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\title{21-241 Matrices and Linear Transformations \\
       Summer 1 2012 \\
       Course notes \\
       June 25}
\author{Paul McKenney}

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\maketitle

Our aim for tomorrow is to prove the following theorem, which I often call \emph{the spectral theorem}.
\begin{thm*}
  If $A$ is a Hermitian matrix, then $A$ is diagonalizable by a unitary matrix.
\end{thm*}

Today we'll develop the background necessary for the proof of this theorem.  It starts with the following definition.
\begin{defn*}
  Let $V_1,\ldots,V_k$ be subspaces of $\CC^n$, such that $V_i\perp V_j$ for $i\neq j$.  Then their \emph{orthogonal sum} is the subspace
  \[
    V_1 \oplus \cdots \oplus V_k = \set{v_1 + \cdots + v_k}{v_1\in V_1,\ldots, v_k\in V_k}
  \]
\end{defn*}

\begin{example}
  Let's take the lines $L = \lspan\{\smat{1 \\ 2 \\ 3}\}$ and $K = \lspan\{\smat{1 \\ 1 \\ -1}\}$ in $\RR^3$.  Then $L\perp K$.  Their orthogonal sum is the plane
  \[
    L\oplus K = \set{v + w}{v\in L\land w\in K}
      = \lspan\left\{\begin{pmatrix}
        1 \\ 2 \\ 3 \\
      \end{pmatrix},
      \begin{pmatrix}
        1 \\ 1 \\ -1 \\
      \end{pmatrix}
    \right\}
  \]
\end{example}

\begin{lemma}
  Suppose $V$ and $W$ are subspaces of $\CC^n$, such that $V\perp W$.  Then $\dim(V\oplus W) = \dim(V) + \dim(W)$, and moreover if $X$ is an orthonormal basis for $V$ and $Y$ an orthonormal basis for $W$, then $X\cup Y$ is an orthonormal basis for $V\oplus W$.
\end{lemma}

\begin{cor}
  If $V_i\perp V_j$ for $i\neq j$, and $X_i$ is an orthonormal basis for $V_i$, then $X_1\cup\cdots\cup X_k$ is an orthonormal basis for $V_1\oplus \cdots \oplus V_k$.
\end{cor}

\begin{cor}
  If $V$ is a subspace of $\CC^n$, then $\dim(V) + \dim(V^\perp) = n$.
\end{cor}

\begin{defn*}
  A subspace $V\subseteq \CC^n$ is \emph{invariant} for a matrix $A\in M_n(\CC)$ if
  \[
    \forall v\in V\quad Av\in V
  \]
\end{defn*}

\begin{lemma}
  If $V$ is invariant for $A$, and $A$ is Hermitian, then so is $V^\perp$.
\end{lemma}
\begin{proof}
  Let $w\in V^\perp$.  Then for all $v\in V$,
  \[
    \brak{Aw}{v} = \brak{w}{A^H v} = \brak{w}{Av} = 0
  \]
  since $Av\in V$.  Hence $Aw\in V^\perp$.
\end{proof}

\begin{note*}
  Let $A\in M_n(\CC)$ be a Hermitian matrix.  Suppose $V$ is an invariant subspace for $A$; thus $V^\perp$ is too by the above lemma.  Let $v_1,\ldots,v_k$ and $w_1,\ldots,w_\ell$ be orthonormal bases for $V$ and $V^\perp$ respectively; then $k + \ell = n$, and
  \[
    v_1,\ldots,v_k,w_1,\ldots,w_\ell
  \]
  is an orthonormal basis for $\CC^n$.  Let $U$ be the matrix whose columns are $v_1,\ldots,v_k,w_1,\ldots,w_\ell$.  Then $U$ is a unitary matrix.  We'd like to look at $U^H A U$.  Its $(i,j)$-entry is
  \[
    \brak{U^H A Ue_j}{e_i} = \brak{AUe_j}{Ue_i}
  \]
  and $Ue_j$ and $Ue_i$ are the $j$th and $i$th columns of $U$, respectively.  Say $i \le k$ and $j > k$.  Then $Ue_i$ is one of the $v$'s, and $Ue_j$ is one of the $w$'s; so in particular, $Ue_i\in V$, whereas $Ue_j\in V^\perp$.  Since $V^\perp$ is invariant for $A$, $AUe_j\in V^\perp$ too, so
  \[
    \brak{AUe_j}{Ue_i} = 0
  \]
  Similarly, if $i > k$ but $j\le k$, then $AUe_j\in V$ and $Ue_i\in V^\perp$, so
  \[
    \brak{AUe_j}{Ue_i} = 0
  \]
  This means that $A$ has the block form
  \[
    A = \begin{pmatrix}
      \hat{A} & 0_{k\times\ell} \\ 0_{\ell\times k} & \check{A} \\
    \end{pmatrix}
  \]
  where $\hat{A}$ is some $k\times k$ matrix, and $\check{A}$ is some $\ell\times\ell$ matrix.
\end{note*}

\begin{lemma}
  If $A$ is a Hermitian matrix, $\lambda_1,\ldots,\lambda_k$ are its eigenvalues (without repetition), and
  \[
    V = V_{\lambda_1}\oplus\cdots\oplus V_{\lambda_k}
  \]
  then $V$ and $V^\perp$ are invariant subspaces for $A$.
\end{lemma}
\begin{proof}
  Let $v\in V$.  Then $v = v_1 + \cdots + v_k$ for some $v_1\in V_{\lambda_1},\ldots,v_k\in V_{\lambda_k}$.  Now,
  \[
    Av = A(v_1 + \cdots + v_k) = Av_1 + \cdots + Av_k = \lambda_1 v_1 + \cdots + \lambda_k v_k
  \]
  Since $v_i\in V_{\lambda_i}$, so is $\lambda_i v_i$.  Hence $Av\in V$.  This proves that $V$ is an invariant subspace for $A$.  That $V^\perp$ is also invariant follows from this and the above lemma.
\end{proof}



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