% vim: tw=50
% 09/11/2022 11AM

\begin{lemma*}
    $V$ $n$ dimensional over $F = \RR, \CC$,
    $\alpha \in \mathcal{L}(V)$. Then
    $\chi_\alpha(t) = (-1)^n t^n + c_{n - 1} t^{n
    - 1} + \cdots + c_0$, $c_0 = \det A = \det
    \alpha$, $c_{n - 1} = (-1)^{n - 1} \Trace A$.
\end{lemma*}

\begin{proof}
    \begin{itemize}
        \item $\chi_\alpha(t) = \det(\alpha - t \id)$
            \[ \implies \chi_\alpha(0) = \det \alpha
            = c_0 \]
        \item Say that $F = \RR$ or $\CC$ (if $F =
            \RR$, we can think of is as having
            complex entries as well). We know that
            $\alpha$ is triangulable over $\CC$,
            so:
            \begin{align*}
                \chi_\alpha(t)
                &= \det
                \begin{pmatrix}
                    a_1 - t & \cdots & * \\
                    \vdots & \ddots & \vdots \\
                    0 & \cdots & a_n - t
                \end{pmatrix} \\
                &= \prod_{i = 1}^n )a_i - 1) \\
                &= (-1)^n t^n + c_{n - 1} t^{n -
                1} + \cdots + c_0
            \end{align*}
            \[ c_{n - 1} = (-1)^{n - 1} \sum{i =
            1}^n a_i = \Trace \alpha \]
    \end{itemize}
\end{proof}

\subsection{Diagonalisation criterion and minimal
polynomial}

\begin{notation*}[polynomial of an endomorphism]
    Pick $p(t)$ polynomial over $F$
    \[ p(t) = a_n t^n + \cdots + a_1 t + a_0,
    \quad a_i \in F \]
    $A \in \mathcal{M}_n(F)$, for all $n$, $A^n
    \in \mathcal{M}_n(F)$. We define:
    \[ p(A) = a_n A^n + \cdots + a_1 A + a_i \id
    \in \mathcal{M}_n(F) \]
    If $\alpha \in \mathcal{L}(V)$, we define
    \[ p(\alpha) = a_n \alpha^n + \cdots + a_1
    \alpha + a_0 \id \]
    where $\alpha = \alpha \circ \cdots \circ
    \alpha \in \mathcal{L}(V)$.
\end{notation*}

\noindent
$\to$ very useful.

\begin{flashcard}[sharp-criterion-of-diagonalisability]
\begin{theorem*}[Sharp criterion of
    diagonalisability]
    \begin{itemize}
        \item $V$ vector space over $F$, $\dim_F V
            < \infty$
        \item $\alpha \in \mathcal{L}(V)$
    \end{itemize}
    \cloze{Then $\alpha$ is diagonalisable if and only if
    there exists a polynomial $p$ which is the
    product of \emph{distinct linear factors}
    such that $p(\alpha) = 0$.}
\end{theorem*}
\end{flashcard}

\begin{align*}
    \alpha \text{ diagonalisable}
    &\iff \exists (\lambda_1, \dots,
    \lambda_n) \text{ distinct }, \lambda_i
    \in F \text{ such that:} \\
    &~~~~~ p(t) = \prod_{i = 1}^k (t - \lambda_i)
    \\
    &~~~~~ \text{ and } p(\alpha) = 0
\end{align*}

