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% 14/11/2022 11AM

\begin{example*}[Jordan block]
    \[ J_\lambda =
    \begin{pmatrix}
        \lambda & 1 & \cdots & 0 \\
        0 & \lambda & \cdots & 0 \\
        \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots & 1 \\
        0 & 0 & \cdots & \lambda
    \end{pmatrix}
    \in \mathcal{M}_n(F) \]
    Check $g_\lambda = 1$, $a_\lambda = n$,
    $c_\lambda = n$
\end{example*}

\begin{lemma*}[characterisation of diagonalisable
    endomorphisms over $F = \CC$]
    $F = \CC$, $\dim_\CC V = n < \infty$, $\alpha
    \in \mathcal{L}(V)$. The following are
    equivalent:
    \begin{enumerate}[(i)]
        \item $\alpha$ diagonalisable
        \item $\forall \lambda$ eigenvalue of
            $\alpha$, $a_\lambda = g_\lambda$
        \item $\forall \lambda$ eigenvalue of
            $\alpha$, $c_\lambda = 1$.
    \end{enumerate}
\end{lemma*}

\begin{proof}
    (i) $\iff$ (iii) done. We need (i) $\iff$
    (ii). Indeed, let $(\lambda_1, \dots,
    \lambda_k)$ be the \emph{distinct} eigenvalues
    of $\alpha$. We showed:
    \[ \alpha \text{ diagonalisable} \iff V =
    \bigoplus_{i = 1}^k V_{\lambda_i} \]
    \begin{align*}
        \dim V
        &= n
        &= \deg \chi_\alpha \\
        &= \sum_{i = 1}^k a_{\lambda_i}
    \end{align*}
    \[ (\chi_\alpha(t) = (-1)^n \prod_{i = 1}^k (t
    - \lambda_i)^{a_i}) \]
    so
    \[ \alpha \text{ diagonalisable} \iff \sum_{i
    = 1}^k a_{\lambda_i} = \sum_{i = 1}^k
    g_{\lambda_i} \tag{$*$} \]
    We know: $\forall 1 \le i \le k$,
    $g_{\lambda_i} \le a_{\lambda_i}$. Hence ($*$)
    holds $\iff \forall 1 \le i \le k$,
    $a_{\lambda_i} = g_{\lambda_i}$.
\end{proof}

\subsection{Jordan normal form}

\begin{note*}
    In this subsection, $F = \CC$.
\end{note*}

\begin{definition*}[Jordan normal form]
    Let $A \in \mathcal{M}_n(\CC)$, we say that
    $A$ is in Jordan Normal Form (JNF) if it is a
    block diagonal matrix:
    \[ A =
    \begin{pmatrix}
        J_{n_1}(\lambda_1) & 0 & \cdots & 0 \\
        0 & J_{n_2}(\lambda_2) & \cdots & 0 \\
        \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots & J_{n_k}(\lambda_k)
    \end{pmatrix}
    \]
    where:
    \begin{itemize}
        \item $k \ge 1$, $k$ integer
        \item $n_1, \dots, n_k$ integers
        \item $\sum_{i = 1}^k n_i = n$
        \item $\lambda_i \in \CC$, $1 \le i \le
            k$: they need \emph{not} be distinct
        \item $m \in \NN$, $m \neq 0$, $\lambda
            \in \CC$, $J_m(\lambda) = \lambda)$ if
            $m = 1$,
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/3549a7be640e11ed.png}
            \end{center}
    \end{itemize}
\end{definition*}

\begin{example*}
    $n = 3$,
    \[ A =
    \begin{pmatrix}
        \lambda & 0 & 0 \\
        0 & \lambda & 0 \\
        0 & 0 & \lambda
    \end{pmatrix}
    =
    \begin{pmatrix}
        J_1(\lambda & 0 & 0 \\
        0 & J_1(\lambda) & 0 \\
        0 & 0 & J_1(\lambda)
    \end{pmatrix}
    \]
    so this is in Jordan Normal Form.
\end{example*}

\begin{theorem*}
    Every matrix $A \in \mathcal{M}_n(\CC)$ is
    similar to a matrix in Jordan Normal Form,
    which is unique up to reordering the Jordan
    block.
\end{theorem*}

\begin{proof}
    Non examinable (in Groups, Rings and Modules
    class). (Proof is in lecturer's notes).
\end{proof}

\begin{example*}
    $n = 2$, possible JNF in this case?
    \[
    \begin{pmatrix}
        \lambda_1 & 0 \\
        0 & \lambda_2
    \end{pmatrix}
    \quad m = (t - \lambda_1)(t - \lambda_2),
    \quad \lambda_1 \neq \lambda_2 \]
    \[
    \begin{pmatrix}
        \lambda & 0 \\
        0 & \lambda
    \end{pmatrix}
    \quad m = (t - \lambda) \]
    \[
    \begin{pmatrix}
        \lambda & 1 \\
        0 & \lambda
    \end{pmatrix}
    \quad m = (t - \lambda)^2 \]
\end{example*}

