio euclideo reale. + Siano $\U$, $\w \in V$. Se $\norm{\U} = \norm{\w}$, allora esiste un sottospazio $W$ di dimensione + $n-1$ per cui la riflessione $\rho_W$ relativa a $\varphi$ è tale che $\rho_W(\U) = \w$. +\end{lemma} + +\begin{proof} Se $\v$ e $\w$ sono linearmente dipendenti, dal momento che $\norm{v} = \norm{w}$, deve valere anche + che $\v = \w$. Sia $\U \neq \vec 0$, $\U \in \Span(\v)^\perp$. Si consideri $U = \Span(\U)$: si osserva che + $\dim U = 1$ e che, essendo $\varphi$ non degenere, $\dim U^\perp = n-1$. Posto allora $W = U^\perp$, si ricava, + sempre perché $\varphi$ è non degenere, che $U = U^\dperp = W^\perp$. Si conclude pertanto che $\rho_W(\v) = + \v = \w$. \\ + + Siano adesso $\v$ e $\w$ linearmente indipendenti e sia $U = \Span(\v - \w)$. Dal momento che $\dim U = 1$ e $\varphi$ è non degenere, $\dim U^\perp = n-1$. Sia allora $W = U^\perp$. Allora, come prima, $U = U^\dperp = W^\perp$. Si consideri dunque la riflessione $\rho_W$: dacché $\v = \frac{\v + \w}{2} + \frac{\v - \w}{2}$, e $\varphi(\frac{\v + \w}{2}, \frac{\v - \w}{2}) = \frac{\norm{\v} - \norm{\w}}{4} = 0$, $\v$ è già decomposto in un elemento di $W$ e in uno di $W^\perp$, per cui si conclude che $\rho_W(\v) = + \frac{\v + \w}{2} - \frac{\v - \w}{2} = \w$, ottenendo la tesi. + +\end{proof} + +\begin{theorem} [di Cartan–Dieudonné] Sia $(V, \varphi)$ uno spazio euclideo reale. + Ogni isometria di $V$ è allora prodotto di al più $n$ riflessioni. +\end{theorem} + +\begin{proof} + Si dimostra la tesi applicando il principio di induzione sulla dimensione $n$ + di $V$. \\ + + \basestep Sia $n = 1$ e sia inoltre $f$ un'isometria di $V$. Sia $\vv 1$ l'unico elemento di una base ortonormale $\basis$ di $V$. Allora $\norm{f(\vv 1)} = \norm{\vv 1} = 1$, da cui si ricava che\footnote{Infatti, detto $\lambda \in \RR$ tale che $f(\vv 1) = \lambda \vv 1$, $\norm{\vv 1} = \norm{f(\vv 1)} = \lambda^2 \norm{\vv 1} \implies \lambda = \pm 1$, ossia $f = \pm \Id$, come volevasi dimostrare.} $f(\vv 1) = \pm \vv 1$, + ossia che $f = \pm \Idv$. Se $f = \Idv$, $f$ è un prodotto vuoto, e già verifica la tesi; altrimenti + $f = \rho_{\zerovecset}$, dove si considera $V = V \oplus^\perp \zerovecset$. Pertanto $f$ è prodotto + di al più una riflessione. \\ + + \inductivestep Sia $\basis = \{ \vv1, \ldots, \vv n \}$ una base di $V$. Sia $f$ un'isometria di $V$. Si + assuma inizialmente l'esistenza di $\vv i$ tale per cui $f(\vv i) = \vv i$. Allora, detto $W = \Span(\vv i)$, si può decomporre $V$ come $W \oplus^\perp W^\perp$. Si osserva che $W^\perp$ è $f$-invariante: infatti, + se $\U \in W^\perp$, $\varphi(\vv i, f(\U)) = \varphi(f(\vv i), f(\U)) = \varphi(\vv i, \U) = 0 \implies + f(\U) \in W^\perp$. Pertanto si può considerare l'isometria $\restr{f}{W^\perp}$. Dacché $\dim W^\perp = n - 1$, + per il passo induttivo esistono $W_1$, ..., $W_k$ sottospazi di $W^\perp$ con $k \leq n-1$ per cui $\rho_{W_1}$, ..., $\rho_{W_k} \in \End(W^\perp)$ sono tali che $\restr{f}{W^\perp} = \rho_{W_1} \circ \cdots \circ \rho_{W_k}$. \\ + + Si considerino allora le riflessioni $\rho_{W_1 \oplus^\perp W}$, ..., $\rho_{W_k \oplus^\perp W}$. + Si mostra che $\restr{\rho_{W_1 \oplus^\perp W} \circ \cdots \circ \rho_{W_k \oplus^\perp W}}{W} = \Idw = \restr{f}{W}$. + Affinché si faccia ciò è sufficiente mostrare che $(\rho_{W_1 \oplus^\perp W} \circ \cdots \circ \rho_{W_k \oplus^\perp W})(\vv i) = \vv i$. Si osserva che $\vv i \in W_i \oplus^\perp W$ $\forall 1 \leq i \leq k$, e + quindi che $\rho_{W_k \oplus^\perp W}(\vv i) = \vv i$. Reiterando l'applicazione di questa identità nel prodotto, + si ottiene infine il risultato desiderato. Infine, si dimostra che $\restr{\rho_{W_1 \oplus^\perp W} \circ \cdots \circ \rho_{W_k \oplus^\perp W}}{W^\perp} = \rho_{W_1} \circ \cdots \circ \rho_{W_k} = \restr{f}{W^\perp}$. Analogamente a prima, + è sufficiente mostrare che $\rho_{W_k \oplus^\perp W}(\U) = \rho_{W_k}(\U)$ $\forall \U \in W^\perp$. + Sia $\U = \rho_{W_k}(\U) + \U'$ con $\U' \in W_k^\perp \cap W^\perp \subseteq (W_k \oplus^\perp W)^\perp$, + ricordando che $W^\perp = W_k \oplus^\perp (W^\perp \cap W_k^\perp)$. + Allora, poiché $\rho_{W_k}(\U) \in W_k \subseteq (W_k \oplus^\perp W)$, si conclude che + $\rho_{W_k \oplus^\perp W}(\U) = \rho_{W_k}(\U)$. Pertanto, dacché vale che $V = W \oplus^\perp W^\perp$ e che $\rho_{W_1 \oplus^\perp W} \circ \cdots \circ \rho_{W_k \oplus^\perp W}$ e $f$, ristretti su $W$ o su $W^\perp$, sono le stesse identiche mappe, allora + in particolare vale l'uguaglianza più generale: + + \[ f = \rho_{W_1 \oplus^\perp W} \circ \cdots \circ \rho_{W_k \oplus^\perp W}, \] + + \vskip 0.05in + + e quindi $f$ è prodotto di $k \leq n-1$ riflessioni. \\ + + Se invece non esiste alcun $\vv i$ tale per cui $f(\vv i) = \vv i$, per il \textit{Lemma 1} esiste + una riflessione $\tau$ tale per cui $\tau(f(\vv i)) = \vv i$. In particolare $\tau \circ f$ è anch'essa + un'isometria, essendo composizione di due isometrie. Allora, da prima, esistono $U_1$, ..., $U_k$ sottospazi + di $V$ con $k \leq n-1$ tali per cui $\tau \circ f = \rho_{U_1} \circ \cdots \circ \rho_{U_k}$, da + cui $f = \tau \circ \rho_{U_1} \circ \cdots \circ \rho_{U_k}$, ossia $f$ è prodotto di al più + $n$ riflessioni, concludendo il passo induttivo. +\end{proof} + +\setcounter{lemma}{0} + +\hr + +\begin{lemma} + Sia $f \in \End(V)$ simmetrico (o hermitiano). Allora $f$ ha solo autovalori reali\footnote{Nel caso + di $f$ simmetrico, si intende in particolare che tutte le radici del suo polinomio caratteristico + sono reali.}. +\end{lemma} + +\begin{proof} + Si assuma che $V$ è uno spazio euclideo complesso, e quindi che $\varphi$ è un prodotto hermitiano. Allora, + se $f$ è hermitiano, sia $\lambda \in \CC$ un suo autovalore\footnote{Tale autovalore esiste sicuramente dal momento + che $\KK = \CC$ è un campo algebricamente chiuso.