
\documentclass[12pt]{amsart}

\usepackage{241}

\title{21-241 Matrices and Linear Transformations \\
       Summer 1 2012 \\
       Course notes \\
       June 20}
\author{Paul McKenney}


\begin{document}

\maketitle

\section{Change of basis continued}

Yesterday I promised to show you an easy way of computing the representation of a given linear transformation $T$ with respect to an arbitrary basis.

\begin{fact}
  Suppose $T : \CC^n\to\CC^n$ is a linear transformation, and $A$ is its standard matrix (ie, its representation with respect to the standard basis).  If $\SB = (b_1,\ldots,b_n)$ is any other (ordered) basis for $\CC^n$, then
  \[
    \rep_\SB(T) = S^{-1}AS
  \]
  where $S$ is the $n\times n$ matrix whose columns are $b_1,\ldots,b_n$.
\end{fact}
\begin{proof}
  Since $A$ is the standard matrix of $T$,
  \[
    T(b_j) = Ab_j
  \]
  Let $\smat{\lambda_{1j} \\ \vdots \lambda_{nj}}$ be the representation of $Ab_j$ in the basis $b_1,\ldots,b_n$, ie the unique sequence of coefficients such that
  \[
    Ab_j = \lambda_{1j} b_1 + \cdots + \lambda_{nj} b_n
  \]
  Then $\rep_\SB(T)$ is the $n\times n$ matrix with entries $\lambda_{ij}$.

  Now, note that $Se_j = b_j$, and hence $S^{-1}b_j = e_j$.  Then,
  \[
    (S^{-1}AS)e_j = S^{-1}Ab_j = S^{-1}(\lambda_{1j} b_1 + \cdots + \lambda_{nj} b_n) = \lambda_{1j} e_1 + \cdots + \lambda_{nj} e_n = \begin{pmatrix} \lambda_{1j} \\ \vdots \\ \lambda_{nj} \end{pmatrix}
  \]
  The leftmost side of the equation above is just the $j$th column of $S^{-1}AS$; the rightmost side is the $j$th column of $\rep_\SB(T)$.
\end{proof}

\begin{example*}
  Let $T : \RR^2\to\RR^2$ be the reflection across the line $L$ which makes an angle of $\theta$ with the $x$-axis.  Find the matrix of $T$ with respect to the standard basis, and the basis
  \[
    \SB = \left(
      \begin{pmatrix}
        \cos\theta \\ \sin\theta \\
      \end{pmatrix},
      \begin{pmatrix}
        -\sin\theta \\ \cos\theta \\
      \end{pmatrix}
    \right)
  \]
\end{example*}
\begin{proof}[Solution]
  In the standard basis the matrix is
  \[
    \begin{pmatrix}
      \cos(2\theta) & \sin(2\theta) \\
      \sin(2\theta) & -\cos(2\theta) \\
    \end{pmatrix}
  \]
  (This was a homework problem from homework 2.)  In the alternate basis it's
  \[
    \begin{pmatrix}
      \cos\theta & -\sin\theta \\
      \sin\theta & \cos\theta \\
    \end{pmatrix}^{-1}
    \begin{pmatrix}
      \cos(2\theta) & \sin(2\theta) \\
      \sin(2\theta) & -\cos(2\theta) \\
    \end{pmatrix}
    \begin{pmatrix}
      \cos\theta & -\sin\theta \\
      \sin\theta & \cos\theta \\
    \end{pmatrix}
  \]
  Which turns out to be
  \[
    \begin{pmatrix}
      1 & 0 \\
      0 & -1 \\
    \end{pmatrix}
  \]
\end{proof}

\section{Diagonalization}

\begin{defn*}
  A matrix $A$ is \emph{diagonalizable} if it is similar to some diagonal matrix $D$.  In this case, if $A = SDS^{-1}$, then we say that $S$ \emph{diagonalizes} $A$.
\end{defn*}
We like matrices that are diagonalizable.

\begin{fact}
  If $D$ is a diagonal matrix with diagonal entries $d_1,\ldots,d_n$, then
  \begin{itemize}
    \item  $\spec(D) = \{d_1,\ldots,d_n\}$,
    \item  $p_D(\lambda) = (\lambda - d_1)\cdots (\lambda - d_n)$, and
    \item  A basis for the eigenspace of $\lambda$ (with respect to $D$) is given by
    \[
      \set{e_i}{\mbox{$i$ is such that $d_i = \lambda$}}
    \]
  \end{itemize}
\end{fact}

\begin{cor}
  If $A$ is diagonalizable, and $\lambda$ is an eigenvalue of $A$, then the algebraic and geometric multiplicities of $\lambda$ (with respect to $A$) are the same.
\end{cor}
\begin{proof}
  By the above fact, this is true for diagonal matrices, and it's preserved by similarity (by a theorem from yesterday). 
\end{proof}

\begin{example*}
  If $n > 1$, then the $n\times n$ shift matrix $S$ is \emph{not} diagonalizable.
\end{example*}
\begin{proof}
  $S$ has only one eigenvalue: $0$.  Its algebraic multiplicity is $n$, whereas its geometric multiplicity is $1$.  Since they're not the same $S$ can't be diagonalizable.
\end{proof}

\end{document}
