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\begin{flashcard}[SimDiagonCommute]
\begin{theorem*}[Simultaneous diagonalisation]
    \begin{itemize}
        \item $\dim_F V < \infty$
        \item $\alpha, \beta \in \mathcal{L}(V)$
            diagonalisable
    \end{itemize}
    Then $\alpha, \beta$ are simultaneously
    diagonalisable ($\exists \mathcal{B}$ basis of
    $V$ in which both $[\alpha]_{\mathcal{B}}$,
    $[\beta]_{\mathcal{B}}$ are diagonal) if and
    only if \cloze{$\alpha$ and $\beta$ commute.}
\end{theorem*}
\end{flashcard}

\begin{proof}
    \begin{enumerate}
        \item[$\Rightarrow$] Exists basis
            $\mathcal{B}$ of $V$ in which
            \[ [\alpha]_{\mathcal{B}} = D_1 \]
            \[ [\beta]_{\mathcal{B}} = D_2 \]
            $D_1, D_2$ both diagonal, then $D_1
            D_2 = D_2 D_1$ so $\alpha\beta = \beta
            \alpha$.
        \item[$\Leftarrow$] Suppose $\alpha, \beta$ are both
            diagonalisable \emph{and} $\alpha\beta
            = \beta\alpha$. Let $\lambda_1, \dots,
            \lambda_k$ be the $k$ distinct
            eigenvalues of $\alpha$. We have
            shown:
            \[ \text{$\alpha$ diagonalisable} \iff
            V = \bigoplus_{i = 1}^k V_{\lambda_i} \]
            $V_{\lambda_i}$ is the eigenspace
            associated to $\lambda_i$. \myskip
            Claim: $V_{\lambda_I}$ stable by
            $\beta$: $\beta(V_{\lambda_i}) \le
            V_{\lambda_i}$. \\
            Indeed, let $v \in V_{\lambda_i}$,
            then
            \[ \alpha \beta(v) = \eta \alpha(v) =
            \beta(\lambda_i v) = \lambda_i
            \beta(v) \]
            so $\beta(v) \in V_{\lambda_i}$.
            \begin{itemize}
                \item We use criterion for
                    diagonalisability: $\beta$ is
                    diagonalisable implies that
                    there exists $p$ with
                    distinct linear factors such
                    that $p(\beta) = 0$.
            \end{itemize}
            Now $B
            |_{V_{\lambda_j}}$
            endomorphism $(\beta :
            V_{\lambda_j} \to
            V_{\lambda_j})$ and
            \[ p(\beta |_{V_{\lambda_j}})
            = 0 \]
            $p$ has distinct linear
            factors, so $\beta |
            V_{\lambda_j}$ is
            diagonalisable. So there
            exists $\mathcal{B}$ basis of
            $V_{\lambda_j}$ in which
            $\beta |_{V_{\lambda_j}}$ is
            diagonal. Then
            \[ V = \bigoplus_{i = 1}^k
            V_{\lambda_i} \]
            so $(\mathcal{B}_1, \dots,
            \mathcal{B}_k) = \mathcal{B}$ is a
            basis of $V$ in which both $\alpha$
            and $\beta$ are in diagonal form.
    \end{enumerate}
\end{proof}

\begin{hiddenflashcard}[commuting-endomorphisms]
Cool fact about commuting endomorphisms? (i.e. if
$\alpha \beta = \beta \alpha$ then\ldots \\
\cloze{
Then $\alpha$ is stable on eigenspaces of $\beta$
and vice versa: if $v \in V_{\lambda}$, i.e.
$\beta v = \lambda v$, then
\[ \beta\alpha v = \alpha\beta v = \alpha \lambda
v = \lambda \beta v \]
so $\beta v \in V_{\lambda}$. \\
This is the main idea behind simultaneous
diagonalisation, but is useful in other contexts
too!
}
\end{hiddenflashcard}

\subsubsection*{Minimal polynomial of an
endormorphism}

\begin{itemize}
    \item Remainder: (Groups, Rings and Modules).
        \\
        Euclidean algorithm for polynomials: $a,
        b$ polynomials over $F$, $b \neq 0$, then
        there exist polynomials $q, r$ over $F$
        with:
        \[ \deg r < \deg b \]
        \[ a = qb + r \]
\end{itemize}

\begin{definition*}[Minimal polynomial]
    $V$ vector space over $F$, $\dim_F V <
    \infty$. Let $\alpha \in \mathcal{L}(V)$. The
    \emph{minimal polynomial} $m_\alpha$ of
    $\alpha$ is the (unique up to a constant) non
    zero polynomial with \emph{smallest degree}
    such that
    \[ m_\alpha(\alpha) = 0 \]
\end{definition*}

