\documentclass[12pt]{amsart}

\usepackage{241}

\title{21-241 Matrices and Linear Transformations \\
       Summer 1 2012 \\
       Course notes \\
       June 11}
\author{Paul McKenney}


\begin{document}

\maketitle

\section{Miscellaneous but useful facts}

The following theorem is useful in more ways than you might think.
\begin{thm}
  If $S$ and $T$ are subspaces of $\CC^n$, and $S\subseteq T$, then $\dim(S) \le \dim(T)$.  Moreover if $\dim(S) = \dim(T)$, then $S = T$.
\end{thm}

You should be able to prove the following two facts using just the above theorem.
\begin{fact}
  If $S$ is a subspace of $\CC^n$, and $\dim(S) = n$, then $S = \CC^n$.
\end{fact}

\begin{fact}
  Let $A$ be an $m\times n$ matrix.  If $\nullity(A) = n$, then $A = 0$.  Equivalently, if $\rank(A) = 0$, then $A = 0$.
\end{fact}

The following facts are also easily proven, this time without any reference to anything but the definitions.
\begin{fact}
  Let $A$ and $B$ be $m\times n$ matrices.  If $T_A = T_B$, then $A = B$.  Equivalently, if $Ax = Bx$ for all $x\in\CC^n$, then $A = B$.
\end{fact}

\begin{fact}
  Let $x,y\in\CC^n$.  Then $\{x,y\}$ is linearly dependent if and only if either $x$ is a scalar multiple of $y$, or vice-versa.
\end{fact}

\section{Eigenstuff}

\begin{defn*}
  Let $A\in M_n(\CC)$.  A complex number $\lambda$ is an \emph{eigenvalue of $A$} if there is some nonzero vector $v\in\CC^n$ such that $Av = \lambda v$.  In this case, $v$ is called an \emph{eigenvector of $A$} with \emph{associated eigenvalue $\lambda$}.  If $\lambda\in\CC$ then
  \[
    V_\lambda = \set{v\in\CC^n}{Av = \lambda v}
  \]
  is called the \emph{eigenspace of $A$ associated to $\lambda$}.  Note that $\lambda$ is an eigenvalue of $A$ if and only if $V_\lambda\neq\{0\}$.  We write $\spec(A)$ for the set of all eigenvalues of $A$.
\end{defn*}

\begin{example*}
  Let $\lambda\in\CC$.  Then $\lambda$ is an eigenvalue of $\lambda I$, and $V_\lambda$ in this case is $\CC^n$.  (So every nonzero vector is an eigenvector of $\lambda I$ with associated eigenvalue $\lambda$.)
\end{example*}

\begin{example*}
  Let $A = \smat{3 & 1 \\ 1 & 3 \\}$.  Show that $\smat{1 \\ 1 \\}$ and $\smat{-1 \\ 1 \\}$ are both eigenvectors for $A$.  What are their associated eigenvalues?  What does this mean, geometrically, about $T_A$?
\end{example*}

\begin{fact}
  If $A$ is a \emph{real} $n\times n$ matrix and $\lambda$ is a \emph{real} eigenvalue of $A$, then there is a \emph{real} eigenvector of $A$ with associated eigenvalue $\lambda$.
\end{fact}

An eigenvector $v$ of a matrix $A$ gives an important bit of geometric information about $A$; if $\lambda$ is its associated eigenvalue, then this tells us that $A$ \emph{stretches} $v$, in the direction of $v$ by a factor of $\lambda$.  Of course, this is much more intelligible when $\lambda\in\RR$.  If $\lambda > 0$, then $A$ stretches $v$ by a factor of $\lambda$ in the direction of $v$, whereas if $\lambda < 0$ then $A$ stretches $v$ by a factor of $|\lambda|$, in the opposite direction.

\begin{example*}
  Let $\theta\in [0,2\pi)$ be given, and let
  \[
    A = \begin{pmatrix}
      \cos\theta & -\sin\theta \\
      \sin\theta &  \cos\theta \\
    \end{pmatrix}
  \]
  (Recall that this matrix implements a counter-clockwise rotation about the origin, by an angle of $\theta$.)
  When (ie, for which values of $\theta$) does $A$ have real eigenvalues?  What are they?  What about complex eigenvalues?
\end{example*}

\begin{example*}
  Let $\theta\in [0,2\pi)$ be given and let
  \[
    A = \begin{pmatrix}
      \cos(2\theta) & \sin(2\theta) \\
      \sin(2\theta) & -\cos(2\theta) \\
    \end{pmatrix}
  \]
  (The reflection across the line which makes an angle of $\theta$ with the $x$-axis.)
  When does $A$ have real eigenvalues?  What are they?  What about complex eigenvalues?
\end{example*}


\begin{lemma}
  $V_\lambda = \nul(\lambda I - A) = \nul(A - \lambda I)$.
\end{lemma}

\begin{thm}
  Let $A\in M_n(\CC)$ and $\lambda\in\CC$.  Then the following are equivalent.
  \begin{enumerate}
    \item  $\lambda$ is an eigenvalue of $A$.
    \item  The nullity of $\lambda I - A$ is nonzero.
    \item  $\det(\lambda I - A) = 0$.
  \end{enumerate}
\end{thm}

