\documentclass{amsart}
\newcommand{\F}{\mathbb F}
\newcommand{\vv}{\vec{v}}
\newcommand{\vu}{\vec{u}}
\newcommand{\vw}{\vec{w}}
\renewcommand{\L}{\mathcal L}
\newcommand{\R}{\mathbb R}
\newcommand{\C}{\mathbb C}
\newcommand{\bilin}[1]{\langle#1\rangle}

\theoremstyle{definition}
\newtheorem{defn}{Definition}
\theoremstyle{plain}
\newtheorem{thm}{Theorem}
\newtheorem{prop}{Proposition}
\newtheorem{lem}{Lemma}
\newtheorem{cor}{Corollary}

\title{Lecture 28: Adjoints and Normal Operators}

\begin{document}
\maketitle
Today's lecture will tie linear operators into our study of Hilbert spaces and discuss an important family of linear operators.

\section*{Adjoints}
We start with a generalization of the Riesz representation theorem.

\begin{thm}[Riesz Representation Theorem]
Let $V$ be a finite dimensional inner product space. Then the map $r\colon V\to V^\ast$ defined by
\[
r(\vv)=\bilin{-,\vv}
\]
is a conjugate-linear isomorphism. The inverse will be denoted $R$
\end{thm}

A word about notation: ``conjugate linear'' means that $r(a\vv)=\bar{a} r(\vv)$. This arises because of the sesquilinearity: the second factor is only conjugate linear.

\begin{proof}
Our argument for bilinear forms can be repeated {\emph{mutatis mutandis}}, so we need not check injectivity. Surjectivity is a little easier in this context. If $f\in V^\ast$, then we can choose a $\vu\in\ker(f)^{\perp}$. Since $V=\bilin{\vu}\oplus\ker(f)$, and since this sum is an orthogonal direct sum, we can easily check that 
\[
R(f)=\frac{\overline{f(\vu)}}{||\vu||^2}\vu.
\]
\end{proof}

This will give us a very important construction which provides a beautiful symmetry for maps between inner product spaces.

\begin{thm}
Let $\tau\in\L(V,W)$ where $V$ and $W$ are finite dimensional inner product spaces. Then there is a unique $\tau^\ast\in L(W,V)$ such that
\[
\bilin{\tau\vv,\vw}=\bilin{\vv,\tau^\ast\vw}.
\]
\end{thm}

\begin{defn}
The linear transformation $\tau^\ast$ is the \underline{adjoint} of $\tau$.
\end{defn}

\begin{proof}
For each $\vw\in W$, we consider the linear functional on $V$ given by
\[
\vv\mapsto\bilin{\tau\vv,\vw}.
\]
This gives us a conjugate linear map $t\colon W\to V^\ast$. By the Riesz representation theorem, we have a conjugate linear map $V^\ast\to V$ that associates to each linear functional its Riesz vector. Thus we let $\tau^*=R\circ t$. Since both $R$ and $t$ are conjugate linear, the composite is linear.

We check that $\tau^\ast$ has the required property:
\[
\bilin{\vv,\tau^\ast\vw}=\bilin{\vv,R(t(\vw))}=t(\vw)(\vv)=\bilin{\tau\vv,\vw},
\]
where the middle equality is the definition of $R$.

For uniqueness, we assume that $S$ is another linear transformation that has the same properties as $\tau^\ast$. Then for all $\vv$ and $\vw$
\[
\bilin{\vv,\tau^\ast\vw}=\bilin{\vv,S\vw},
\]
and so $S=\tau^\ast$.
\end{proof}

How can we make this more concrete? Add in having a basis.

\begin{prop}
If $A$ is an $n\times m$-matrix representing a linear transformation $T\colon V\to W$, then $T^\ast$ is represented by $\overline{A}^t$.
\end{prop}

Why would this be true? Assume that our bases are orthonormal bases. Then the inner product of $\vv$ and $\vw$ is $\vv^t\cdot\overline{\vw}$. Thus 
\[
\bilin{A\vv,\vw}=(A\vv)^t\cdot\overline{\vw}=\vv^t\cdot A^t\cdot\overline{\vw}=\vv^t\cdot \overline{(\bar{A}^t\vw)}=\bilin{\vv,\bar{A}^t\vw}.
\]
The operation of ``conjugate transpose'' is clearly compatible with conjugation by an invertible matrix, so this also tells us the general case.

Passage to adjoints is a very nice operation. The map that sends $\tau$ to $\tau^\ast$ is conjugate linear, and moreover, the conjugate symmetry of the inner products shows that
\[
\tau^{\ast\ast}=\tau
\] 
for any linear operator.

There is a very close connection between adjunctions, orthogonality, and kernels and images. 

\begin{prop}
Let $T\in\L(V,W)$ where everything in sight is a finite dimensional inner product space. 
\begin{enumerate}
\item $\ker(T^\ast)=Im(T)^{\perp}$
\item $\ker(T^\ast T)=\ker(T)$
\end{enumerate}
\end{prop}
\begin{proof}
Let $\vu\in\ker(T^\ast)$. This is the same statement as $T^\ast\vu=0$, and so in particular for all $\vv\in V$, we see
\[
0=\bilin{T^\ast\vu,\vv}=\bilin{\vu,T\vv}.
\]
In particular, we see that $\vu$ is orthogonal to $T\vv$ for all $\vv\in V$, and therefore $\vu\in Im(T)^{\perp}$.

