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\subsection{Eigenvectors, eigenvalues and trigonal
matrices}

First step towards the diagonalisation of
endomorphisms.

\begin{itemize}
    \item $V$ vector space over $F$, $\dim_F V =
        n < \infty$. $\alpha : V \to V$ linear
        (endomorphism of $V$). General problem:
        Can we find a basis $\mathcal{B}$ of $V$
        such that in this basis,
        \[ [\alpha]_{\mathcal{B}} =
        [\alpha]_{\mathcal{B}, \mathcal{B}} \]
        has a ``nice'' form.
\end{itemize}
Reminder: $\mathcal{B}'$ another basis of $V$, $P$
= change of basis matrix,
\[ [\alpha]_{\mathcal{B}'} = P^{-1}
[\alpha]_{\mathcal{B}} P \]
Equivalently: given a matrix $A \in
\mathcal{M}_n(F)$, is it conjugated to a matrix
with a ``simple'' form?

\begin{definition*}
    \begin{enumerate}[(i)]
        \item $\alpha \in \mathcal{L}(V)$ ($\alpha
            : V \to V$ linear) is
            \emph{diagonalisable} if there exists
            a basis $\mathcal{B}$ of $V$ such that
            $[\alpha]_{\mathcal{B}}$ in
            $\mathcal{B}$ is diagonal:
            \[ [\alpha]_{\mathcal{B}} =
            \begin{pmatrix}
                \lambda_1 & 0 & \cdots & 0 \\
                0 & \lambda_2 & \cdots & 0 \\
                \vdots & \vdots & \ddots & \vdots \\
                0 & 0 & \cdots & \lambda_n
            \end{pmatrix}
            \]
        \item $\alpha \in \mathcal{L}(V)$ is
            \emph{triangulable} if there exists
            $\mathcal{B}$ basis of $V$ such that
            $[\alpha]_{\mathcal{B}}$ is
            triangular:
            \[ [\alpha]_{\mathcal{B}} =
            \begin{pmatrix}
                \lambda_1 & * & \cdots & * \\
                0 & \lambda_2 & \cdots & * \\
                \vdots & \vdots & \ddots & \vdots \\
                0 & 0 & \cdots & \lambda_n
            \end{pmatrix}
            \]
    \end{enumerate}
\end{definition*}

\begin{remark*}
    A matrix is diagonalisable (respectively
    triangulable) if and only if it is conjugated
    to a diagonal (respectively triangular)
    matrix.
\end{remark*}

\begin{definition*}[eigenvalue, eigenvector,
    eigenspace]
    \begin{enumerate}[(i)]
        \item $\lambda \in F$ is an eigenvalue of
            $\alpha \in \mathcal{L}(V)$ if and
            only if there exists $v \in V
            \setminus \{0\}$ such that $\alpha(v)
            = \lambda v$.
        \item $v \in V$ is an eigenvector of
            $\alpha \in \mathcal{L}(v)$ if and
            only if $v \neq 0$ and there exists
            $\lambda \in F$ such that $\alpha(v) =
            \lambda v$.
        \item $V_\lambda = \{v \in V \mid
            \alpha(v) = \lambda v\} \le V$ is the
            eigenspace associated to $\lambda \in
            F$.
    \end{enumerate}
\end{definition*}

\begin{remark*}
    Once can write evalue, evectors, espace.
\end{remark*}

\begin{lemma*}
    $\alpha \in L(v)$, $\lambda \in F$, then
    \[ \text{$\lambda$ eigenvalue} \iff
    \det(\alpha - \lambda \id) = 0 \]
\end{lemma*}

\begin{proof}
    \begin{align*}
        \text{$\lambda$ eigenvalue}
        &\iff \exists v \in V \setminus \{0\} \mid
        \alpha(v) = \lambda v \\
        &\iff \exists v \in V \setminus \{0\} \mid
        (\alpha - \lambda \id)(v) = 0 \\
        &= \Ker (\alpha - \lambda \id) \neq \{0\}
        \\
        &= \alpha - \lambda \id \text{ not
        injective} \\
        &\iff \alpha - \lambda \id \text{ not
        surjective} \\
        &\iff \alpha - \lambda \id \text{ not
        bijective} \\
        &\iff \det(\alpha - \lambda \id) = 0
    \end{align*}
\end{proof}

\begin{remark*}
    If $\alpha(v_j) = \lambda v_j$, $v_j \neq 0$.
    I can complete it to a basis $(v_1, \dots,
    v_{j - 1}, v_j, \dots, v_n)$ of $V$. Then
    \[ [\alpha]_{\mathcal{B}} = (\mid \cdots \mid
    \text{($\lambda$ in $j$-th entry)} \mid
    \cdots) \]
\end{remark*}

