
\documentclass[12pt]{amsart}

\usepackage{241}

\title{21-241 Matrices and Linear Transformations \\
       Summer 1 2012 \\
       Course notes \\
       June 27}
\author{Paul McKenney}

\begin{document}

\maketitle

Here's some neat facts about diagonalizable matrices.

\begin{fact}
  If $A = S^{-1} D S$, where $D$ is diagonal and $S$ is invertible, then the diagonal values of $D$ are exactly the eigenvalues of $A$, and the columns of $S$ are eigenvectors of $A$ with associated eigenvalue the corresponding diagonal entry of $D$.
\end{fact}

\begin{fact}
  Suppose $A$ is diagonalizable, and its only eigenvalue is $\lambda$.  Then $A = \lambda I$.
\end{fact}
\begin{proof}
  We have $A = S^{-1}DS$, where $D$ is diagonal, and its entries are the eigenvalues of $A$.  Since $\lambda$ is the only one, $D = \lambda I$.  But then,
  \[
    A = S^{-1} (\lambda I) S = \lambda (S^{-1} I S) = \lambda S^{-1}S = \lambda I
  \]
\end{proof}

\begin{lemma}
  Suppose $A = S^{-1} D S$, where $S$ is invertible.  Then $A^k = S^{-1}D^k S$ for any $k$.
\end{lemma}

The following theorem is known as the Spectral Mapping Theorem, and it holds for all square matrices $A$.  I'll only be able to prove it for the diagonalizable ones.
\begin{fact}
  Let $A\in M_n(\CC)$, and let $p(z)$ be some polynomial in $z$.  Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$, possibly with multiplicity.  Then $p(A)$ (the result of substituting $A$ for $z$, and $I$ for the constant $1$, in $p(z)$) has eigenvalues $p(\lambda_1),\ldots,p(\lambda_n)$.
\end{fact}
\begin{proof}[Proof for diagonalizable $A$]
  Say $p(z) = c_k z^k + c_{k-1} z^{k-1} + \cdots + c_1 z + c_0$.  Let $A = S^{-1}DS$, where $S$ is invertible and $D$ diagonal, with diagonal entries $\lambda_1,\ldots,\lambda_n$.  Then
  \begin{align*}
    p(A) & = c_k A^k + c_{k-1} A^{k-1} + \cdots + c_1 A + c_0 I \\
         & = c_k S^{-1}D^k S + c_{k-1} S^{-1}D^{k-1}S + \cdots + c_1 S^{-1}DS + c_0 I \\
         & = S^{-1} (c_k D^k + c_{k-1}D^{k-1} + \cdots + c_1 D + c_0 I) S = S^{-1} p(D) S
  \end{align*}
  Now note that since $D$ is diagonal, $D^t$ is just the diagonal matrix whose entries are those of $D$, taken to the power $t$.  Hence the matrix
  \[
    c_k D^k + c_{k-1}D^{k-1} + \cdots + c_1 D + c_0 I
  \]
  is diagonal, and its $i$th diagonal entry is
  \[
    c_k \lambda_i^k + c_{k-1} \lambda_i^{k-1} + \cdots + c_1 \lambda_i + c_0\cdot 1 = p(\lambda_i)
  \]
  Then these are the eigenvalues of $p(A)$, since $p(A) = S^{-1} p(D) S$.
\end{proof}

This corollary is known as the Cayley-Hamilton theorem.  Again, it's known to be true for all square matrices $A$, but I'll only prove it for the diagonalizable ones.
\begin{cor*}
  Let $A$ be a square matrix, and let $p_A$ be its characteristic polynomial.  Then $p_A(A)$ is the zero matrix.
\end{cor*}
\begin{proof}[Proof for diagonalizable $A$]
  By the spectral mapping theorem, the eigenvalues of $p_A(A)$ are $p_A(\lambda_1),\ldots,p_A(\lambda_n)$ where $\lambda_1,\ldots,\lambda_n$.  But $p_A(\lambda_i) = 0$, for all $i$!  So the only eigenvalue of $p_A(A)$ is $0$.  Now, if $A$ is diagonalizable, then it's easy to see that $p_A(A)$ is, too (the same calculation we did above for the spectral mapping theorem).  Hence $p_A(A) = 0I = 0_{n\times n}$.
\end{proof}

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