For the second part, it is obvious that $\ker(T)\subset\ker(T^\ast T)$. Now let $\vu\in\ker(T^\ast T)$. Then
\[
0=\bilin{T^\ast T\vu,\vu}=\bilin{T\vu,T\vu},
\]
and so $T\vu=0$.
\end{proof}

Combining these, we also see
\[
Im(T^\ast T)=\ker(T^\ast T)^{\perp}=\ker(T)^{\perp}=Im(T^\ast).
\]

We learn a number of things on top of this. We see that $T$ is surjective iff $T^\ast$ is injective, and $T$ is injective iff $T^\ast$ is surjective. This should remind you of the dual of short exact sequences being short exact. Here we've replaced the dual of a linear operator with the adjoint.

Now we can focus on a few specific kinds of special linear transformations.

\begin{defn}
A linear operator $T\colon V\to V$ is
\begin{enumerate}
\item \underline{Normal} if $T^\ast T=TT^\ast$
\item \underline{self-adjoint} if $T^\ast=T$ (\underline{Hermitian} if $\F=\C$ and \underline{symmetric} if $\F=\R$)
\item \underline{skew-self-adjoint} if $T^\ast=-T$
\item \underline{unitary} if $T^\ast=T^{-1}$
\end{enumerate}
\end{defn}

\begin{prop}
If $T$ is a normal operator and $p(x)$ is any polynomial, then $p(T)$ is a normal operator. In particular, $T-\lambda I$ is normal.
\end{prop}

\section*{Normal Operators}

We finish the class tying together this theory of normal operators with Jordan form.

\begin{thm}
If $T$ is a normal operator, then $T$ is diagonalizable. Moreover, the eigenspaces are orthogonal.
\end{thm}

We'll quickly see that $T$ is diagonalizable. Let $\lambda_1,\dots,\lambda_k$ be the eigenvalues of $T$. We know that we can put $T$ into Jordan form. The algorithm starts with $B=T-\lambda_i I$, and then it considers the ascending chain
\[
\ker(B)\subset\ker(B^2)\subset\dots\subset\ker(B^r)=\dots.
\]
The smallest value of $r$ such that $\ker(B^r)=\ker(B^{r+1})$ is the size of the largest Jordan block. Thus the diagonalizablity is equivalent to $r=1$, since then all of the Jordan blocks are $1\times 1$ blocks. We will have proved the theorem if we can see $\ker(B)=\ker(B^r)$ for all $r\geq 1$.

\begin{lem}
If $T$ is normal, then $\ker(T)=\ker(T^r)$ for all $r\geq 1$.
\end{lem}
\begin{proof}
Let $S=T^\ast T$. This is a self-adjoint linear operator. We can see that $\ker(S)=\ker(S^r)$ for all $r\geq 1$. Assume $\vu\in\ker(S^r)$. Then we have
\[
0=\bilin{S^r(\vu),S^{r-2}(\vu)}=\bilin{S^{r-1}(\vu),S^{r-1}(\vu)},
\]
and we therefore know that $S^{r-1}(\vu)=0$. By downward induction on $r$, we see that $\vu\in\ker(S)$.

Now let $\vv\in\ker(T^r)$. Then since $S^r=(T^\ast)^r T^r$,
\[
\vv\in\ker(T^r)\subset\ker(S^r)=\ker(S)=\ker(T),
\]
and so $\ker(T)=\ker(T^r)$ for all $r\geq 1$.
\end{proof}

Now we can check that the eigenspaces are orthogonal.

\begin{lem}
If $T$ is normal, then 
\[
\bilin{T\vv,T\vw}=\bilin{T^\ast\vv,T^\ast\vw},
\]
so $\ker(T)=\ker(T^\ast)$.
\end{lem}
\begin{proof}
We use normality to see this:
\[
\bilin{T\vv,T\vw}=\bilin{T^\ast T\vv,\vw}=\bilin{TT^\ast\vv,\vw}=\bilin{T^\ast\vv,T^\ast\vw}.
\]

The second part follows from the first by considering
\[
\bilin{T\vv,T\vv}=\bilin{T^\ast\vv,T^\ast\vv}.
\]
\end{proof}

In particular, this shows us that the eigenvalues of $T^\ast$ are the conjugates of those of $T$, and the eigenvectors are the same.

\begin{prop}
If $\vv$ and $\vu$ are eigenvectors for $T$ for distinct eigenvalues, then $\vv\perp\vu$.
\end{prop}
\begin{proof}
Let $\lambda$ be the eigenvalue for $\vv$ and let $\mu$ be the eigenvalue for $\vu$. Then
\[
\lambda\bilin{\vv,\vu}=\bilin{T\vv,\vu}=\bilin{\vv,T^\ast\vu}=\bilin{\vv,\bar{\mu}\vu}=\mu\bilin{\vv,\vu}.
\]
Since $\lambda\neq\mu$, we only have such an equality if $\bilin{\vv,\vu}=0$, and so $\vv\perp\vu$.
\end{proof}

\end{document}