} e sia $\v \in V_\lambda$. Allora $\varphi(\v, f(\v)) = + \varphi(f(\v), \v) = \conj{\varphi(\v, f(\v))} \implies \varphi(\v, f(\v)) \in \RR$. Inoltre vale + la seguente identità: + + \[ \varphi(\v, f(\v)) = \varphi(\v, \lambda \v) = \lambda \varphi(\v, \v), \] + + da cui, ricordando che $\varphi$ è non degenere e che $\varphi(\v, \v) \in \RR$, si ricava che: + + \[ \lambda = \frac{\varphi(\v, f(\v))}{\varphi(\v, \v)} \in \RR. \] + + \vskip 0.05in + + Sia ora invece $V$ è uno spazio euclideo reale e $\varphi$ è un prodotto scalare. Allora, $(V_\CC, \varphi_\CC)$ + è uno spazio euclideo complesso, e $f_\CC$ è hermitiano. Sia $\basis$ una base di $V$. Allora, come visto all'inizio di questa + dimostrazione, $f_\CC$ ha solo autovalori reali, da cui si ricava che il polinomio caratteristico + di $f_\CC$ è completamente riducibile in $\RR$. Si osserva inoltre che $p_f(\lambda) = \det(M_\basis(f) - \lambda I_n) = \det(M_\basis(f_\CC) - \lambda I_n) = p_{f_\CC}(\lambda)$. Si conclude dunque che + anche $p_f$ è completamente riducibile in $\RR$. +\end{proof} + +\begin{remark} + Dal lemma precedente consegue immediatamente che se $A \in M(n, \RR)$ è simmetrica (o se appartiene a + $M(n, \CC)$ ed è hermitiana), considerando l'operatore simmetrico $f_A$ indotto da $A$ in $\RR^n$ (o $\CC^n$), + $f_A$ ha tutti autovalori reali, e dunque così anche $A$. +\end{remark} + +\begin{lemma} + Sia $f \in \End(V)$ simmetrico (o hermitiano). Allora se $\lambda$, $\mu$ sono due autovalori distinti + di $f$, $V_\lambda \perp V_\mu$. +\end{lemma} + +\begin{proof} + Siano $\v \in V_\lambda$ e $\w \in V_\mu$. Allora\footnote{Si osserva che non è stato coniugato $\lambda$ + nei passaggi algebrici, valendo $\lambda \in \RR$ dallo scorso lemma.} $\lambda \varphi(\v, \w) = \varphi(\lambda \v, \w) = \varphi(f(\v), \w) = \varphi(\v, f(\w)) = \varphi(\v, \mu \w) = \mu \varphi(\v, \w)$. + Pertanto vale la seguente identità: + + \[ (\lambda - \mu) \varphi(\v, \w) = 0. \] + + \vskip 0.05in + + In particolare, valendo $\lambda - \mu \neq 0$ per ipotesi, $\varphi(\v, \w) = 0 \implies V_\lambda \perp V_\mu$, + da cui la tesi. +\end{proof} + +\begin{lemma} + Sia $f \in \End(V)$ simmetrico (o hermitiano). Se $W \subseteq V$ è $f$-invariante, allora anche + $W^\perp$ lo è. +\end{lemma} + +\begin{proof} + Siano $\w \in W$ e $\v \in W^\perp$. Allora $\varphi(\w, f(\v)) = \varphi(\underbrace{f(\w)}_{\in \, W}, \v) = 0$, da cui si ricava che $f(\v) \in W^\perp$, ossia la tesi. +\end{proof} + +\begin{theorem} [spettrale reale] + Sia $(V, \varphi)$ uno spazio euclideo reale (o complesso) e sia $f \in \End(V)$ simmetrico (o hermitiano). Allora esiste una base ortogonale $\basis$ di $V$ composta di autovettori per $f$. +\end{theorem} + +\begin{proof} + Siano $\lambda_1$, ..., $\lambda_k$ tutti gli autovalori reali di $f$. Sia inoltre + $W = V_{\lambda_1} \oplus \cdots \oplus V_{\lambda_k}$. Per i lemmi precedenti, + vale che: + + \[ W = V_{\lambda_1} \oplus^\perp \cdots \oplus^\perp V_{\lambda_k}. \] + + \vskip 0.05in + + Sicuramente $W \subset V$. Si assuma però che $W \subsetneq V$. Allora $V = W \oplus^\perp W^\perp$. In particolare, per il lemma + precedente, $W^\perp$ è $f$-invariante. Quindi $\restr{f}{W^\perp}$ è un endomorfismo + di uno spazio di dimensione non nulla. Si osserva che $\restr{f}{W^\perp}$ è chiaramente + simmetrico (o hermitiano), essendo solo una restrizione di $f$. Allora $\restr{f}{W^\perp}$ ammette + autovalori reali per i lemmi precedenti; tuttavia questo è un assurdo, dal momento che ogni autovalore di $\restr{f}{W^\perp}$ è anche autovalore di $f$ e si era supposto che\footnote{Infatti tale autovalore $\lambda$ + non può già comparire tra questi autovalori, altrimenti, detto $i \in \NN$ tale che $\lambda = \lambda_i$, $V_{\lambda_i} \cap W^\perp \neq \zerovecset$, violando la somma diretta supposta.} $\lambda_1$, ..., $\lambda_k$ fossero + tutti gli autovalori di $f$, \Lightning. Quindi $W = V$. Pertanto, detta $\basis_i$ una base ortonormale + di $V_{\lambda_i}$, $\basis = \cup_{i=1}^k \basis_i$ è una base ortonormale di $V$, da cui la tesi. +\end{proof} + +\begin{corollary} [teorema spettrale per le matrici] + Sia $A \in M(n, \RR)$ simmetrica (o appartenente a $M(n, \CC)$ ed hermitiana). Allora + $\exists P \in O_n$ (o $P \in U_n$) tale che $P\inv A P = P^\top A P$ (o $P\inv A P = P^* A P$ nel caso hermitiano) + sia una matrice diagonale reale. +\end{corollary} + +\begin{proof} + Si consideri $f_A$, l'operatore indotto dalla matrice $A$ in $\RR^n$ (o $\CC^n$). Allora + $f_A$ è un operatore simmetrico (o hermitiano) sul prodotto scalare (o hermitiano) standard. + Pertanto, per il teorema spettrale reale, esiste una base ortonormale $\basis = \{ \vv 1, \ldots, \vv n\}$ composta di autovettori + di $f_A$. In particolare, detta $\basis'$ la base canonica di $\RR^n$ (o $\CC^n$), vale + la seguente identità: + + \[ M_\basis(f) = M_{\basis'}^{\basis}(\Id)\inv M_{\basis'}(f) M_{\basis'}^{\basis}(\Id), \] + + dove $M_{\basis'}(f) = A$, $M_\basis(f)$ è diagonale, essendo $\basis$ composta di autovettori, e $P = M_{\basis'}^{\basis}$ + si configura nel seguente modo: + + \[ M_{\basis'}^{\basis}(f) = \Matrix{ \vv 1 & \rvline & \cdots & \rvline & \vv n }. \] + + Dacché $\basis$ è ortogonale, $P$ è anch'essa ortogonale, da cui la tesi. +\end{proof} + +\begin{remark}\nl + \li Un importante risultato che consegue direttamente dal teorema spettrale per le matrici riguarda + la segnatura di un prodotto scalare (o hermitiano). Infatti, detta $A = M_\basis(\varphi)$, + $D = P^\top A P$, e dunque $D \cong A$. Allora, essendo $D$ diagonale, l'indice di positività + è esattamente il numero di valori positivi sulla diagonale, ossia il numero di autovalori + positivi di $A$. Analogamente l'indice di negatività è il numero di autovalori negativi, + e quello di nullità è la molteplicità algebrica di $0$ come autovalore (ossia esattamente + la dimensione di $V^\perp_\varphi = \Ker a_\varphi$). +\end{remark} + +\begin{theorem} [di triangolazione con base ortonormale] + Sia $f \in \End(V)$, dove $(V, \varphi)$ è uno spazio euclideo su $\KK$. Allora, + se $p_f$ è completamente riducibile in $\KK$, esiste una base ortonormale $\basis$ + tale per cui $M_\basis(f)$ è triangolare superiore (ossia esiste una base ortonormale + a bandiera per $f$). +\end{theorem} + +\begin{proof} + Per il teorema di triangolazione, esiste una base $\basis$ a bandiera per $f$. Allora, + applicando l'algoritmo di ortogonalizzazione di Gram-Schmidt, si può ottenere da $\basis$ + una nuova base $\basis'$ ortonormale e che mantenga le stesse bandiere. Allora, + se $\basis' = \{ \vv1, \ldots, \vv n \}$ è ordinata, dacché $\Span(\vv 1, \ldots, \vv i)$ è $f$-invariante, + $f(\vv i) \in \Span(\vv 1, \ldots, \vv i)$, e quindi $M_{\basis'}(f)$ è triangolare superiore, da cui la tesi. +\end{proof} + +\begin{corollary} + Sia $A \in M(n, \RR)$ (o $M(n, \CC)$) tale per cui $p_A$ è completamente riducibile. + Allora $\exists P \in O_n$ (o $U_n$) tale per cui + $P\inv A P = P^\top A P$ (o $P\inv A P = P^* A P$) è triangolare superiore. +\end{corollary} + +\begin{proof} + Si consideri l'operatore $f_A$ indotto da $A$ in $\RR^n$ (o $\CC^n$). Sia $\basis$ la base canonica di $\RR^n$ (o di $\CC^n$). Allora, per il teorema + di triangolazione con base ortonormale, esiste una base ortonormale $\basis' = \{ \vv1, \ldots, \vv n \}$ di $\RR^n$ (o di $\CC^n$) + tale per cui $T = M_{\basis'}(f_A)$ è triangolare superiore. Si osserva inoltre che $M_{\basis}(f_A) = A$ e che $P = M_{\basis}^{\basis'} (f_A) = \Matrix{\vv 1 & \rvline & \cdots & \rvline & \vv n}$ è ortogonale (o unitaria), dacché le sue colonne + formano una base ortonormale. Allora, dalla formula del cambiamento di base per la applicazioni lineari, + si ricava che: + + \[ A = P T P\inv \implies T = P\inv T P, \] + + da cui, osservando che $P\inv = P^\top$ (o $P\inv = P^*$), si ricava la tesi. +\end{proof} + +\begin{definition} [operatore normale] + Sia $(V, \varphi)$ uno spazio euclideo reale. Allora $f \in \End(V)$ si dice \textbf{normale} + se commuta con il suo trasposto (i.e.