
\documentclass[12pt]{amsart}

\usepackage{241}

\title{21-241 Matrices and Linear Transformations \\
       Summer 1 2012 \\
       Course notes \\
       June 19}
\author{Paul McKenney}


\begin{document}

\maketitle


Let $\{x_1,\ldots,x_k\}$ be a basis for some subspace $V$ of $\CC^n$.  We've seen before that any vector $v$ in $V$ can be written uniquely as a linear combination of $x_1,\ldots,x_k$;
\[
  v = \lambda_1 x_1 + \cdots + \lambda_k x_k
\]
\begin{defn*}
  Let $\SB = \{b_1,\ldots,b_n\}$ be a basis for $\CC^n$.  If $x\in\CC^n$, then the \emph{representation of $x$ in the basis $b_1,\ldots,b_n$} is the unique vector $\smat{\lambda_1 \\ \vdots \\ \lambda_n}$ in $\CC^n$ such that
  \[
    x = \lambda_1 b_1 + \cdots + \lambda_n b_n
  \]
  We refer to this vector as $\rep_\SB(x)$.
\end{defn*}

\begin{defn*}
  Let $\SB = \{b_1,\ldots,b_n\}$ be a basis for $\CC^n$, and let $T : \CC^n\to\CC^n$ be a linear transformation.  The \emph{matrix of $T$ with respect to $b_1,\ldots,b_n$} is the unique $n\times n$ matrix $A$ such that, for all $x\in\CC^n$,
  \[
    \rep_\SB(T(x)) = A\rep_\SB(x)
  \]
  Equivalently, $A$ is the matrix with entries
  \[
    A_{ij} = (\rep_\SB(b_j))_i
  \]
  We refer to this matrix as $\rep_\SB(T)$.
\end{defn*}

\begin{example*}
  Let $L$ be the subspace of $\RR^2$ spanned by $\smat{1 \\ 1}$.  Compute the matrix of $\PP_L$ with respect to the (ordered) standard basis $\SE = (e_1,e_2)$, and the basis
  \[
    \SB = 
    \left(
      \frac{1}{\sqrt{2}}\begin{pmatrix}
        1 \\ 1 \\
      \end{pmatrix},
      \frac{1}{\sqrt{2}}\begin{pmatrix}
        1 \\ -1 \\
      \end{pmatrix}
    \right)
  \]
\end{example*}
\begin{proof}[Solution]
  The first is
  \[
    \rep_\SE(\PP_L) = \begin{pmatrix}
      1/2 & 1/2 \\
      1/2 & 1/2 \\
    \end{pmatrix}
  \]
  The second is simply
  \[
    \rep_\SB(\PP_L) = \begin{pmatrix}
      1 & 0 \\
      0 & 0 \\
    \end{pmatrix}
  \]
\end{proof}
How are these matrices related?  They both represent the same linear transformation $\PP_L$, but how does this fact present itself in the algebra of the matrices themselves?

\begin{defn*}
  Two square matrices $A$ and $B$ are \emph{similar}, written $A\sim_s B$, if there is some invertible $S$ such that $A = SBS^{-1}$.
\end{defn*}

\begin{fact}
  $\sim_s$ is an equivalence relation.
\end{fact}


\begin{thm}
  Let $A$ and $B$ be square matrices, of the same size.  Then the following are equivalent.
  \begin{enumerate}
    \item  $A$ and $B$ are similar.
    \item  $A$ and $B$ represent the same linear transformation.
  \end{enumerate}
\end{thm}

\begin{example*}
  Let $T : \RR^3\to\RR^3$ be the linear transformation
  \[
    T\begin{pmatrix}
      x \\ y \\ z \\
    \end{pmatrix}
    = \begin{pmatrix}
      y + z \\ x + z \\ x + y \\
    \end{pmatrix}
  \]
  Find the matrix of $T$ with respect to the standard basis, and the basis
  \[
    \left(
      \begin{pmatrix}
        1 \\ -1 \\ 0 \\
      \end{pmatrix},
      \begin{pmatrix}
        1 \\ 1 \\ -2 \\
      \end{pmatrix},
      \begin{pmatrix}
        1 \\ 1 \\ 1 \\
      \end{pmatrix}
    \right)
  \]
\end{example*}
\begin{proof}[Solution]
  In the standard basis, the matrix of $T$ is
  \[
    \begin{pmatrix}
      0 & 1 & 1 \\
      1 & 0 & 1 \\
      1 & 1 & 0 \\
    \end{pmatrix}
  \]
  whereas in the alternate basis, the matrix of $T$ is
  \[
    \begin{pmatrix}
      -1 &  0 & 0 \\
       0 & -1 & 0 \\
       0 &  0 & 2 \\
    \end{pmatrix}
  \]
\end{proof}

In physics one often studies ``coordinate-free'' properties of linear transformations; ie, properties that are \emph{invariant} under a change of basis.  The idea is that physical laws should be independent of the observer.
\begin{thm}
  \label{invariant}
  If $A$ and $B$ are similar, then $\spec(A) = \spec(B)$.  Moreover, the (algebraic and geometric) multiplicity of each eigenvalue with respect to $A$ is the same as that with respect to $B$.
\end{thm}

I'll prove the theorem below, but first I need a lemma.

\begin{lemma}
  If $S$ is an invertible matrix then $\det(S^{-1}) = \det(S)^{-1}$.
\end{lemma}
\begin{proof}
  We have $\det(S)\det(S^{-1}) = \det(SS^{-1}) = \det(I) = 1$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{invariant}]
  Say $A = SBS^{-1}$.  Then,
  \begin{align*}
    p_A(\lambda) & = \det(\lambda I - A) = \det(\lambda SIS^{-1} - SBS^{-1}) = \det(S(\lambda I - B)S^{-1}) \\
       & = \det(S)\det(\lambda I - B)\det(S)^{-1} = \det(\lambda I - B) = p_B(\lambda)
  \end{align*}
  So $A$ and $B$ have the same characteristic polynomial.  This shows that they have the same eigenvalues and their eigenvalues have the same algebraic multiplicities.  As for the geometric multiplicities, notice that for all $\lambda$,
  \[
    \col(\lambda I - B) = \col(S(\lambda I - B)) \qquad \row(S(\lambda I - B)) = \row(S(\lambda I - B)S^{-1})
  \]
  Hence, $\rank(\lambda I - B) = \rank(\lambda I - A)$.  By the rank-nullity theorem, this implies $\nullity(\lambda I - B) = \nullity(\lambda I - A)$.  But the nullity of $\lambda I - B$ is simply the geometric multiplicity of $\lambda$ with respect to $B$, and analogously for $A$.
\end{proof}
Of course, the eigenspaces may change under a change of basis.   Both of the examples from above are typical in this regard.

\end{document}
