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\subsection{Spectral theory for self adjoint maps}

\begin{itemize}
    \item \textbf{Spectral theory} $\equiv$ study
        of the spectrum of operators
        \begin{itemize}
            \item[$\to$] mathematics
            \item[$\to$] physics (QUANTUM
                MECHANICS)
            \item[$\Rightarrow$] INFINITE
                DIMENSIONAL. Finite dimension
                $\to$ infinite dimension.
                Linear maps $\to$ Hilbert space /
                compact operator.
        \end{itemize}
    \item Adjoint operator: $V$, $W$ finite
        dimensional inner product spaces, $\alpha
        \in \mathcal{L}(V, W)$, then the adjoint
        $\alpha^* \in \mathcal{L}(W, V)$ such that
        $\forall (v, w) \in V \times W$,
        \[ \langle \alpha(v), w \rangle_W =
        \langle v, \alpha^*(w) \rangle_V \]
        We defined:
        \begin{itemize}
            \item[-] Self adjoint maps, $V = W$,
                $\alpha = \alpha^*$,
                \[ \iff \forall (v, w) \in V
                \times V, \quad \langle \alpha v,
                w \rangle = \langle v, \alpha w
                \rangle \]
            \item[-] isometries $V = W$, $\alpha^*
                = \alpha^{-1}$
                \[ \iff \forall (v, w) \in V
                \times V, \quad \langle \alpha
                v, \alpha w \rangle = \langle v, w
                \rangle \]
            \item $\RR$: orthogonal group
            \item $\CC$: unitary group.
        \end{itemize}
\end{itemize}

\subsubsection*{Spectral theory for self adjoint
operators}

\begin{lemma*}
    Let $V$ be a finite dimensional inner product
    space. Let $\alpha \in \mathcal{L}(V)$ be self
    adjoint: ($\alpha = \alpha^*$). Then:
    \begin{enumerate}[(i)]
        \item $\alpha$ has real eigenvalues
        \item eigenvectors of $\alpha$ with
            respect to \emph{different}
            eigenvalues are orthogonal.
    \end{enumerate}
\end{lemma*}

\begin{proof}
    \begin{enumerate}[(i)]
        \item $v \in V \setminus \{0\}$, $\lambda
            \in \CC$ such that $\alpha v = \lambda
            v$. Then
            \begin{align*}
                \lambda \|v\|^2
                &= \langle \lambda v, v \rangle \\
                &= \langle \alpha v, v \rangle \\
                &= \langle v, \alpha^* v \rangle
                \\
                &= \langle v, \alpha v \rangle \\
                &= \langle v, \lambda v \rangle \\
                &= \ol{\lambda} \|v\|^2
            \end{align*}
            So $(\lambda - \ol{\lambda})\|v\|^2 =
            0$. But $\|v\|^2 \neq 0$ since $v \neq
            0$ so $\lambda = \ol{\lambda}$ so
            $\lambda \in \RR$.
        \item $\alpha v = \lambda v$, $\lambda \in
            \RR$, $v \neq 0$. $\alpha w = \mu w$,
            $\mu \in \RR$, $w \neq 0$. Also
            $\lambda \neq \mu$. Then
            \begin{align*}
                \lambda \langle v, w \rangle
                &= \langle \lambda v, w \rangle \\
                &= \langle \alpha v, w \rangle \\
                &= \langle v, \alpha^* w \rangle
                \\
                &= \langle v, \alpha w \rangle \\
                &= \langle v, \mu w \rangle \\
                &= \ol{\mu} \langle v, w \rangle
                \\
                &= \mu \langle v, w \rangle
            \end{align*}
            So $(\lambda - \mu) \langle v, w
            \rangle = 0$. But $\lambda \neq \mu$
            so $\langle v, w \rangle = 0$.
    \end{enumerate}
\end{proof}

\begin{flashcard}[self-adjoint-orthonormal-basis]
\begin{theorem*}
    Let $V$ be a finite dimensional inner product
    space. Let $\alpha \in \mathcal{L}(V)$ be self
    adjoint ($\alpha = \alpha^*$). Then $V$ has an
    \emph{orthonormal} basis made of eigenvectors
    of $\alpha$.
\end{theorem*}

\noindent
$\to$ We also say: $\alpha$ can be diagonalised in
an orthonormal basis for $V$.