~se $f f^\top = f^\top f$). Analogamente, + se $(V, \varphi)$ è uno spazio euclideo complesso, allora $f$ si dice normale se commuta con il suo + aggiunto (i.e.~se $f f^* = f^* f$). +\end{definition} + +\begin{definition} [matrice normale] + Una matrice $A \in M(n, \RR)$ (o $M(n, \CC)$) si dice \textbf{normale} se $A A^\top = A^\top A$ (o $A A^* = A^* A$). +\end{definition} + +\begin{remark}\nl + \li Se $A \in M(n, \RR)$ e $A$ è simmetrica ($A = A^\top$), antisimmetrica ($A = -A^\top$) o + ortogonale ($A A^\top = A^\top A = I_n$), sicuramente $A$ è normale. \\ + \li Se $A \in M(n, \CC)$ e $A$ è hermitiana ($A = A^*$), antihermitiana ($A = -A^*$) o + unitaria ($A A^* = A^* A = I_n$), sicuramente $A$ è normale. \\ + \li $f$ è normale $\iff$ $M_\basis(f)$ è normale, con $\basis$ ortonormale di $V$. \\ + \li $A$ è normale $\iff$ $f_A$ è normale, considerando che la base canonica di $\CC^n$ è già + ortonormale rispetto al prodotto hermitiano standard. \\ + \li Se $V$ è euclideo reale, $f$ è normale $\iff$ $f_\CC$ è normale. Infatti, se $f$ è normale, $f$ e $f^\top$ + commutano. Allora anche $f_\CC$ e $(f^\top)_\CC = (f_\CC)^*$ commutano, e quindi $f_\CC$ è normale. + Ripercorrendo i passaggi al contrario, si osserva infine che vale anche il viceversa. +\end{remark} + +\setcounter{lemma}{0} + +\begin{lemma} + Sia $A \in M(n, \CC)$ triangolare superiore e normale (i.e.~$A A^* = A^* A$). Allora + $A$ è diagonale. +\end{lemma} + +\begin{proof} + Se $A$ è normale, allora $(A^*)_i A^i = \conj{A}\,^i A^i$ deve essere uguale a + $A_i (A^*)^i = A_i \conj{A}_i$ $\forall 1 \leq i \leq n$. Si dimostra per induzione + su $i$ da $1$ a $n$ che tutti gli elementi, eccetto per quelli diagonali, delle + righe $A_1$, ..., $A_i$ sono nulli. \\ + + \basestep Si osserva che valgono le seguenti identità: + + \begin{gather*} + \conj{A}\,^1 A^1 = \abs{a_{11}}^2, \\ + A_1 \conj{A}_1 = \abs{a_{11}}^2 + \abs{a_{12}}^2 + \ldots + \abs{a_{1n}}^2. + \end{gather*} + + Dovendo vale l'uguaglianza, si ricava che $\abs{a_{12}}^2 \ldots + \abs{a_{1n}}^2$, + e quindi che $\abs{a_{1i}}^2 = 0 \implies a_{1i} = 0$ \, $\forall 2 \leq i \leq n$, + dimostrando il passo base\footnote{Gli altri elementi sono infatti già nulli per ipotesi, essendo + $A$ triangolare superiore}. \\ + + \inductivestep Analogamente a prima, si considerano le seguenti identità: + + \begin{gather*} + \conj{A}\,^i A^i = \abs{a_{1i}}^2 + \ldots + \abs{a_{ii}}^2 = \abs{a_{ii}}^2, \\ + A_i \conj{A}_i = \abs{a_{ii}}^2 + \abs{a_{i(i+1)}}^2 + \ldots + \abs{a_{in}}^2, + \end{gather*} + + dove si è usato che, per il passo induttivo, tutti gli elementi, eccetto per quelli diagonali, delle + righe $A_1$, ..., $A_{i-1}$ sono nulli. Allora, analogamente a prima, si ricava che + $a_{ij} = 0$ \, $\forall i < j \leq n$, dimostrando il passo induttivo, e quindi la tesi. +\end{proof} + +\begin{remark}\nl + \li Chiaramente vale anche il viceversa del precedente lemma: se infatti $A \in M(n, \CC)$ è diagonale, + $A$ è anche normale, dal momento che commuta con $A^*$. \\ + \li Reiterando la stessa dimostrazione del precedente lemma per $A \in M(n, \RR)$ triangolare superiore e normale reale (i.e.~$AA^\top = A^\top A$) si può ottenere una tesi analoga. +\end{remark} + +\begin{theorem} + Sia $(V, \varphi)$ uno spazio euclideo complesso. Allora $f$ è un operatore normale $\iff$ esiste + una base ortonormale $\basis$ di autovettori per $f$. +\end{theorem} + +\begin{proof} Si dimostrano le due implicazioni separatamente. \\ + + \rightproof Poiché $\CC$ è algebricamente chiuso, $p_f$ è sicuramente riducibile. Pertanto, + per il teorema di triangolazione con base ortonormale, esiste una base ortonormale $\basis$ + a bandiera per $f$. In particolare, $M_\basis(f)$ è sia normale che triangolare superiore. + Allora, per il \textit{Lemma 1}, $M_\basis(f)$ è diagonale, e dunque $\basis$ è anche una + base di autovettori per $f$. \\ + + \leftproof Se esiste una base ortonormale $\basis$ di autovettori per $f$, $M_\basis(f)$ è + diagonale, e dunque anche normale. Allora, poiché $\basis$ è ortonormale, anche $f$ + è normale. +\end{proof} + +\begin{corollary} + Sia $A \in M(n, \CC)$. Allora $A$ è normale $\iff$ $\exists U \in U_n$ tale che $U\inv A U = U^* A U$ + è diagonale. +\end{corollary} + +\begin{proof} Si dimostrano le due implicazioni separatamente. \\ + + \rightproof Sia $\basis$ la base canonica di $\CC^n$. + Si consideri l'applicazione lineare $f_A$ indotta da $A$ su $\CC^n$. Se $A$ è normale, allora + anche $f_A$ lo è. Pertanto, per il precedente teorema, esiste una base ortonormale $\basis' = \{ \vv 1, \ldots, \vv n \}$ di + autovettori per $f_A$. In particolare, $U = M_{\basis}^{\basis'}(\Id) = \Matrix{\vv 1 & \rvline & \cdots & \rvline & \vv n}$ è unitaria ($U \in U_n$), dacché le colonne di $U$ sono ortonormali. Si osserva inoltre che + $M_{\basis}(f_A) = A$ e che $D = M_{\basis'}(f_A)$ è diagonale. Allora, per la formula del cambiamento di base per le applicazioni lineari, + si conclude che: + + \[ A = U D U\inv \implies D = U\inv A U = U^* A U, \] + + ossia che $U^* A U$ è diagonale. \\ + + \leftproof Sia $D = U^* A U$. Dacché $D$ è diagonale, $D$ è anche normale. Pertanto $D D^* = D^* D$. + Sostituendo, si ottiene che $U^* A U U^* A^* U = U^* A^* U U^* A U$. Ricordando che $U^* U = I_n$ e + che $U \in U_n$ è sempre invertibile, si conclude che $A A^* = A^* A$, ossia che $A$ è normale a + sua volta, da cui la tesi. +\end{proof} + +\begin{remark}\nl + \li Si può osservare mediante l'applicazione dell'ultimo corollario che, se $A$ è hermitiana (ed è dunque + anche normale), + $\exists U \in U_n \mid U^* A U = D$, dove $D \in M(n, \RR)$, ossia tale + corollario implica il teorema spettrale in forma complessa. Infatti + $\conj{D} = D^* = U^* A^* U = U^* A U = D \implies D \in M(n, \RR)$. \\ + + \li Se $A \in M(n, \RR)$ è una matrice normale reale (i.e.