\begin{example*}
  Let $A$ be the rotation matrix from the previous example.  Let $\lambda\in\CC$ be given.  Then,
  \[
    \det(\lambda I - A) = (\lambda - \cos\theta)^2 + \sin^2\theta = \lambda^2 - 2(\cos\theta)\lambda + 1
  \]
  This quadratic is zero if and only if
  \[
    \lambda = \cos\theta \pm \sqrt{\cos^2\theta - 1} = \cos\theta \pm i\sin\theta
  \]
  Hence $A$ has eigenvalues $\cos\theta + i\sin\theta$ and $\cos\theta - i\sin\theta$.  The eigenvectors can be found by reducing
  \[
    \begin{pmatrix}
      (\cos\theta + i\sin\theta) - \cos\theta & +\sin\theta \\
      -\sin\theta & (\cos\theta + i\sin\theta) - \cos\theta \\
    \end{pmatrix}
      =
    \begin{pmatrix}
      i\sin\theta & +\sin\theta \\
      -\sin\theta & i\sin\theta \\
    \end{pmatrix}
  \]
  and
  \[
    \begin{pmatrix}
      (\cos\theta - i\sin\theta) - \cos\theta & +\sin\theta \\
      -\sin\theta & (\cos\theta - i\sin\theta) - \cos\theta \\
    \end{pmatrix}
      =
    \begin{pmatrix}
      -i\sin\theta & +\sin\theta \\
      -\sin\theta & -i\sin\theta \\
    \end{pmatrix}
  \]
  If $\sin\theta \neq 0$, then these reduce to, respectively,
  \[
    \begin{pmatrix}
      1 & -i \\
      0 &  0 \\
    \end{pmatrix}
    \qquad\mbox{and}\qquad
    \begin{pmatrix}
      1 & i \\
      0 & 0 \\
    \end{pmatrix}
  \]
  So the first gives an eigenvector of $\smat{i \\ 1 \\}$, and the second $\smat{-i \\ 1 \\}$.
\end{example*}

\begin{defn*}
  The \emph{characteristic polynomial} of an $n\times n$ matrix $A$ is the polynomial, in variable $z$, described by
  \[
    p_A(z) = \det(z I - A)
  \]
\end{defn*}

\begin{fact}
  If $p$ is a degree-$n$ polynomial with complex coefficients, then $p$ can be factored into $n$-many linear terms, with a constant;
  \[
    p(z) = \mu (z - \lambda_1) \cdots (z - \lambda_n)
  \]
  Moreover, $\mu$ is simply the coefficient of $z^n$ in $p$.
\end{fact}

\begin{example*}
  Let $A$ be the matrix
  \[
    A = \begin{pmatrix}
      1 & 0 \\
      1 & 1 \\
    \end{pmatrix}
  \]
  What are the eigenvalues and eigenvectors of $A$?  What does this mean geometrically about $T_A$?
\end{example*}

\begin{thm}
  The eigenvalues of a square matrix $A$ are simply the roots to its characteristic polynomial $p_A$.
\end{thm}

\begin{fact}
  If $A$ is $n\times n$ then $p_A$ has degree $n$.
\end{fact}

\begin{thm}
  If $A\in M_n(\CC)$ then $A$ can have at most $n$ eigenvalues.
\end{thm}

\begin{example*}
  Find the eigenvalues and eigenvectors of the matrix
  \[
    A = 
    \begin{pmatrix}
      0 & 1 & 0 \\
      -1 & 0 & 1 \\
      0 & -1 & 0 \\
    \end{pmatrix}
  \]
\end{example*}
\begin{proof}[Solution]
  The matrix $\lambda I - A$ is
  \[
    \begin{pmatrix}
      \lambda & -1 & 0 \\
      1 & \lambda & -1 \\
      0 & 1 & \lambda \\
    \end{pmatrix}
  \]
  and its determinant is thus
  \[
    \lambda^3 + 0 + 0 - \lambda(1)(-1) - (1)(-1)\lambda - 0 = \lambda^3 + 2\lambda
  \]
  The roots of the polynomial $z^3 + 2z$ are $z = 0$ and $z = \pm \sqrt{2}i$, hence these are the eigenvalues of $A$.  Let's look at the eigenspace associated to eigenvalue $0$;
  \[
    0I - A =
    \begin{pmatrix}
      0 & -1 &  0 \\
      1 &  0 & -1 \\
      0 &  1 &  0 \\
    \end{pmatrix}
  \]
  This matrix has nullity $1$, and its null space is spanned by the single vector $\smat{1 \\ 0 \\ 1 \\}$.

  For eigenvalue $\sqrt{2}i$, we get
  \[
    \sqrt{2}i I - A = 
    \begin{pmatrix}
      \sqrt{2}i & -1 &  0 \\
      1 &  \sqrt{2}i & -1 \\
      0 &  1 &  \sqrt{2}i \\
    \end{pmatrix}
  \]
  This reduces to
  \[
    \begin{pmatrix}
      1 & 0 &    1      \\
      0 & 1 & \sqrt{2}i \\
      0 & 0 &    0      \\
    \end{pmatrix}
  \]
  hence the vector $\smat{-1 \\ -\sqrt{2}i \\ 1 \\}$ is an eigenvector for this eigenvalue.  (And it spans the eigenspace.)

  Finally, for eigenvalue $-\sqrt{2}i$, similar work shows that $\smat{1 \\ -\sqrt{2}i \\ 1 \\}$ is an eigenvector.  (And, again, it spans the eigenspace.)
\end{proof}

\end{document}