\begin{proof}
    \cloze{
    $F = \RR$ or $\CC$. We argue by induction on
    the dimension of $V$, $\dim_F V = n$.
    \begin{itemize}
        \item $n = 1$ $\to$ trivial.
        \item $n - 1 \to n$. $\mathcal{B}$ any
            orthonormal basis of $V$ say $A =
            [\alpha]_{\mathcal{B}}$. By the
            fundamental Theorem of Algebra, we
            know that $\chi_A(t)$ ($\equiv$
            characteristic polynomial of $A$) has
            a \emph{complex} root. This root is an
            eigenvalue of $\alpha$ and $\alpha =
            \alpha^*$ $\implies$ this root is
            \emph{real}. Let us call $\lambda \in
            \RR$ this eigenvalue, pick an
            eigenvector $v_1 \in V \setminus
            \{0\}$ such that $\|v_1\| = 1$,
            $\alpha v_1 = \lambda v_1$. Let $U =
            \langle v_1 \rangle^\perp \le V$. Then
            KEY OBSERVATION: $U$ stable by
            $\alpha$, i.e. $\alpha(U) \le U$.
            Indeed, let $u \in U$, then:
            \begin{align*}
                \langle \alpha u, v_1 \rangle
                &= \langle u, \alpha v_1^* \rangle
                \\
                &= \langle u, \alpha v_1 \rangle
                \\
                &= \langle u, \lambda v_1 \rangle
                \\
                &= \lambda \langle u, v_1 \rangle
                \\
                &= 0
            \end{align*}
            So $\alpha(u) \in U$. This implies: we
            may consider $\alpha|_U \in
            \mathcal{L}(U)$ and self adjoint, and
            then $n = \dim V = \dim U + 1$, so
            $\dim U = n - 1$ so by induction
            hypothesis there exists $(v_2, \dots,
            v_n)$ orthonormal basis of
            eigenvectors for $\alpha |_U$. Then $V
            = \langle v_1 \rangle \orthoplus U$ so
            $(v_1, \dots, v_n)$ orthonormal basis
            of $V$ made of eigenvectors of
            $\alpha$.
    \end{itemize}
}
\end{proof}
\end{flashcard}

\begin{remark*}
    If you want to think in terms of matrices for
    the proof of (ii), then the choice of $U$
    means that $[A]$ is written as
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/8c753dd66f5411ed.png}
    \end{center}
\end{remark*}

\begin{corollary*}
    $V$ finite dimensional inner product space. If
    $\alpha \in \mathcal{L}(V)$ is self adjoint,
    then $V$ is the \emph{orthogonal direct sum}
    of all the eigenspaces of $\alpha$.
\end{corollary*}

\subsubsection*{Spectral theory for unitary maps}

\begin{lemma*}
    $V$ be a \emph{complex} inner product space
    (Hermitian sesquilinear structure). Let
    $\alpha \in \mathcal{L}(V)$ be unitary
    ($\alpha^* = \alpha^{-1}$). Then
    \begin{enumerate}[(i)]
        \item all the eigenvalues of $\alpha$ lie
            on the unit circle
        \item eigenvectors corresponding to
            \emph{distinct} eigenvalues are
            \emph{orthogonal}.
    \end{enumerate}
\end{lemma*}