~$A A^\top = A^\top A$) con + $p_A$ completamente riducibile in $\RR$, allora è possibile reiterare la dimostrazione + del precedente teorema per concludere che $\exists O \in O_n \mid O^\top A O = D$ con + $D \in M(n, \RR)$, ossia che $A = O D O^\top$. + Tuttavia questo implica che $A^\top = (O D O^\top) = O D^\top O^\top = O D O^\top = A$, + ossia che $A$ è simmetrica. In particolare, per il teorema spettrale reale, vale + anche il viceversa. Pertanto, se $A \in M(n, \RR)$, $A$ è una matrice normale reale con $p_A$ completamente + riducibile in $\RR$ $\iff$ $A = A^\top$. +\end{remark} + +\begin{exercise} + Sia $V$ uno spazio dotato del prodotto $\varphi$. Sia + $W \subseteq V$ un sottospazio di $V$. Sia $\basis_W= \{ \ww 1, \ldots, \ww k \}$ + una base di $W$ e sia $\basis = \{ \ww 1, ..., \ww k, \vv{k+1}, ..., \vv n \}$ una base di $V$. + Sia $A = M_\basis(\varphi)$. Si dimostrino allora i seguenti risultati. + + \begin{enumerate}[(i)] + \item $W^\perp = \{ \v \in V \mid \varphi(\v, \ww i) = 0 \}$, + \item $W^\perp = \{ \v \in V \mid A_{1,\ldots,k} [\v]_\basis = 0 \} = [\cdot]_\basis\inv (\Ker A_{1,\ldots,k})$, + \item $\dim W^\top = \dim V - \rg(A_{1,\ldots,k})$, + \item Se $\varphi$ è non degenere, $\dim W + \dim W^\perp = \dim V$. + \end{enumerate} +\end{exercise} + +\begin{proof}[Soluzione] + Chiaramente vale l'inclusione $W^\perp \subseteq \{ \v \in V \mid \varphi(\v, \ww i) = 0 \}$. Sia + allora $\v \in V \mid \varphi(\v, \ww i) = 0$ $\forall 1 \leq i \leq k$ e sia $\w \in W$. Allora esistono $\alpha_1$, ..., $\alpha_k$ tali + che $\w = \alpha_1 \ww 1 + \ldots + \alpha_k \ww k$. Pertanto si conclude che $\varphi(\v, \alpha_1 \ww 1 + \ldots + \alpha_k \ww k) = \alpha_1 \varphi(\v, \ww 1) + \ldots + \alpha_k \varphi(\v, \ww k) = 0 \implies \v \in W^\top$. Pertanto $W^\top = \{ \v \in V \mid \varphi(\v, \ww i) = 0 \}$, dimostrando (i). \\ + + Si osserva che $\varphi(\v, \ww i) = 0 \iff \varphi(\ww i, \v) = 0$. Se $\varphi$ è scalare, allora + $\varphi(\ww i, \v) = 0 \defiff [\ww i]_\basis^\top A [\v]_\basis = (\e i)^\top A [\v]_\basis = A_i [\v]_\basis = 0$. Pertanto $\v \in W^\top \iff A_i [\v]_\basis = 0$ $\forall 1 \leq i \leq k$, ossia se + $A_{1, \ldots, k} [\v]_\basis = 0$ e $[\v]_\basis \in \Ker A_{1, \ldots, k}$, dimostrando (ii). Analogamente + si ottiene la tesi se $\varphi$ è hermitiano. + Applicando la formula delle dimensioni, si ricava dunque che $\dim W^\top = \dim \Ker A_{1, \ldots, k} = + \dim V - \rg A_{1, \ldots, k}$, dimostrando (iii). \\ + + Se $\varphi$ è non degenere, $A$ è invertibile, dacché $\dim V^\perp = \dim \Ker A = 0$. Allora + ogni minore di taglia $k$ di $A$ ha determinante diverso da zero. Dacché ogni minore di taglia $k$ + di $A_{1,\ldots,k}$ è anche un minore di taglia $k$ di $A$, si ricava che anche ogni minore di taglia + $k$ di $A_{1, \ldots, k}$ ha determinante diverso da zero, e quindi che $\rg(A_{1,\ldots,k}) \geq k$. + Dacché deve anche valere $\rg(A_{1,\ldots,k}) \leq \min\{k,n\} = k$, si conclude che $\rg(A_{1,\ldots,k})$ + vale esattamente $k = \dim W$. Allora, dal punto (iii), vale che $\dim W^\perp + \dim W = \dim W^\perp + \rg(A_{1,\ldots,k}) = \dim V$, dimostrando il punto (iv). +\end{proof} + +\begin{exercise} + Sia $V$ uno spazio dotato del prodotto $\varphi$. Sia + $U \subseteq V$ un sottospazio di $V$. Si dimostrino allora i seguenti due + risultati. + + \begin{enumerate}[(i)] + \item Il prodotto $\varphi$ + induce un prodotto $\tilde \varphi : V/U \times V/U \to \KK$ tale che + $\tilde \varphi(\v + U, \v' + U) = \varphi(\v, \v')$ se e soltanto se $U \subseteq V^\perp$, ossia + se e solo se $U \perp V$. + + %TODO: controllare che debba valere $U = V^\perp$ + \item Se $U = V^\perp$, allora il prodotto $\tilde \varphi$ è non degenere. + + \item Sia $\pi : V \to V/V^\perp$ l'applicazione lineare di proiezione al quoziente. Allora + $U^\perp = \{ \v \in V \mid \tilde \varphi(\pi(\v), \pi(\U)) = 0 \, \forall \U \in U \} = \pi\inv(\pi(U)^\perp)$. + + \item Vale la formula delle dimensioni per il prodotto $\varphi$: $\dim U + \dim U^\perp = \dim V + \dim (U \cap V^\perp)$. + \end{enumerate} +\end{exercise} + +\begin{proof}[Soluzione] + Sia $\w = \v + \uu 1 \in \v + U$, con $\uu 1 \in U$. + Se $\tilde \varphi$ è ben definito, allora deve valere l'uguaglianza $\varphi(\v, \v') = \varphi(\w, \v') = + \varphi(\v + \uu 1, \v') = \varphi(\v, \v') + \varphi(\uu 1, \v')$, ossia $\varphi(\uu 1, \v') = 0$ $\forall \v' \in V \implies \uu 1 \in V^\perp \implies U \subseteq V^\perp$. Viceversa, se $U \subseteq V^\perp$, + sia $\w' = \v' + \uu 2 \in \v' + U$, con $\uu 2 \in U$. Allora vale la seguente identità: + + \[ \varphi(\w, \w') = \varphi(\v + \uu 1, \v' + \uu 2) = \varphi(\v, \v') + \underbrace{\varphi(\v, \uu 2) + \varphi(\uu 1, \v') + \varphi(\uu 1, \uu 2)}_{=\,0}. \] + + Pertanto $\tilde \varphi$ è ben definito, dimostrando (i). \\ + + Sia ora $U = V/V^\perp$. Sia $\v + U \in (V/U)^\perp = \Rad(\tilde \varphi)$. Allora, $\forall \v' + U \in V/U$, + $\tilde \varphi(\v + U, \v' + U) = \varphi(\v, \v') = 0$, ossia $\v \in V^\perp = U$. Pertanto + $\v + U = U \implies \Rad(\tilde \varphi) = \{ V^\perp \}$, e quindi $\tilde \varphi$ è non degenere, + dimostrando (ii). \\ + + Si dimostra adesso l'uguaglianza $U^\perp = \pi\inv(\pi(U)^\perp)$. Sia $\v \in U^\perp$. Allora + $\tilde \varphi(\pi(\v), \pi(\U)) = \tilde \varphi(\v + V^\perp, \U + V^\perp) = \varphi(\v, \U) = 0$ $\forall + \U \in U$, da cui si ricava che vale l'inclusione $U^\perp \subseteq \pi\inv(\pi(U)^\perp)$. Sia + ora $\v \in \pi\inv(\pi(U)^\perp)$, e sia $\U \in U$. Allora $\varphi(\v, \U) = \tilde \varphi(\v + V^\perp, \U + V^\perp) = \tilde \varphi(\pi(\v), \pi(\U)) = 0$, da cui vale la doppia inclusione, e dunque l'uguaglianza + desiderata, dimostrando (iii). \\ + + Dall'uguaglianza del punto (iii), l'applicazione della formula delle dimensioni e l'identità + ottenuta dal punto (iv) dell'\textit{Esercizio 2} rispetto al prodotto $\tilde \varphi$ non degenere, si ricavano + le seguenti identità: + + \[ \system{ \dim \pi(U) = \dim U - \dim (U \cap \Ker \pi) = \dim U - \dim (U \cap V^\perp), \\ \dim \pi(U)^\perp = \dim V/V^\perp - \dim \pi(U) = \dim V - \dim V^\perp - \dim \pi(U), \\ \dim U^\perp = \dim \pi(U)^\perp + \dim \Ker \pi = \dim \pi(U)^\perp + \dim V^\perp, } \] + + dalle quali si ricava la seguente identità: + + \[ \dim U^\perp = \dim V - \dim V^\perp - (\dim U - \dim(U \cap V^\perp)) + \dim V^\perp, \] + + \vskip 0.05in + + da cui si ricava che $\dim U + \dim U^\perp = \dim V + \dim(U \cap V^\perp)$, dimostrando (iv). +\end{proof} + +\begin{exercise} Sia $V$ uno spazio vettoriale dotato del