\begin{example*}
    $n = 3$,
    \[
    \begin{pmatrix}
        \lambda_1 & 0 & 0 \\
        0 & \lambda_2 & 0 \\
        0 & 0 & \lambda_3
    \end{pmatrix}
    \quad (t - \lambda_1)(t - \lambda_2)(t -
    \lambda_3) \quad \lambda_1, \lambda_2,
    \lambda_3 \text{ distinct} \]
    \[
    \begin{pmatrix}
        \lambda_1 & 0 & 0 \\
        0 & \lambda_2 & 0 \\
        0 & 0 & \lambda_2
    \end{pmatrix}
    \quad (t - \lambda_1)(t - \lambda_2) \]
    \[
    \begin{pmatrix}
        \lambda_1 & 0 & 0 \\
        0 & \lambda_2 & 1 \\
        0 & 0 & \lambda_2
    \end{pmatrix}
    \quad (t - \lambda_1)(t - \lambda_2)^2 \]
    \[
    \begin{pmatrix}
        \lambda & 0 & 0 \\
        0 & \lambda & 0 \\
        0 & 0 & \lambda
    \end{pmatrix}
    \quad (t - \lambda) \]
    \[
    \begin{pmatrix}
        \lambda & 0 & 0 \\
        0 & \lambda & 1 \\
        0 & 0 & \lambda
    \end{pmatrix}
    \quad (t - \lambda)^2 \]
    \[
    \begin{pmatrix}
        \lambda & 1 & 0 \\
        0 & \lambda & 1 \\
        0 & 0 & \lambda
    \end{pmatrix}
    \quad (t - \lambda)^3 \]
\end{example*}

\noindent
Useful observation: which explains why JNF is
unique. $\to$ we can directly compute in the JNF
the quantities $a_\lambda, g_\lambda, c_\lambda$.
Indeed, let $M \ge 2$ and let $J_m(\lambda)$. Then
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/9657be14640f11ed.png}
\end{center}
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/afcd15a6640f11ed.png}
\end{center}
By induction we can show:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c0cd77e2640f11ed.png}
\end{center}
for $k \le m$, and is 0 for $k = m$. We say that
the matrix $(J_m - \lambda \id)$ is
\emph{nilpotent} of order $m$. ($u^m = 0$ and
$u^{m - 1} \neq 0$). So
\[ a_\lambda \equiv \text{sum of sizes of blocks
with eigenvalue $\lambda$} \equiv \text{number of
$\lambda$ on the diagonal} \]
\[ g_\lambda = \dim \Ker(A - \lambda \id) =
\text{number of blocks with eigenvalue $\lambda$} \]
\[ c_\lambda J_m(\lambda) \to (t - \lambda)^m \text{kills
it} \]
(because $(J_m - \lambda \id)$ is nilpotent of order
exactly $m$) so
\[ c_\lambda \equiv \text{size of the largest
block with eigenvalue $\lambda$} \]

\begin{example*}
    \[ A =
    \begin{pmatrix}
        0 & -1 \\
        1 & 2
    \end{pmatrix}
    \]
    Find a basis in which $A$ is Jordan Normal
    Form?
    \begin{enumerate}[(i)]
        \item $\chi_A(t) = (t - 1)^2$ eigenvalue
            $\lambda = 1$. $A - \id \neq 0$
            implies $m_A(t) = (t - 1)^2$, and
            Jordan Normal Form
            \[
            \begin{pmatrix}
                1 & 1 \\
                0 & 1
            \end{pmatrix}
            \]
        \item Eigenvectors:
            \[ A - \id =
            \begin{pmatrix}
                -1 & -1 \\
                1 & 1
            \end{pmatrix}
            \]
            $\Ker(A - \id) = \langle v_1 \rangle$,
            $v_1 = (1, -1)^\top$. I look for a
            (non-unique!) $v_2$ such that
            \[ (A - \id) v_2 = v_1 \]
            $v_2 = (-1, 0)^\top$ works.
            \[ [A]_{\mathcal{B}} = J_1(1) \]
            for $\mathcal{B} = (v_1, v_2)$.
            \[ P^{-1} =
            \begin{pmatrix}
                1 & -1 \\
                -1 & 0
            \end{pmatrix}
            \]
            \[ A =
            \ub{\begin{pmatrix}
                1 & -1 \\
                -1 & 0
            \end{pmatrix}}_{P^{-1}}
            \ub{\begin{pmatrix}
                1 & 1 \\
                0 & 1
            \end{pmatrix}}_{J}
            \ub{\begin{pmatrix}
                1 & -1 \\
                -1 & 0
            \end{pmatrix}^{-1}}_{P}
            \]
    \end{enumerate}
\end{example*}

\noindent
Exercise:
\[ A =
\begin{pmatrix}
    3 & -2 & 0 \\
    1 & 0 & 0 \\
    1 & 0 & 1
\end{pmatrix}
\]
Find a basis in which $A$ is JNF.