\noindent
Existence and uniqueness follow from the following
observations:

\begin{itemize}
    \item $\dim_F V = n$, $\alpha \in
        \mathcal{L}(V)$. We know:
        \[ \dim_F \mathcal{L}(V) = n^2 \]
        \[ \implies \id, \alpha, \dots,
        \alpha^{n^2} \text{ cannot be free} \]
        \[ \implies a_{n^2} \alpha^{n^2} + \cdots
        + a_1 \alpha + a_0 = 0 \]
        \[ \implies \exists p \in F[t] \mid
        p(\alpha) = 0, p \neq 0 \]
        That is, there does exist a polynomial $p$
        that kills $\alpha$.
    \item Lemma: $\alpha \in \mathcal{L}(V)$, $p
        \in F[t]$. Then $p(\alpha) = 0$ if and
        only if $m_\alpha$ is a factor of $p$. \\
        Proof: $p \in F[t]$, $p(\alpha) = 0$,
        $m_\alpha$ is minimum polynomial of
        $\alpha$. So $\deg m_\alpha \le \deg p$.
        By Euclidean division:
        \[ p = m_\alpha q + r \]
        \[ \deg r < \deg m_\alpha \]
        Then
        \[ p(\alpha) = 0 = m_\alpha q(\alpha)
        r(\alpha) \]
        so $r(\alpha) = 0$. If $r \neq 0$, then
        this would contradict the definition of
        $m_\alpha$. So $r \equiv 0$. So $p =
        m_\alpha q$, that is, $m_\alpha$ divides
        $p$.
    \item If $m_1, m_2$ are both polynomial with
        smallest degree which kill $\alpha$ then
        by the above lemma, $m_1 \mid m_2$, $m_2
        \mid m_1$ so $m_2 = cm_1$, $c \in F$. That
        is, the minimal polynomial is unique up to
        a constant.
\end{itemize}

\begin{example*}
    $V = \RR^2$
    \[ A =
    \begin{pmatrix}
        1 & 0 \\
        0 & 1
    \end{pmatrix}
    \quad B = 
    \begin{pmatrix}
        1 & 1 \\
        0 & 1
    \end{pmatrix}
    \]
    \begin{itemize}
        \item Let $p(t) = (t - 1)^2$, then $p(A) =
            p(B) = 0$. So minimal polynomial is
            either $t - 1$ or $(t - 1)^2$.
        \item Check: $m_A = t - 1$, $m_B = (t -
            1)^2$. So $A$ is diagonalisable but
            $B$ is not.
    \end{itemize}
\end{example*}

\subsection{Cayley Hamiton Theorem and
multiplicity of eiganvalues}

\begin{flashcard}[cayley-hamilton]
\begin{theorem*}[Cayley Hamilton]
    Let $V$ be an $F$ vector space, $\dim_F V <
    \infty$. Let $\alpha \in \mathcal{L}(V)$ with
    characteristic polynomial $\chi_\alpha(t) =
    \det(\alpha - t\id)$. Then
    $\chi_\alpha(\alpha) = 0$.
\end{theorem*}

\fcscrap{
\begin{corollary*}
    $m_\alpha \mid \chi_\alpha$.
\end{corollary*}
}

\begin{proof}
    $F = \CC$ (general proof is in the notes).
    \cloze{
    $\alpha \in \mathcal{L}(V)$, $n = \dim_\CC V$.
    Exists basis $\mathcal{B} = \{v_1, \dots,
    v_n\}$ such that
    \[ [\alpha]_{\mathcal{B}} =
    \begin{pmatrix}
        a_1 & \cdots & * \\
        \vdots & \ddots & \vdots \\
        0 & \cdots & a_n
    \end{pmatrix}
    \]
    (triangulable). Let $U_j = \langle v_1, \dots,
    v_j\rangle$. Then because of the triangular
    form, $(\alpha - a_j \id) U_j \le U_{j - 1}$.
    \[ \chi_\alpha(t) = \prod_{i = 1}^n (a_i - t) \]
    \[ (\alpha - a_1 \id) \cdots (\alpha - a_{n -
    1} \id)(\alpha - a_n \id) V \]
    \[ \le (\alpha - a_1\id) \cdots (\alpha - a_{n
    - 1} \id) U_{n - 1} \]
    \[ \vdots \]
    \[ \le 0 \]
    So $\chi_\alpha(\alpha) = 0$. For the general
    case, see the notes.
}
\end{proof}
\end{flashcard}