prodotto $\varphi$. Si dimostri allora che $(W^\perp)^\perp = W + V^\perp$. +\end{exercise} + +\begin{proof}[Soluzione] + Sia $\v = \w' + \v' \in W + V^\perp$, con $\w' \in W$ e $\v' \in V^\perp$. Sia inoltre $\w \in W^\perp$. + Allora $\varphi(\v, \w) = \varphi(\w' + \v', \w) = \varphi(\w', \w) + \varphi(\v', \w) = 0$, + dove si è usato che $\w' \perp \w$ dacché $\w \in W^\perp$ e $\w' \in W$ e che $\v' \in V^\perp$. Allora + vale l'inclusione $W + V^\perp \subseteq (W^\perp)^\perp$. \\ + + Applicando le rispettive formule delle dimensioni a $W^\perp$, $(W^\perp)^\perp$ e $W + V^\perp$ si ottengono le seguenti identità: + + \[ \system{ \dim W^\perp = \dim V + \dim (W \cap V^\perp) - \dim W, \\ \dim (W^\perp)^\perp = \dim V + \dim (W^\perp \cap V^\perp) - \dim W^\perp, \\ \dim (W + V^\perp) = \dim W + \dim V^\perp - \dim (W \cap V^\perp), } \] + + \vskip 0.05in + + da cui si ricava che: + + \[ \dim (W^\perp)^\perp = \dim W + \dim V^\perp - \dim (W \cap V^\perp) = \dim (W + V^\perp). \] + + Dal momento che vale un'inclusione e l'uguaglianza dimensionale, si conclude che $(W^\perp)^\perp = W + V^\perp$, + da cui la tesi. +\end{proof} + +\begin{exercise} Sia $A \in M(n, \CC)$ anti-hermitiana (i.e.~$A = -A^*$). Si dimostri allora che $A$ + è normale e che ammette solo autovalori immaginari. +\end{exercise} + +\begin{proof}[Soluzione] + Si mostra facilmente che $A$ è normale. Infatti $A A^* = A(-A) = -A^2 = (-A)A = A^* A$. Sia allora + $\lambda \in \CC$ un autovalore di $A$ e sia $\v \neq \vec 0$, $\v \in V_\lambda$. Si consideri il prodotto hermitiano + standard $\varphi$ su $\CC^n$. Allora vale la seguente identità: + \begin{gather*} + \lambda \, \varphi(\v, \v) = \varphi(\v, \lambda \v) = \varphi(\v, A \v) = \varphi(A^* \v, \v) = \\ + \varphi(-A\v, \v) = \varphi(-\lambda \v, \v) = -\conj{\lambda} \, \varphi(\v, \v). + \end{gather*} + + Dacché $\varphi$ è definito positivo, $\varphi(\v, \v) \neq 0 \implies \lambda = -\conj{\lambda}$. Allora + $\Re(\lambda) = \frac{\lambda + \conj{\lambda}}{2} = 0$, e quindi $\lambda$ è immaginario, da cui la tesi. +\end{proof} + +\begin{exercise} + Sia $V$ uno spazio vettoriale dotato del prodotto $\varphi$. Siano $U$, $W \subseteq V$ due sottospazi + di $V$. Si dimostrino allora le due seguenti + identità. + + \begin{enumerate}[(i)] + \item $(U + W)^\perp = U^\perp \cap W^\perp$, + \item $(U \cap W)^\perp \supseteq U^\perp + W^\perp$, dove vale l'uguaglianza insiemistica se $\varphi$ + è non degenere. + \end{enumerate} +\end{exercise} + +\begin{proof}[Soluzione] + Sia $\v \in (U + W)^\perp$ e siano $\U \in U \subseteq U + W$, $\w \in W \subseteq U + W$. Allora + $\varphi(\v, \U) = 0 \implies \v \in U^\perp$ e $\varphi(\v, \w) = 0 \implies \v \in W^\perp$, + da cui si conclude che $(U + W)^\perp \subseteq U^\perp \cap W^\perp$. Sia adesso + $\v \in U^\perp \cap W^\perp$ e $\v' = \U + \w \in U + W$ con $\U \in V$ e $\w \in W$. Allora + $\varphi(\v, \v') = \varphi(\v, \U) + \varphi(\v, \w) = 0 \implies \v \in (U + W)^\perp$, da cui + si deduca che vale la doppia inclusione, e quindi che $(U + W)^\perp = U^\perp \cap W^\perp$, + dimostrando (i). \\ + + Sia ora $\v' = \U' + \w' \in U^\perp + W^\perp$ con $\U' \in U^\perp$ e $\w' \in W^\perp$. Sia + $\v \in U \cap W$. Allora $\varphi(\v, \v') = \varphi(\v, \U') + \varphi(\v, \w') = 0 \implies + \v' \in (U \cap W)^\perp$, da cui si deduce che $(U \cap W)^\perp \supseteq U^\perp + W^\perp$. + Se $\varphi$ è non degenere, $\dim (U^\perp + W^\perp) = \dim U^\perp + \dim W^\perp - \dim (U^\perp \cap W^\perp) = 2 \dim V - \dim U - \dim W - \dim (U+W)^\perp = \dim V - \dim U - \dim W + \dim (U + W) = + \dim V - \dim (U + W) = \dim (U + W)^\perp$. Valendo pertanto l'uguaglianza dimensionale, si + conclude che in questo caso $(U \cap W)^\perp = U^\perp + W^\perp$, dimostrando (ii). + +\end{proof} \ No newline at end of file diff --git a/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/logo.png b/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/logo.png new file mode 100644 index 0000000000000000000000000000000000000000..822d40e180950097cf3da6cc96dce63fc5dbf295 GIT binary patch literal 95066 zcmXV1Wl$ST*DVDKl;Tj_t$1;Fcei3K?he6Sg1bv_cX!(2?k>UI-TBhz{eEP2l9^Hd;UG32JA|VKYVyElok_KbJIV`g7Z>SzZ*Dj zwUa?-Ns~SR8va>tuWFCb!;}&VJ8pYhV6@Uqhh^XT;YUGIUS57&DZ9@)WK3 zW(X}@NXdGry3YR&c4tLEfRQRWT0h@sO}jtqJa4+cT0dXozSmXx8qsH{EhrzA(*1zYjTOQLQd{~6o?P4}pgK=!EjR=@Gair>jc4|!vqc1_-b zVXYRA$Ys8QkKr$-d>2iQB08G0sL9H(QqHrIPRw#!xzz|v3*ahOZ8&Qw3G=<0n2*NZ zSJBd|^is`gDYF)Y#dk(u^JB4W751_soqbD}`x;}ae#nQunbrR9!-aNwrv*-?xqwR; z4sb}u0l6-7^nEYne|=_UUVlrx4(`yqvNPP(FGZ)>39Y_`KM26@9GB5b_@r<<=g(M7 zR<3y`@?q`XlBzgiW>iOAzHSuH#$(M$ENsok9Q~mGI`aNi;|8+I!og!Ffy9Pa;geQy z0h4^I5`pMdNQ6IH4Q_cYPqelvt%Vo1R-CrOLJ--@rxv}LOb(`(amfW02W-8jciE9i z_GD|?v4hTIO3ta3+K5F{edSNQL}?%8)$i^i<`NUuZj>Qgb~VyJu-GfAHw7B;V8Q(uOD(^nD}R-l{rM5Ojhlu9M9SrZX|!AmjAu})_VCljk;8;;g#{fQZ)P5MwRRwrlC z8rI=>iKu?7FIvTyTzmn4SN8zHP5wSdzWT{P(?R`+Nsx8r4O}=)AdJ{^w1^wl6T)UmQ`)Ut?xL?cLnKK|P7ILa?~bgz{2&d&3utdlT2j~yD&nf9P$ddpq9qd8aFMw|n4 z%sLegY{C%iOj}d`IgGEi&gm<4NaU@+G&Brj>lo?kDm=Af?3!z`rv1wxuXWt7yy#0r zWvc6AhgQtRm&!HSDPP|n=3l*7ahsQ$z5bpuuQNjoUn6$+Bd|IliP&I4Fy7tZDMH6;ki>Fqd z<9Ud?+#Zwpf>FU>%p`b=*+DynG{K_)LmN^-6WDbB{=$i`qBnGdtqCO4O@AaMl)|LP zDn?87Lr$aa61<3r(KV1?KjHQ#ZYrk8x0WFP84d52)cPwcmMA{t!R(@Cqc$t~b_N7v zg<6-((K2ClNQV`=S>5zSf8#Xj?{^xM&NZ0d$_JcisdfU^IL#L^33yQkp5mSW(ww(e z>blb>35yVH@`=`(vVts4sLG#x`^GftCxfJ(jamdooXDG|faM?ioH2v87oru%#DUZV z>@#F41SvLuQfKeV3XIvOj%k~DyPkxo0P=AO8OUo552~g*vvqc+*o0zsb%IJ+OM0xR zU2^QzIcRp}#J{=pzhEzKUjU-TV5(&oCr%KGXXrwi6~_-GXb zMP4ZW1OFH6UwINLvjm*8nx&J zu__M!Ejnyz9`LM}NxX1i@b=2}ypRkEFm(2f?6V}?Dns<(D)Je8R6hcUMjO%J3jZ*C5*9jCGK@vOxe-RaIhU$PC0mlWHgcf3RboB) z>~>&jq6ao$98T((PT9Xlx^!bj^l1XuklP`;e$y- zJ7hl*a1y0d%`RGXbsO+|oweRh3Ecdu;2J7cZR!in${F(|L1U#r&8A&W!jLtjJ7&BE zcYnv~gs33Qtl!Dp(%kjTA?wrHvbSo44+NY2s4!$Ike!L)rH$baOh$V-q7SM^q z?G0ZK#7Q1m^*I%Dtbwq6d!Z)uuVH_S2=R zCk%wS=0_dD)OJs2LfL&yt8~r9cuR4J1?sb{dCn7@agB@**K%uhmtiOhV2p z2gbh^pUeO?WLzxulJlnUoms--DHJTPm-YPXZk#a9Z_m7#C|pAVGC0wEDttzTu}{LL z$?udgAQf=ze7z-kaY%-*Cy!hBxG=di6|QVHP!!e1$l$R)e%<^bDtHy91`{5Mc!LbK*0nW@d7=>^w;&P32DjYj}oGWJu z@l%sk303==dqKd^cAOiwb&;ftws(YJbzV15ez64lW6b?u2r8Sf{`%LwmnwFU>5tJF z>4{HZ0v3z_WMapZsKRjH^6Ck;bukpY>j9p=ZyZruox<+qoych)s8wfB`QSh|=GA$k ziC-rWj^S{@rKlF#VmB5x@?$A!8}l_kekW?@eB>$s^e!CbkBAX?dhKLt_3aXuXdB@p zJ}HhLvmF$j;jfI($v<5y&kCpM2|2|}PY|^lnKu&R!TG0udcAO&)NI|OOxBrfwfABg z^ZNW`pxcPR6-^zdkz#(Q7t+;&=iZ~Sh%JylcwKguT6i9ZjgwZFBj`E_<3`=_y?7dv{pqyU3{Z5=77fSRd ziI&drsPPyBZAq4A6WmJgHjIFy)g!%>n>xAv^>4ZV+5rbXnRS{Pe^a($U$bzgPh8lQ zkQXZGP2%{{z0)dMH|0X!g@w!w-~Q*<0;i_PbRMKGGqgL`ge-=u6+LR)twtlZvQ-~c zD_+_ykDhU-w2ir+cML+L481vr$3tU{N|ULb#28thg?-#IxAE-?