\begin{proof}
    \begin{enumerate}[(i)]
        \item $\alpha v = \lambda v$, $v \neq 0$,
            $\lambda \in \CC$.
            \begin{itemize}
                \item $\lambda \neq 0$: $\alpha$
                    unitary implies $\alpha$
                    invertible.
                \item \begin{align*}
                    \lambda \|v\|^2
                    &= \lambda \langle v, v
                    \rangle \\
                    \langle \lambda v, v \rangle
                    \\
                    &= \langle \alpha v, v \rangle
                    \\
                    &= \langle v, \alpha^* v
                    \rangle \\
                    &= \langle v, \alpha^{-1} v
                    \rangle \\
                    &= \left\langle v,
                    \frac{1}{\lambda} v
                    \right\rangle \\
                    &= \frac{1}{\ol{\lambda}} \|v\|^2
                \end{align*}
                    So $\lambda \|v\|^2 =
                    \frac{1}{\ol{\lambda}} \|v\|^2$ so
                    since $v \neq 0$, $\lambda
                    \ol{\lambda} = 1$, i.e. $|\lambda|
                    = 1$.
                \item $\alpha v = \lambda v$,
                    $\alpha w = \mu w$, $\lambda,
                    \mu \neq 0$, $\mu \neq
                    \lambda$. Then
                    \begin{align*}
                        \lambda \langle v, w
                        \rangle
                        &= \langle \lambda v, w
                        \rangle \\
                        &= \langle \alpha v, w
                        \rangle \\
                        &= \langle v, \alpha^* w
                        \rangle \\
                        &= \langle v, \alpha^{-1}
                        w \rangle \\
                        &= \left\langle v,
                        \frac{1}{\mu} w
                        \right\rangle \\
                        &= \frac{1}{\ol{\mu}}
                        \langle v, w \rangle \\
                        &= \mu \langle v, w
                        \rangle
                    \end{align*}
                    (by (i)). So $(\lambda - \mu)
                    \langle v, w \rangle = 0$. But
                    $\lambda \neq \mu$ so $\langle
                    v, w \rangle = 0$.
            \end{itemize}
    \end{enumerate}
\end{proof}

\begin{flashcard}[spectral-theory-for-unitary-maps]
\begin{theorem*}[Spectral theory for unitary maps]
    Let $V$ be a finite dimensional \emph{complex}
    inner product space. Let $\alpha \in
    \mathcal{L}(V)$ be unitary ($\alpha^* =
    \alpha^{-1}$). Then $V$ has an orthonormal
    basis made of eigenvectors of $\alpha$.
\end{theorem*}

\noindent
$\to$ Equivalently, $\alpha$ unitary on $V$
Hermitian can be diagonalised in an orthonormal
basis.

\begin{proof}
    \cloze{
    Pick $\mathcal{B}$ any orthonormal basis of
    $V$. $A = [\alpha]_{\mathcal{B}}$. Then
    $\chi_A(t)$ ($\equiv$ characteristic
    polynomial of $A$) has a \emph{complex} root.
    So $\alpha$ has a \emph{complex} eigenvalue.
    Fix $v_1 \in V \setminus \{0\}$ with $\|v_1\|
    \neq 0$, $\alpha v_1 = \lambda v_1$. Let $U =
    \langle v_1 \rangle^\perp$. Then: KEY
    OBSERVATION: $\alpha(U) \le U$. Indeed: $u \in
    U$, then
    \begin{align*}
        \langle \alpha u, v_1 \rangle
        &= \langle u, \alpha^* v_1 \rangle \\
        &= \langle u, \alpha^{-1} v_1 \rangle \\
        &= \left\langle u, \frac{1}{\lambda} v_1
        \right\rangle \\
        &= \frac{1}{\ol{\lambda}} \langle u, v_1
        \rangle \\
        &= 0
    \end{align*}
    $\implies \alpha u \in U$, so $\alpha(U) \le
    U$. We argue by induction on $\dim_\CC V = n$.
    We consider $\alpha |_U \in \mathcal{L}(U)$
    which is unitary, and by the induction
    hypothesis, $\alpha|_U$ is diagonalisable in
    an orthonormal basis $(v_2, \dots, v_n)$ of
    $U$ $\implies (v_1, \dots, v_n)$ is an
    orthonormal basis of $V$, made of eigenvectors
    of $\alpha$.
}
\end{proof}
\end{flashcard}

\begin{warning*}
    We used the complex structure to make sure
    that there is an eigenvalue (which is a priori
    complex valued). \\
    In general, a real valued orthonormal matrix
    ($AA^\top = \id$) \emph{cannot} be
    diagonalised over $\RR$.
\end{warning*}

\begin{example*}[Rotation map in $\RR^2$]
    \[ A =
    \begin{pmatrix}
        \cos\alpha & - \sin \alpha \\
        \sin \alpha & \cos\alpha
    \end{pmatrix}
    \]
    \[ \chi_A(\lambda) = (\cos \alpha - \lambda)^2
    + \sin^2 \alpha \]
    Then the eigenvalues are $\lambda = e^{\pm
    i\alpha}$ ($\not\in$ generally). (Hence
    diagonalisable in $\CC$ but not $\RR$).
\end{example*}