\begin{theorem*}[Generalised eigenspace
    decomposition]
    \begin{itemize}
        \item $\dim_\CC V = n < \infty$
        \item $\alpha \in \mathcal{L}(V)$.
        \item $m_\alpha(t) = (t - \lambda_1)^{c_1}
            \cdots (t - \lambda_k)^{c_k}$
        \item $\lambda_1, \dots, \lambda_k$
            distinct eigenvalues of $\alpha$.
    \end{itemize}
    Then
    \[ V = \bigoplus_{j = 1}^k V_j \]
    \[ V_j = \Ker [(\alpha - \lambda \id)^{c_j}] \]
    ($V_j$ is generalised eigenspace)
\end{theorem*}

\begin{remark*}
    $\alpha$ diagonalisable, $c_j = 1$. THen $V_j$
    eigenspace associated to $\lambda_j$.
\end{remark*}

\begin{proof}
    projectors onto $V_j$ are \emph{explicit}.
    Indeed, let
    \[p_j(t) = \prod_{i \neq j} (t -
    \lambda_i)^{c_i}  \]
    Then the $p_j$ have no common factor, so by
    Euclid's algorithm, we can find $q_1, \dots,
    q_k$ polynomials such that
    \[ \sum_{i = 1}^k p_i q_i = 1 \tag{$*$} \]
    We define
    \[ \pi_j = q_j p_j(\alpha) \]
    \begin{enumerate}[(i)]
        \item By ($*$),
            \[ \id = \sum_{j = 1}^k q_j p_j
            (\alpha) = \sum_{j = 1}^k \pi_j \]
            \[ \implies \forall v \in V, v =
            \sum_{j = 1}^k \pi_j(v) \]
        \item $m_\alpha(\alpha) = 0$, $m_\alpha =
            \prod_{j = 1}^k (t -
            \lambda_j)^{c_j}$
            \[ \implies (\alpha - \lambda_j
            \id)^{c_j} \pi_j = (\alpha - \lambda_j
            \id)^{c_j} q_j p_j(\alpha) = 0\]
            \[ \implies \forall v \in V, \pi(v)
            \in V_j \]
            \[ V_j = \Ker(\alpha - \lambda_j
            \id)^{c_j} \]
            Hence $\forall v \in V$
            \[ v = \sum_{j = 1}^k \pi_j(v) \]
            \[ \implies V = \sum_{j = 1}^k
            V_{\lambda_j} \]
        \item Show that:
            \[ \sum_{j = 1}^k V_{\lambda_j} =
            \bigoplus_{j = 1}^k V_{\lambda_j} \]
            Indeed, $\pi_i \pi_j = 0$ if $i \neq
            j$ and so $\pi_i = \pi_i \left( \sum_{j =
            1}^k \pi_j \right) = \pi_i^2$.
            \[ \implies \pi_i |_{V_{\lambda_i}} =
            \id \]
            $\implies$ direct sum projection
            follows:
            \[ v = V_{\lambda_i} \cap \left(
            \sum_{i \neq j} V_{\lambda_i} \right) \]
            \[ v = \sum_{i \neq j} v_j, \quad v_j
            \in V_{\lambda_j} \]
            If apply $\pi_i$ and use:
            \[ \pi_i |_{V_{\lambda_i}} = \id \]
            \[ \pi_i |_{V_{\lambda_j}} = 0 \text{
            for $j \neq i$} \]
            so $v = 0$.
    \end{enumerate}
\end{proof}

\myskip
\[ V = \bigoplus_{i = 1}^k V_{\lambda_i} \]
\[ V_{\lambda_i} = \Ker (\alpha - \lambda_i
\id)^{c_{\lambda_i}} \]
By definition
\[ (\alpha - \lambda_i \id) |_{V_{\lambda_i}} \]
is nilpotent, since
\[ (\alpha - \lambda_i \id)^{c_{\lambda_i}}
|_{V_{\lambda_i}} = 0 \]
$\implies$ all I need to do is to find JNF
nilpotent endomorphism.
\begin{itemize}
    \item $\alpha \in \mathcal{L}(V)$, $\dim_\CC V
        = n$, $a^k = 0$, $a^{k - 1} \neq 0$.
        $\stackrel{?}{\implies}$ JNF with blocks
        $J_m(0)$. $\to$ By induction on the
        dimension.
        \[ \alpha^k = 0, \alpha^{k - 1} \neq 0 \]
        \[ \implies \exists x \in V, (x,
        \alpha(x), \dots, \alpha^{k - 1}(x)) \]
        free.
\end{itemize}

\noindent
Question: $F = \Span \langle x, \alpha(x), \dots,
\alpha^{k - 1}(x) \rangle$. Can I find $G$ such
that:
\[ V = F \oplus G \]
$G$ stable by $\alpha$? \\
$\to$ done.