>kHc-kxC874XVK; 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z*U-Y5#~qJ_hw1Ssql?2~$R<#QS!qadc$%D`ZetHB?7qpQ!k|H&nwo1Go7Foc7y#G4 NI`!!@?9}0%{u_geXr2H7 literal 0 HcmV?d00001 diff --git a/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/main.tex b/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/main.tex new file mode 100644 index 0000000..4ed14dd --- /dev/null +++ b/Geometria 1/Prodotto scalare e hermitiano/I prodotti di uno spazio vettoriale/main.tex @@ -0,0 +1,39 @@ +\PassOptionsToPackage{main=italian}{babel} +\documentclass[11pt]{scrbook} +\usepackage[utf8]{inputenc} +\usepackage[italian]{babel} +\usepackage{recap} + +\begin{document} + + \title{I prodotti di uno spazio vettoriale} + \subtitle{Dispense del corso di Geometria 1} + \author{Gabriel Antonio Videtta} + \date{A.A. 2022/2023 \\ \vskip 1in \includegraphics[scale=0.3]{logo.png}} + \maketitle + + \newpage + \thispagestyle{empty} + ~\newpage + + \tableofcontents + + \newpage + \thispagestyle{empty} + ~\newpage + + \include{1. Introduzione al prodotto scalare} + + \newpage + \thispagestyle{empty} + ~\newpage + + \include{2. Proprietà e teoremi principali sul prodotto scalare} + + \newpage + \thispagestyle{empty} + ~\newpage + + \include{3. Prodotti hermitiani, spazi euclidei e teorema spettrale} + +\end{document} diff --git a/Geometria 1/Spazi affini/2023-04-26, Azioni di un gruppo e introduzione agli spazi affini/main.pdf b/Geometria 1/Spazi affini/2023-04-26, Azioni di un gruppo e introduzione agli spazi affini/main.pdf index 56d316f4f359ec03cc06d71189c3148ae00d618c..3c67f6c39e88dc0aa1f6795f00e3e8a71b6d78e0 100644 GIT binary patch delta 161 zcmZ2*n{UBwzJ?aY7N#xC{)@DX4b6=Vj4ZSbjMNPb)HS*Eee+XX5=&AQG+eBJ62=Cm z22ds2D;F^z=SpyMb~AT2H8pZFwXifbaxrnVa5gqDGqE%>c64=dGBUAKupy`97W ctGFbwsHCDOHI2*2*uu!tgiBS`)!&T^0KxbvX8-^I delta 161 zcmZ2*n{UBwzJ?aY7N#xC{)@B>4b4pqOboOQjMNPb)HS*Eee+XX5=&AQG+eBV42+Bo zObwt)wpT797W ctGFbwsHCDOHI2*2*uu!tgiBS`)!&T^0JzO5MF0Q* diff --git a/Geometria 1/Spazi affini/2023-04-26, Azioni di un gruppo e introduzione agli spazi affini/main.tex b/Geometria 1/Spazi affini/2023-04-26, Azioni di un gruppo e introduzione agli spazi affini/main.tex index 18cfa69..2a6c95a 100644 --- a/Geometria 1/Spazi affini/2023-04-26, Azioni di un gruppo e introduzione agli spazi affini/main.tex +++ b/Geometria 1/Spazi affini/2023-04-26, Azioni di un gruppo e introduzione agli spazi affini/main.tex @@ -90,7 +90,7 @@ tramite la congruenza, ossia tramite la mappa $(P, A) \mapsto P^\top A P$. L'orbita $\Orb(A)$ è la classe di congruenza delle matrice simmetria $A \in \Sym(n, \KK)$. Analogamente si può costruire un'azione per le matrici hermitiane. - \item Se $G = O_n$, il gruppo delle matrici ortogonali di taglia $n$ su $\KK$, $G$ opera su $\RR^n$ tramite la mappa $O \cdot \vec v \mapsto O \vec v$. L'orbita $\Orb(\vec v)$ è in particolare la sfera $n$-dimensionale di raggio $\norm{x}$. + \item Se $G = O_n$, il gruppo delle matrici ortogonali di taglia $n$ su $\KK$, $G$ opera su $\RR^n$ tramite la mappa $O \cdot \vec v \mapsto O \vec v$. L'orbita $\Orb(\vec v)$ è in particolare la sfera $n$-dimensionale di raggio $\norm{v}$. \end{enumerate} \end{example} diff --git a/tex/latex/style/recap.sty b/tex/latex/style/recap.sty new file mode 100644 index 0000000..ab1da82 --- /dev/null +++ b/tex/latex/style/recap.sty @@ -0,0 +1,615 @@ +\ProvidesPackage{recap} + +\usepackage{amsmath,amssymb} +\usepackage{amsfonts} +\usepackage{amsthm} +\usepackage{amssymb} +\usepackage{amsopn} +\usepackage{mathtools} +\usepackage{stmaryrd} +\usepackage{marvosym} +\usepackage{float} +\usepackage{enumerate} +\usepackage{scalerel} +\usepackage{stackengine} +\usepackage{wasysym} +\usepackage{iftex} +\usepackage[minimal]{yhmath} + +\PassOptionsToPackage{usenames,svgnames,dvipsnames,table}{xcolor} +\usepackage{xcolor} +\usepackage[colorlinks=true]{hyperref} +\hypersetup{urlcolor=RubineRed,linkcolor=RoyalBlue,citecolor=ForestGreen} +\usepackage[nameinlink]{cleveref} + +\usepackage{amsthm} +\usepackage{thmtools} +\usepackage[framemethod=TikZ]{mdframed} + +\usepackage{listings} +\usepackage{mathrsfs} +\usepackage{textcomp} + +\usepackage[shortlabels]{enumitem} +\usepackage[obeyFinal,textsize=scriptsize,shadow]{todonotes} +\usepackage{textcomp} +\usepackage{multirow} +\usepackage{ellipsis} +\usepackage{mathtools} +\usepackage{microtype} +\usepackage{xstring} +\usepackage{wrapfig} + +\usepackage[headsepline]{scrlayer-scrpage} + +\hfuzz=\maxdimen +\tolerance=10000 +\hbadness=10000 + +\newcommand{\system}[1]{\begin{cases} #1 \end{cases}} + +\newcommand{\wip}{\begin{center}\textit{Questo avviso sta ad indicare che questo documento è ancora una bozza e non è + da intendersi né completo, né revisionato.}\end{center}} + +\newcommand\hr{\vskip 0.05in \par\vspace{-.5\ht\strutbox}\noindent\hrulefill\par} + +% Modalità matematica/fisica +\let\oldvec\vec +\renewcommand{\vec}[1]{\underline{#1}} + +\newcommand{\E}{\text{ e }} +\newcommand{\altrimenti}{\text{altrimenti}} +\newcommand{\se}{\text{se }} +\newcommand{\tc}{\text{ t.c. }\!} +\newcommand{\epari}{\text{ è pari}} +\newcommand{\edispari}{\text{ è dispari}} + +\newcommand{\nl}{\ \\} + +\newcommand{\lemmaref}[1]{\textit{Lemma \ref{#1}}} +\newcommand{\thref}[1]{\textit{Teorema \ref{#1}}} + +\newcommand{\li}[0]{$\blacktriangleright\;\;$} + +\newcommand{\tends}[1]{\xrightarrow[\text{$#1$}]{}} +\newcommand{\tendsto}[1]{\xrightarrow[\text{$x \to #1$}]{}} +\newcommand{\tendstoy}[1]{\xrightarrow[\text{$y \to #1$}]{}} +\newcommand{\tendston}[0]{\xrightarrow[\text{$n \to \infty$}]{}} + +\setlength\parindent{0pt} + +% Principio di induzione e setup dimostrativi. +\newcommand{\basestep}{\mbox{(\textit{passo base})}\;} +\newcommand{\inductivestep}{\mbox{(\textit{passo induttivo})}\;} + +\newcommand{\rightproof}{\mbox{($\implies$)}\;} +\newcommand{\leftproof}{\mbox{($\impliedby$)}\;} + +% Spesso utilizzati al corso di Fisica 1. +\newcommand{\dx}{\dot{x}} +\newcommand{\ddx}{\ddot{x}} +\newcommand{\dv}{\dot{v}} + +\newcommand{\del}{\partial} +\newcommand{\tendstot}[0]{\xrightarrow[\text{$t \to \infty$}]{}} + +\newcommand{\grad}{\vec{\nabla}} +\DeclareMathOperator{\rot}{rot} + +\newcommand{\ihat}{\hat{i}} +\newcommand{\jhat}{\hat{j}} +\newcommand{\khat}{\hat{k}} + +\newcommand{\der}[1]{\frac{d#1}{dx}} +\newcommand{\parx}{\frac{\del}{\del x}} +\newcommand{\pary}{\frac{\del}{\del y}} +\newcommand{\parz}{\frac{\del}{\del z}} + +% Spesso utilizzati al corso di Analisi 1. +%\newcommand{\liminf}{\lim_{x \to \infty}} +\newcommand{\liminfty}{\lim_{x \to \infty}} +\newcommand{\liminftym}{\lim_{x \to -\infty}} +\newcommand{\liminftyn}{\lim_{n \to \infty}} +\newcommand{\limzero}{\lim_{x \to 0}} +\newcommand{\limzerop}{\lim_{x \to 0^+}} +\newcommand{\limzerom}{\lim_{x \to 0^-}} + +\newcommand{\cc}{\mathcal{C}} + +\newcommand{\xbar}{\overline{x}} +\newcommand{\ybar}{\overline{y}} +\newcommand{\tbar}{\overline{t}} +\newcommand{\zbar}{\overline{z}} +\newcommand{\RRbar}{\overline{\RR}} + +% Spesso utilizzati al corso di Geometria 1. +\newcommand{\Aa}{\mathcal{A}} +\DeclareMathOperator{\An}{\mathcal{A}_n} +\DeclareMathOperator{\AnK}{\mathcal{A}_n(\KK)} +\DeclareMathOperator{\Giac}{Giac} + +\DeclareMathOperator{\IC}{IC} +\DeclareMathOperator{\Aff}{Aff} +\DeclareMathOperator{\Orb}{Orb} +\DeclareMathOperator{\Stab}{Stab} +\DeclareMathOperator{\Gr}{Gr} +\newcommand{\vvec}[1]{\overrightarrow{#1}} + +\newcommand{\conj}[1]{\overline{#1}} + +\let\imm\Im +\let\Im\undefined +\DeclareMathOperator{\Im}{Im} +\DeclareMathOperator{\Rad}{Rad} +\newcommand{\restr}[2]{ + #1\arrowvert_{#2} +} + +\newcommand{\innprod}[1]{\langle #1 \rangle} + +\newcommand{\zerovecset}{\{\vec 0\}} +\newcommand{\bigzero}{\mbox{0}} +\newcommand{\rvline}{\hspace*{-\arraycolsep}\vline\hspace*{-\arraycolsep}} + +\newcommand{\Idv}{\operatorname{Id}_V} +\newcommand{\Idw}{\operatorname{Id}_W} +\newcommand{\IdV}[1]{\operatorname{Id}_{#1}} +\DeclareMathOperator{\CI}{CI} +\DeclareMathOperator{\Bil}{Bil} +\DeclareMathOperator{\Mult}{Mult} +\DeclareMathOperator{\sgn}{sgn} +\DeclareMathOperator{\Ann}{Ann} +\DeclareMathOperator{\adj}{adj} +\DeclareMathOperator{\Cof}{Cof} +\DeclareMathOperator{\pr}{pr} +\DeclareMathOperator{\Sp}{sp} + +\newcommand{\dperp}{{\perp\perp}} + +\newcommand{\Eigsp}[0]{V_{\lambda}} +\newcommand{\Gensp}[0]{\widetilde{V_{\lambda}}} +\newcommand{\eigsp}[1]{V_{\lambda_{#1}}} +\newcommand{\gensp}[1]{\widetilde{V_{\lambda_{#1}}}} +\newcommand{\genspC}[1]{\widetilde{V_{#1}}} + +\DeclareMathOperator{\val}{val} +\DeclareMathOperator{\Span}{Span} +\newcommand{\charpoly}[1]{p_{#1}} +\newcommand{\charpolyrestr}[2]{p_{#1\arrowvert_#2}\hspace{-1pt}(\lambda)} +\newcommand{\minpoly}[1]{\varphi_{#1}} +\newcommand{\valf}{\val_f} +\newcommand{\valfv}{\val_{f,\V}} +\newcommand{\e}[1]{\vec{e_{#1}}} +\newcommand{\V}{\vec{v}} +\newcommand{\VV}[1]{\vec{v_{#1}}} +\newcommand{\basisdual}{\dual{\basis}} +\newcommand{\vecdual}[1]{\vec{\dual{#1}}} +\newcommand{\vecbidual}[1]{\vec{\bidual{#1}}} +\newcommand{\Matrix}[1]{\begin{pmatrix} #1 \end{pmatrix}} +\newcommand{\Vector}[1]{\begin{pmatrix} #1 \end{pmatrix}} + +\DeclareMathOperator{\rg}{rg} + +\let\v\undefined +\newcommand{\v}{\vec{v}} +\newcommand{\vv}[1]{\vec{v_{#1}}} +\newcommand{\w}{\vec{w}} +\newcommand{\U}{\vec{u}} +\newcommand{\ww}[1]{\vec{w_{#1}}} +\newcommand{\uu}[1]{\vec{u_{#1}}} + +\newcommand{\mapstoby}[1]{\xmapsto{#1}} +\newcommand{\oplusperp}{\oplus^\perp} + +% Comandi personali. + +\newcommand{\card}[1]{\left|#1\right|} +\newcommand{\nsqrt}[2]{\!\sqrt[#1]{#2}\,} +\newcommand{\zeroset}{\{0\}} +\newcommand{\setminuszero}{\setminus \{0\}} + +\newenvironment{solution} +{\textit{Soluzione.}\,} + +\theoremstyle{definition} + +\let\abstract\undefined +\let\endabstract\undefined + +\newcommand{\basis}{\mathcal{B}} +\newcommand{\BB}{\mathcal{B}} + +\newcommand{\basisC}{\mathcal{B}} + +\newcommand{\HH}{\mathbb{H}} + +\newcommand{\FFp}[1]{\mathbb{F}_p} +\newcommand{\FFpx}[1]{\mathbb{F}_p[x]} + +\newcommand{\CCx}{\mathbb{C}[x]} + +\newcommand{\KK}{\mathbb{K}} +\newcommand{\KKx}{\mathbb{K}[x]} + +\newcommand{\QQx}{\mathbb{Q}[x]} +\newcommand{\RRx}{\mathbb{R}[x]} + +\newcommand{\ZZi}{\mathbb{Z}[i]} +\newcommand{\ZZp}{\mathbb{Z}_p} +\newcommand{\ZZpx}{\mathbb{Z}_p[x]} +\newcommand{\ZZx}{\mathbb{Z}[x]} + +\newcommand{\ii}{\mathbf{i}} +\newcommand{\jj}{\mathbf{j}} +\newcommand{\kk}{\mathbf{k}} + +\newcommand{\bidual}[1]{#1^{**}} +\newcommand{\dual}[1]{#1^{*}} +\newcommand{\LL}[2]{\mathcal{L} \left(#1, \, #2\right)} % L(V, W) + +\newcommand{\nsg}{\mathrel{\unlhd}} % sottogruppo normale + +% evan.sty original commands +\newcommand{\cbrt}[1]{\sqrt[3]{#1}} +\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor} +\newcommand{\ceiling}[1]{\left\lceil #1 \right\rceil} +\newcommand{\mailto}[1]{\href{mailto:#1}{\texttt{#1}}} +\newcommand{\eps}{\varepsilon} +\newcommand{\vocab}[1]{\textbf{\color{blue}\sffamily #1}} +\providecommand{\alert}{\vocab} +\newcommand{\catname}{\mathsf} +\providecommand{\arc}[1]{\wideparen{#1}} + +% From H113 "Introduction to Abstract Algebra" at UC Berkeley +\newcommand{\CC}{\mathbb C} +\newcommand{\FF}{\mathbb F} +\newcommand{\NN}{\mathbb N} +\newcommand{\QQ}{\mathbb Q} +\newcommand{\RR}{\mathbb R} +\newcommand{\ZZ}{\mathbb Z} +\DeclareMathOperator{\Aut}{Aut} +\DeclareMathOperator{\Inn}{Inn} +\DeclareMathOperator{\Syl}{Syl} +\DeclareMathOperator{\Gal}{Gal} +\DeclareMathOperator{\GL}{GL} +\DeclareMathOperator{\SL}{SL} + +% From Kiran Kedlaya's "Geometry Unbound" +\newcommand{\abs}[1]{\left\lvert #1 \right\rvert} +\newcommand{\norm}[1]{\left\lVert #1 \right\rVert} +\newcommand{\dang}{\measuredangle} %% Directed angle +\newcommand{\ray}[1]{\overrightarrow{#1}} +\newcommand{\seg}[1]{\overline{#1}} + +% From M275 "Topology" at SJSU +\newcommand{\Id}{\mathrm{Id}} +\newcommand{\id}{\mathrm{id}} +\newcommand{\taking}[1]{\xrightarrow{#1}} +\newcommand{\inv}{^{-1}} + +\DeclareMathOperator{\ord}{ord} +\newcommand{\defeq}{\overset{\mathrm{def}}{=}} +\newcommand{\defiff}{\overset{\mathrm{def}}{\iff}} + +% From the USAMO .tex files +\newcommand{\dg}{^\circ} + +\newcommand{\liff}{\leftrightarrow} +\newcommand{\lthen}{\rightarrow} +\newcommand{\opname}{\operatorname} +\newcommand{\surjto}{\twoheadrightarrow} +\newcommand{\injto}{\hookrightarrow} + +% Alcuni degli operatori più comunemente utilizzati. + +\DeclareMathOperator{\Char}{char} +\DeclareMathOperator{\Dom}{Dom} +\DeclareMathOperator{\Fix}{\textit{Fix}\,} +\DeclareMathOperator{\End}{End} +\DeclareMathOperator{\existsone}{\exists !