\begin{proof}
    \begin{enumerate}
        \item[$\Rightarrow$] Suppose $\alpha$ is
            diagonalisable, with $\lambda_1,
            \dots, \lambda_k$ the distinct
            eigenvalues. Let $p(t) = \prod_{i =
            1}^k (t - \lambda_i)$. Let
            $\mathcal{B}$ be the basis of $V$ made
            of eigenvectors of $\alpha$ (it is
            precisely the basis in which
            $[\alpha]_{\mathcal{B}}$ is diagonal).
            Then $v \in \mathcal{B}$, then
            $\alpha(v) = \lambda_i(v)$ for some $i
            \in \{1, \dots, k\}$, implies $(a -
            \lambda_i \id)(v) = 0$, implies
            \[ p(\alpha) = \left[ \prod_{j = 1}^k
            (\alpha - \lambda_j \id) \right] (v) =
            0 \]
            but the terms in the product commute,
            i.e.
            \[ (\alpha - \lambda_j \id)(\alpha - \lambda_k
            \id) = (\alpha - \lambda_k \id)(\alpha
            - \lambda_j \id) \]
            so for all $v \in \mathcal{B}$,
            $p(\alpha)(v) = 0$, so $p(\alpha)(v) =
            0$ for all $v \in V$ (since
            $\mathcal{B}$ is a basis for $V$). So
            $p(\alpha) = 0$.
        \item[$\Leftarrow$] (Kernel lemma,
            Bezout's theorem for prime
            polynomials)
            \begin{itemize}
                \item Suppose $p(\alpha) = 0$ for:
                    \[ p(t) = \prod_{i = 1}^k (t -
                    \lambda_i) \]
                    $\lambda_i \neq \lambda_j$, $i
                    \neq j$.
                \item Let $V_{\lambda_i} = \Ker
                    (\alpha - \lambda_i \id)$, we
                    claim:
                    \[ V = \bigoplus_{i = 1}^k
                    V_{\lambda_i} \]
                    Indeed let:
                    \[ q_j(t) = \prod_{\substack{i
                    = 1\\i \neq j}}^k \left(
                    \frac{t - \lambda_i}{\lambda_j -
                    \lambda_i} \right), \quad 1
                    \le j \le k \]
                    Then
                    \[ q_j(\lambda_i) = \begin{cases}
                        1 & \text{if $i = j$} \\
                        0 & \text{if $i \neq j$}
                    \end{cases} \]
                    Hence let us consider:
                    \[ q(t) = \sum_{j = 1}^k
                    q_j(t) \]
                    Then $\deg q_j \le k - 1$, so
                    $\deg q \le k - 1$. Also
                    $q(\lambda_j) = 1$ for all $1
                    \le j \le k$. So the
                    polynomial $q(t) - 1$ has
                    degree $\le k - 1$ and at
                    least $k$ roots, so for all
                    $t$, $q(t) = 1$. So for all
                    $t$,
                    \[ q_1(t) + \cdots + q_k(t) =
                    1 \]
                \item Let us define the
                    \emph{projector}
                    \[ \i_j q_j(\alpha) \in
                    \mathcal{L}(V) \]
                    Then
                    \begin{align*}
                        \sum_{j = 1}^k \pi_j
                        &= \sum_{j = 1}^k
                        q_j(\alpha) \\
                        &= \left( \sum_{j = 1}^k
                        q_j \right) (\alpha) \\
                        &= \id
                    \end{align*}
                    This means for all $v \in V$,
                    \[ v = q(\alpha)(v) = \sum_{j
                    = 1}^k \pi_j(v) = \sum_{j =
                    1}^k q_j(\alpha)(v) \]
                    Observe: pick $j \in \{1
                    \dots, k\}$,
                    \[ (\alpha - \lambda_j \id)
                    q_j(\alpha)(v) =
                    \frac{1}{\prod_{i \neq j}
                    (\lambda_i -
                    \lambda_j)}p(\alpha)(v) = 0 \]
                    so
                    \[ \forall j \in \{1, \dots,
                    k\}, \quad (\alpha - \lambda_j
                    \id) \pi_j(v) = 0 \]
                    \[ \implies \forall j \in \{1,
                    \dots, k\} \pi_j(v) \in
                    V_{\lambda_j} \]
                    ($\pi_j$ is a projector on
                    $V_{\lambda_j}$) Now for all
                    $v \in V$,
                    \[ v = \sum_{j = 1}^k \pi_j(v) \]
                    hence
                    \[ V = \sum_{j = 1}^k
                    V_{\lambda_j} \]
                    We need to prove that the sum
                    is \emph{direct}. Indeed, let
                    $v \in V_{\lambda_j} \cap
                    \left( \sum_{i \neq j}
                    V_{\lambda_i} \right)$.
                    \begin{itemize}
                        \item $v \in
                            V_{\lambda_j}$. Then
                            \begin{align*}
                                \pi_j(v)
                                &= q_j(\alpha)(v) \\
                                &= \prod_{i = 1,
                                i\neq j}^k
                                \frac{(\alpha -
                                \lambda_i
                                \id)}{\lambda_i -
                                \lambda_j} (v) \\
                                &= \left[ \prod_{i
                                = 1,i\neq j}^k
                                \frac{(\lambda_i -
                                \lambda_j)}{\lambda_i
                                - \lambda_j}
                                \right] (v) \\
                                &= v
                            \end{align*}
                            so $\pi_j
                            \mid_{V_{\lambda_j}} =
                            \id$.
                        \item By assumption $v \in
                            \sum_{i \neq j}
                            V_{\lambda_i}$. Now,
                            $i_0 \neq j$, $v \in
                            V_{\lambda_{i_0}}$,
                            $\alpha(v) =
                            \lambda_{i_0} v$,
                            \begin{align*}
                                \implies \pi_j(v)
                                &= q_j(\alpha)(v)
                                \\
                                &= \left[ \prod_{i
                                \neq j}
                                \frac{(\alpha -
                                \lambda_i
                                \id)}{\lambda_i -
                                \lambda_j}\right]
                                (v) \\
                                &= \left(\prod_{i \neq
                                j}
                                \frac{\lambda_{i_0} -
                                \lambda_j}{\lambda_i -
                                \lambda_j} \right)
                                \\
                                &= 0
                            \end{align*}
                            so $\pi_j
                            \mid_{V_{\lambda_i}} =
                            0$ for $i \neq j$.
                    \end{itemize}
            \end{itemize}
            As a conclusion:
            $v \in V_{\lambda_j} \cap \left(
            \sum_{i \neq j} V_{\lambda_i} \right)$
            \begin{enumerate}[(1)]
                \item $v \in V_{\lambda_i}$,
                    implies $\pi_j(v) = v$
                \item $v \in \sum_{i \neq j}
                    V_{\lambda_i}$ implies
                    $\pi_j(v) = 0$
            \end{enumerate}
            so $v = 0$. We have proved:
            \[ V = \bigoplus_{j = 1}^k
            V_{\lambda_j} \]
            \[ \pi_j \mid_{V_{\lambda_j}} = \id \]
            \[ \pi_i \mid_{V_{\lambda_j}} = 0 \]
            for $i \neq j$.
            \begin{remark*}
                We have proved the following: if $\lambda_1,
                \dots, \lambda_k$ are $k$ distinct eigenvalues
                of $\alpha$, then
                \[ \sum_{i = 1}^k V_{\lambda_i} = \bigoplus_{i
                = 1}^k V_{\lambda_i} \]
                (always true) (and we know the projectors)
            \end{remark*}
            This means that the only way
            diagonalisation fails is if:
            \[ \bigoplus_{i = 1}^k V_{\lambda_i} =
            \sum_{j = 1}^k V_{\lambda_j} \neq V \]
    \end{enumerate}
\end{proof}