\subsubsection*{Elementary facts about polynomials}

We will study $P(\alpha)$, $P$ polynomial. $\alpha
\in \mathcal{L}(V)$.
\begin{itemize}
    \item $F$ field,
        \[f(t) = a_n t^n + a_{n - 1} t^{n - 1} +
        \cdots + a_1 t + a_0  \]
        $a_i \in F$. $n \equiv$ the largest
        exponent such that $a_n \neq 0$, $n = \deg
        f$.
    \item $\deg (f + g) \le \max\{\deg f, \deg
        g\}$, $\deg(fg) = \deg f + \deg g$
    \item $F[t] = \{\text{polynomials with
        coefficients in $F$}\}$
    \item $\lambda$ root of $f(t) \iff f(\lambda) =
        0$.
\end{itemize}

\begin{lemma*}
    $\lambda$ is a root of $f$, then $t - \lambda$
    divides $f$:
    \[ f(t) = (t - \lambda) g(t), \quad g(t) \in
    F(t) \]
\end{lemma*}

\begin{proof}
    $f(t) = a_n t^n + \cdots + a_1 t + a_0$,
    $f(\lambda) = a_n \lambda^n + \cdots + a_1
    \lambda + a_0 = 0$.
    \begin{align*}
        f(t)
        &= f(t) - f(\lambda) \\
        &= a_n (t^n - \lambda^n) + \cdots + a_1 (t
        - \lambda) \\
        &= a_n(t - \lambda)(t^{n - 1} + \lambda
        t^{n - 2} + \cdots \lambda^{n - 2} t +
        \lambda^{n - 1}) + \cdots
    \end{align*}
\end{proof}

\begin{corollary*}
    A \emph{nonzero} polynomial of degree $n$
    ($\ge 0$) has \emph{at most} $n$ roots
    (counted with multiplicity).
\end{corollary*}

\begin{proof}
    Induction on the degree. (Exercise)
\end{proof}

\begin{corollary*}
    $f_1, f_2$ polynomials of degree $< n$ such
    that $f_1(t_i) = f_2(t_i)$, $(t_i)_{1 \le i
    \le n}$ $n$ distinct values. Then $f_1 \equiv
    f_2$.
\end{corollary*}

\begin{proof}
    $f_1 - f_2$ has degree $< n$ and at least
    $n$ roots so $f_1 - f_2 \equiv 0$.
\end{proof}

\begin{theorem*}
    Any $f \in \CC[t]$ of positive degree has a
    (complex) root (hence exactly $\deg f$ roots
    when counted with multiplicity).
\end{theorem*}

\noindent
So $f \in \CC[t]$,
\[  f(t) = c \prod_{i = 1}^r (t -
\lambda_i)^\alpha_i \quad c, \lambda_i \in \CC,
\alpha_i \in \NN \]
$\to$ complex analysis.

\begin{definition*}[characteristic polynomial]
    Let $\alpha \in \mathcal{L}(V)$, the
    characteristic polynomial of $\alpha$ is
    \[ \chi_\alpha(t) = \det (A - t\id) \]
\end{definition*}

\begin{remark*}
    The fact that $\det(A - \lambda \id)$ is a
    polynomial in $\lambda$ follows from the very
    definition of $\det$.
\end{remark*}

\begin{remark*}
    Conjugate matrices have the same
    characteristic polynomial.
    \begin{align*}
        \det(P^{-1}AP - \lambda \id)
        &= \det(P^{-1}(A - \lambda \id)P) \\
        &= \det(A - \lambda \id)
    \end{align*}
    So we can define
    \[ \chi_\alpha(t) = \det(A - \lambda \id) \]
    where $A = [\alpha]_{\mathcal{B}}$, and the
    polynomial does not depend on the choice of
    basis.
\end{remark*}

\begin{flashcard}[triangulable-iff]
\begin{theorem*}
    $\alpha \in \mathcal{L}(V)$ is triangulable if
    and only $\chi_\alpha$ can be written as a
    product of linear factors over $F$:
    \[ \chi_\alpha(t) = c \prod_{i = 1}^n (t -
    \lambda_i) \]
\end{theorem*}