} +\DeclareMathOperator{\Hom}{Hom} +\DeclareMathOperator{\Imm}{Imm} +\DeclareMathOperator{\Ker}{Ker} +\DeclareMathOperator{\rank}{rank} +\DeclareMathOperator{\MCD}{MCD} +\DeclareMathOperator{\Mor}{Mor} +\DeclareMathOperator{\mcm}{mcm} +\DeclareMathOperator{\Sym}{Sym} +\DeclareMathOperator{\tr}{tr} + +% Reimposta alcuni simboli presenti di default in LaTeX con degli analoghi +% più comuni. + +\let\oldemptyset\emptyset +\let\emptyset\varnothing + +% Trasforma alcuni simboli in operatori matematici. + +\let\oldcirc\circ +\let\circ\undefined +\DeclareMathOperator{\circ}{\oldcirc} + +\let\oldexists\exists +\let\exists\undefined +\DeclareMathOperator{\exists}{\oldexists} + +\let\oldforall\forall +\let\forall\undefined +\DeclareMathOperator{\forall}{\oldforall} + +\let\oldnexists\nexists +\let\nexists\undefined +\DeclareMathOperator{\nexists}{\oldnexists} + +\let\oldland\land +\let\land\undefined +\DeclareMathOperator{\land}{\oldland} + +\let\oldlnot\lnot +\let\lnot\undefined +\DeclareMathOperator{\lnot}{\oldlnot} + +\let\oldlor\lor +\let\lor\undefined +\DeclareMathOperator{\lor}{\oldlor} + +\DeclareOption{physics}{ + \let\vec\oldvec +} + +\ProcessOptions\relax + +\title{} +\author{Gabriel Antonio Videtta} +\date{\today} + +\mdfdefinestyle{mdbluebox}{% + roundcorner=10pt, + linewidth=1pt, + skipabove=12pt, + innerbottommargin=9pt, + skipbelow=2pt, + linecolor=blue, + nobreak=true, + backgroundcolor=TealBlue!5, +} +\declaretheoremstyle[ +headfont=\sffamily\bfseries\color{MidnightBlue}, +mdframed={style=mdbluebox}, +headpunct={\\[3pt]}, +postheadspace={0pt} +]{thmbluebox} + +\mdfdefinestyle{mdbluebox2}{% + roundcorner=10pt, + linewidth=1pt, + skipabove=12pt, + innerbottommargin=9pt, + skipbelow=2pt, + linecolor=blue, + nobreak=true, + backgroundcolor=BlueViolet!9, +} +\declaretheoremstyle[ +headfont=\sffamily\bfseries\color{RoyalPurple}, +mdframed={style=mdbluebox2}, +headpunct={\\[3pt]}, +postheadspace={0pt} +]{thmbluebox2} + +\mdfdefinestyle{mdredbox}{% + linewidth=0.5pt, + skipabove=12pt, + frametitleaboveskip=5pt, + frametitlebelowskip=0pt, + skipbelow=2pt, + frametitlefont=\bfseries, + innertopmargin=4pt, + innerbottommargin=8pt, + nobreak=true, + backgroundcolor=Salmon!5, + linecolor=RawSienna, +} +\declaretheoremstyle[ +headfont=\bfseries\color{RawSienna}, +mdframed={style=mdredbox}, +headpunct={\\[3pt]}, +postheadspace={0pt}, +]{thmredbox} + +\mdfdefinestyle{mdredbox2}{% + roundcorner=10pt, + linewidth=1pt, + skipabove=12pt, + innerbottommargin=9pt, + skipbelow=2pt, + linecolor=red, + nobreak=true, + backgroundcolor=WildStrawberry!5, +} +\declaretheoremstyle[ +headfont=\sffamily\bfseries\color{Maroon}, +mdframed={style=mdredbox2}, +headpunct={\\[3pt]}, +postheadspace={0pt} +]{thmredbox2} + +\mdfdefinestyle{mdgreenbox}{% + skipabove=8pt, + linewidth=2pt, + rightline=false, + leftline=true, + topline=false, + bottomline=false, + linecolor=ForestGreen, + backgroundcolor=ForestGreen!5, +} +\declaretheoremstyle[ +headfont=\bfseries\sffamily\color{ForestGreen!70!black}, +bodyfont=\normalfont, +spaceabove=2pt, +spacebelow=1pt, +mdframed={style=mdgreenbox}, +headpunct={ --- }, +]{thmgreenbox} + +\mdfdefinestyle{mdgreenbox2}{% + roundcorner=10pt, + linewidth=1pt, + skipabove=12pt, + innerbottommargin=9pt, + skipbelow=2pt, + linecolor=ForestGreen, + nobreak=true, + backgroundcolor=ForestGreen!5, +} +\declaretheoremstyle[ +headfont=\sffamily\bfseries\color{ForestGreen!70!black}, +mdframed={style=mdgreenbox2}, +headpunct={\\[3pt]}, +postheadspace={0pt} +]{thmgreenbox2} + +\mdfdefinestyle{mdblackbox}{% + skipabove=8pt, + linewidth=3pt, + rightline=false, + leftline=true, + topline=false, + bottomline=false, + linecolor=black, + backgroundcolor=RedViolet!5!gray!10, +} +\declaretheoremstyle[ +headfont=\bfseries, +bodyfont=\normalfont\small, +spaceabove=0pt, +spacebelow=0pt, +mdframed={style=mdblackbox} +]{thmblackbox} + +\mdfdefinestyle{mdblackbox2}{% + skipabove=8pt, + linewidth=3pt, + rightline=false, + leftline=true, + topline=false, + bottomline=false, + linecolor=gray, + backgroundcolor=RedViolet!5!gray!10, +} +\declaretheoremstyle[ +headfont=\bfseries, +bodyfont=\normalfont\small, +spaceabove=0pt, +spacebelow=0pt, +mdframed={style=mdblackbox2} +]{thmblackbox2} + +\newcommand{\listhack}{$\empty$\vspace{-2em}} + +\declaretheorem[% +style=thmbluebox,name=Teorema,numberwithin=section]{theorem} + +\declaretheorem[style=thmgreenbox2,name=Lemma,sibling=theorem]{lemma} +\declaretheorem[style=thmredbox2,name=Proposizione,sibling=theorem]{proposition} +\declaretheorem[style=thmbluebox2,name=Corollario,sibling=theorem]{corollary} +\declaretheorem[style=thmbluebox,name=Teorema,numbered=no]{theorem*} +\declaretheorem[style=thmgreenbox2,name=Lemma,numbered=no]{lemma*} +\declaretheorem[style=thmredbox2,name=Proposizione,numbered=no]{proposition*} +\declaretheorem[style=thmbluebox2,name=Corollario,numbered=no]{corollary*} +\declaretheorem[style=thmgreenbox,name=Nota,numbered=no]{note} +\declaretheorem[style=thmgreenbox,name=Claim,sibling=theorem]{claim} +\declaretheorem[style=thmgreenbox,name=Claim,numbered=no]{claim*} +\declaretheorem[style=thmredbox,name=Esempio,sibling=theorem]{example} +\declaretheorem[style=thmredbox,name=Esempio,numbered=no]{example*} +\declaretheorem[style=thmblackbox, name=Definizione,sibling=theorem]{definition} +\declaretheorem[style=thmblackbox, name=Definizione,numbered=no]{definition*} +\declaretheorem[style=thmblackbox2,name=Osservazione,sibling=theorem]{remark} +\declaretheorem[style=thmblackbox2,name=Osservazione,numbered=no]{remark*} +\declaretheorem[name=Congettura,sibling=theorem]{conjecture} +\declaretheorem[name=Congettura,numbered=no]{conjecture*} +\declaretheorem[name=Esercizio,sibling=theorem]{exercise} +\declaretheorem[name=Esercizio,numbered=no]{exercise*} +\declaretheorem[name=Asserzione,sibling=theorem]{fact} +\declaretheorem[name=Asserzione,numbered=no]{fact*} +\declaretheorem[name=Problema,sibling=theorem]{problem} +\declaretheorem[name=Problema,numbered=no]{problem*} +\declaretheorem[name=Domanda,sibling=theorem]{ques} +\declaretheorem[name=Domanda,numbered=no]{ques*} + +\Crefname{claim}{Claim}{Claim} +\Crefname{conjecture}{Congettura}{Congetture} +\Crefname{exercise}{Esercizio}{Esercizi} +\Crefname{fact}{Asserzione}{Asserzioni} +\Crefname{problem}{Problema}{Problemi} +\Crefname{ques}{Domanda}{Domande} + +\addtokomafont{partprefix}{\rmfamily} +\renewcommand*{\partformat}{\color{purple} + \scalebox{2.5}{\thepart}\enlargethispage{2em}} + +\@ifundefined{chapter}{}{ + \addtokomafont{partprefix}{\rmfamily} + \renewcommand*{\partformat}{\color{purple} + \scalebox{2.5}{\thepart}\enlargethispage{2em}} + \addtokomafont{chapterprefix}{\raggedleft} + \RedeclareSectionCommand[beforeskip=0.5em]{chapter} + \renewcommand*{\chapterformat}{\mbox{% + \scalebox{1.5}{\chapappifchapterprefix{\nobreakspace}}% + \scalebox{2.718}{\color{purple}\thechapter}\enskip}} +} + +\renewcommand*{\sectionformat}% +{\color{purple}\S\thesection\enskip} +\renewcommand*{\subsectionformat}% +{\color{purple}\S\thesubsection\enskip} +\renewcommand*{\subsubsectionformat}% +{\color{purple}\S\thesubsubsection\enskip} +\KOMAoptions{numbers=noenddot} + +\lstset{basicstyle=\ttfamily\scriptsize, + backgroundcolor=\color{green!2!white}, + breakatwhitespace=true, + breaklines=true, + commentstyle=\color{green!70!black}, + 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