\begin{hiddenflashcard}[proof-of-sharp-criterion-of-diagonalisability]
Proof of the sharp criterion of diagonalisability?
\\
\cloze{
If $\alpha$ is diagonalisable, then consider
$[\alpha]_{\mathcal{B}}$ where $\mathcal{B}$ is a
basis in which $\alpha$ is diagonal. Then easily
find non-zero polynomial $p$ which is a product of
distinct linear factors for which $p(\alpha) = 0$.

\myskip
For the other direction, $p(\alpha) = 0$ with
$p(t) = \prod_{i = 1}^k (t - \lambda_i)$ with
$\lambda_i$ distinct. Let $V_{\lambda_i} =
\ker(\alpha - \lambda_i \id)$. Define
\[ q_j(t) = \prod_{\substack{1 \le i \le k\\i
\neq j}} \left( \frac{t - \lambda_i}{\lambda_j-
\lambda_i} \right), \qquad \pi_j = q_j(\alpha) \]
Note $q_j$ has degree $k - 1$ and $q_j(\lambda_i) = \delta_{ij}$,
so $(\sum q_j) - 1$ has $k$ roots but is degree
$\le k - 1$ hence identically $0$. So $\sum \pi_j
= \id$. Now note $(\alpha - \lambda_j \id) \pi_j
(v) = 0$ since $p(\alpha)$ annihilates $v$. So $\pi_j(v) \in
V_{\lambda_j}$.
\[ v = \ub{\pi_1(v)}_{\in V_{\lambda_1}} + \cdots
+ \ub{\pi_k(v)}_{\in V_{\lambda_k}} \]
so $V = \sum_{i = 1}^k V_{\lambda_k}$ hence there
is a generating set of eigenvectors, which can be
reduced to a basis. Then in this basis, $\alpha$
is diagonal.
}
\end{hiddenflashcard}

\begin{example*}
    $A \in \mathcal{M}_n(F)$, $F = \CC$. $A$ has
    finite order. (there exists $m \in \NN$ such
    that $A^m = \id$). Then $A$ is diagonalisable
    (over $\CC$).
    TODO..
\end{example*}