\noindent
$\to$ If $F = \CC$, any matrix is triangulable.

\begin{proof}
    \begin{enumerate}
        \item[$\Rightarrow$] \cloze{Suppose $\alpha$
            triangulable, then
            \[ [\alpha]_{\mathcal{B}} =
            \begin{pmatrix}
                a_1 & * & \cdots & * \\
                0 & a_2 & \cdots & * \\
                \vdots & \vdots & \ddots & \vdots \\
                0 & 0 & \cdots & a_n
            \end{pmatrix}
            \]
            so
            \[ \chi_\alpha(t) = \det
            \begin{pmatrix}
                a_1 & * & \cdots & * \\
                0 & a_2 & \cdots & * \\
                \vdots & \vdots & \ddots & \vdots \\
                0 & 0 & \cdots & a_n
            \end{pmatrix}
            = \prod_{i = 1}^n (a_i - t) \]}
        \item[$\Leftarrow$] \cloze{We argue by induction on $n = \dim
            V$.
            \begin{itemize}
                \item $n = 1$ easy.
                \item $n > 1$. By assumption, let
                    $\chi_\alpha(t)$ which has a
                    root $\lambda$. Then
                    $\chi_\alpha(\lambda) = 0$ if
                    and only if $\lambda$ is an
                    eigenvalue of $\alpha$. Let $U
                    = V_\lambda$ be associated
                    eigenspace. Let $(v_1, \dots,
                    v_k)$ be a basis of $U$. We
                    complete to $(v_{k + 1},
                    \dots, v_n)$ of $V$
                    \[ \Span (v_{k + 1}, \dots,
                    v_n) = W \]
                    \[ V = U \oplus W \]
                \item $[\alpha]_{\mathcal{B}} =$
                    \begin{center}
                        \includegraphics[width=0.6\linewidth]
                        {images/cf9b46e65e9211ed.png}
                    \end{center}
                    $\alpha$ induces an
                    endormorphism $\ol{\alpha} : V
                    / U \to V / U$,
                    \[ C =
                    [\ol{\alpha}]_{\ol{\mathcal{B}}},
                    \ol{\mathcal{B}} = (v_{k + 1}
                    + U, \dots, v_n + U) \]
                    Then: (block product)
                    \[ \det(\alpha - \id) =  \]
                    \begin{center}
                        \includegraphics[width=0.6\linewidth]
                        {images/2282c3c05e9311ed.png}
                    \end{center}
                    \[ = (\lambda - t)^k \det(C -
                    t \id) = c\prod_{i = 1}^n (t -
                    a_i) \]
                    \[ \implies \det(C - t\id) =
                    \tilde{c} \prod_{i = k + 1}^n
                    (t - \tilde{a}_i) \]
                    so use the induction because
                    $\dim V / U = \dim V - \dim U
                    < \dim V$. So
                    $\tilde{\mathcal{B}} =
                    (\tilde{v}_{k + 1}, \dots,
                    \tilde{v}_n)$ basis of $W$
                    where:
                    \[ [C]_W =
                    \begin{pmatrix}
                        \tilde{a}_1 & * & \cdots & * \\
                        0 & \tilde{a}_2 & \cdots & * \\
                        \vdots & \vdots & \ddots & \vdots \\
                        0 & 0 & \cdots & \tilde{a}_n
                    \end{pmatrix}
                    \]
                    implies $V = U \oplus W$,
                    \[ \hat{\mathcal{B}} = (v_1,
                    \dots, v_k, \tilde{v}_{k + 1},
                    \dots, \tilde{v}_n) \]
                    basis of $V$ in which
                    \begin{center}
                        \includegraphics[width=0.6\linewidth]
                        {images/bbfaad2e5e9311ed.png}
                    \end{center}
                    $\to$ triangular form.
            \end{itemize}
            }
    \end{enumerate}
\end{proof}
\end{flashcard}