\begin{definition*}[algebraic / geometric
    multiplicity]
    $\dim_F V < \infty$, $\alpha \in
    \mathcal{L}(V)$. Let $\lambda$ eigenvalue of
    $\alpha$. Then
    \[ \chi_\alpha(t) = (t - \lambda)^{a_\lambda}
    q(t) \]
    \[ q \in F[t], \quad q(\lambda) \neq 0 \]
    \begin{itemize}
        \item $a_\lambda$ is the algebraic
            multiplicity of $\lambda$.
        \item $g_\lambda$ is the geometric
            multiplicity of $\lambda$, and
            $g_\lambda = \dim \Ker(\alpha -
            \lambda \id)$.
    \end{itemize}
\end{definition*}

\begin{remark*}
    $\lambda$ eigenvalue $\iff$ $\alpha - \lambda
    \iff$ singular $\iff$ $\det(\alpha - \lambda
    \id) = \chi_\alpha(\lambda) = 0$
\end{remark*}

\begin{lemma*}
    $\lambda$ eigenvalue of $\alpha \in
    \mathcal{L}(V)$, then $1 \le g_\lambda \le
    a_\alpha$.
\end{lemma*}

\begin{proof}
    \begin{itemize}
        \item $g_\lambda = \dim \Ker(\alpha -
            \lambda \id) \ge 1$ since $\lambda$ is
            an eigenvalue.
        \item Let us show that $g_\lambda \le
            a_\lambda$. Indeed, let $(v_1, \dots,
            v_{g_\lambda})$ basis of $V_\lambda =
            \Ker (\alpha - \lambda\id)$, and
            compute $\mathcal{B} = (v_1, \dots,
            v_{\lambda_g}, v_{g_\lambda + 1},
            \dots, v_n)$ of $V$. Then
            \[ [\alpha]_{\mathcal{B}} =
            \begin{pmatrix}
                \lambda \id_{g_{\lambda}} & * \\
                0 & A_1
            \end{pmatrix}
            \]
            \[ \implies \det[\alpha - t\id] = \det
            \begin{pmatrix}
                (\lambda - t)\id_{g_\lambda} & * \\
                0 & A_1 - t\id
            \end{pmatrix}
            = (\lambda - t)^{g_\lambda}
            \chi_{A_1}(t) \]
            \[ \implies g_\lambda \le a_\lambda \]
    \end{itemize}
\end{proof}

\begin{lemma*}
    $\lambda$ eigenvalue of $\alpha \in \mathcal{L}(V)$.
    Let:
    \[ c_\lambda \equiv \text{multiplicity of
    $\lambda$ as a root of $m_\alpha$ (minimal
    polynomial)} \]
    Then $1 \le c_\lambda \le a_\lambda.$
\end{lemma*}

\begin{proof}
    \begin{itemize}
        \item Cayley-Hamilton implies $m_\alpha
            \mid \chi_\alpha$. So $c_\lambda \le
            a_\lambda$.
        \item $c_\lambda \ge 1$. Indeed, there
            exists $b \neq 0$ such that $\alpha(v)
            = \lambda v$ so then for all $p \in
            F[t]$, $p(\alpha)(v) = (p(\lambda))v$
            ($\alpha^n(v) = \lambda^n v$) so
            $m(\alpha)(v) = (m(\lambda))v$ so
            $m(\lambda) = 0$ so $c_\lambda \ge 1$.
    \end{itemize}
\end{proof}

\begin{example*}
    \[ A =
    \begin{pmatrix}
        1 & 0 & -2 \\
        0 & 1 & 1 \\
        0 & 0 & 2
    \end{pmatrix}
    \]
    $m_A$?
    \begin{itemize}
        \item $\chi_A(t) = (t - 1)^2(t - 2)$
        \item So $m_\alpha$ is either $(t - 1)^2(t
            - 2)$ or $(t - 1)(t - 2)$. Check $(A -
            I)(A - 2I) = 0$, so $m_\alpha = (t -
            1)(t - 2)$ so $A$ is diagonalisable.
    \end{itemize}
\end{example*}
