\documentclass[12pt]{amsart}

\setlength{\parsep}{3pc}
\setlength{\itemsep}{0.2in}


\usepackage{fullpage}
\usepackage{epsfig}

\newtheorem{thm}{Theorem}


\newcommand{\Z}{\mathbb Z}
\newcommand{\F}{\mathbb F}
\newcommand{\R}{\mathbb R}
\newcommand{\C}{\mathbb C}

\title[Problem Set 17 Solutions]{Problem Set 17 Solutions.}

\begin{document}

\maketitle

\begin{enumerate}
\item Consider the matrix
$$
A=\left[\begin{array}{rr}1&0\\-1&2\end{array}\right].
$$
\begin{enumerate}
\item Find the eigenvalues of $A$.
\item Find the eigenvectors of $A$.
\item Diagonalize $A$: write it as $A=PDP^{-1}$.
\end{enumerate}\vskip 5pt
\noindent {\bf  ANSWER:}
    \begin{enumerate}

    \item We just have to solve the characteristic equation
    \[
    \det \left[\begin{array}{rr}
         1 - \lambda &     0 \\
        -1           &   2 - \lambda
    \end{array}\right] \ = (1 - \lambda)(2 - \lambda) = 0.
    \]
    So $\lambda = 1,2$. Note that in general the eigenvalues of an
    upper triangular or lower triangular matrix are the diagonal
    entries.

    \item To find the eigenvector corresponding to $\lambda$, we have to find the
    null space of $A - \lambda I$. There are two cases.
    \begin{itemize}
        \item $\lambda = 1$. Here we have to solve
        \[
        \left[\begin{array}{rr}
                 0   &   0 \\
                -1   &   1
            \end{array}\right]
        \left[\begin{array}{r}
                 a  \\
                 b  \end{array}\right] =
        \left[\begin{array}{r}
                 0  \\
                 0  \\
        \end{array}\right]
        \]
        So the eigenvector is
        \[
        \left[\begin{array}{r}
                 1  \\
                 1  \end{array}\right]
        \]
        \item $\lambda = 2$. Here we have to solve
        \[
        \left[\begin{array}{rr}
                -1   &   0 \\
                -1   &   0
            \end{array}\right]
        \left[\begin{array}{r}
                 a  \\
                 b  \end{array}\right] =
        \left[\begin{array}{r}
                 0  \\
                 0  \\
        \end{array}\right]
        \]
        So the eigenvector is
        \[
        \left[\begin{array}{r}
                 0  \\
                 1  \end{array}\right]
        \]
    \end{itemize}
\end{enumerate}

\item Consider the matrix
$$
A=\left[\begin{array}{rrr}
    -3  &  4  &  -4  \\
    -3  &  5  &  -3  \\
    -1  &  2  &   0
    \end{array}\right].
$$
\begin{enumerate}
\item Find the eigenvalues of $A$.  Just kidding!  The eigenvalues are
$-1$, $1$ and $2$.  Two eigenvectors are
$$
\begin{array}{ccc}
v=\left[\begin{array}{r}
    2  \\
    1  \\
    0  \end{array}\right]
& \& &
w=\left[\begin{array}{r}
     1  \\
     0  \\
    -1  \end{array}\right].
\end{array}
$$
Check that these are eigenvectors.  What are the corresponding
eigenvalues? \vskip 5pt
    \noindent {\bf  ANSWER:}
    \[
    \left[\begin{array}{rrr}
        -3  &  4  &  -4  \\
        -3  &  5  &  -3  \\
        -1  &  2  &   0
        \end{array}\right]
    \left[\begin{array}{r}
        2  \\
        1  \\
        0
        \end{array}\right]
    =
    \left[\begin{array}{r}
        -2  \\
        -1  \\
         0
        \end{array}\right]
    \]
So the eigenvalue corresponding to $v$ is $-1$.
    \[
    \left[\begin{array}{rrr}
        -3  &  4  &  -4  \\
        -3  &  5  &  -3  \\
        -1  &  2  &   0
        \end{array}\right]
    \left[\begin{array}{r}
         1  \\
         0  \\
        -1
        \end{array}\right]
    =
    \left[\begin{array}{r}
          1  \\
          0  \\
         -1
        \end{array}\right]
    \]
So the eigenvalue corresponding to $w$ is $1$.
\item Find a third eigenvector corresponding to the third eigenvalue.
\end{enumerate}\vskip 5pt
    \noindent {\bf  ANSWER:}
    We need to find an element of the null space of $A - 2 I$, in
    other words a solution to
    \[
    \left[\begin{array}{rrr}
            -5  &  4  &  -4  \\
            -3  &  3  &  -3  \\
            -1  &  2  &  -2
    \end{array}\right]
    \left[\begin{array}{r}
         a  \\
         b  \\
         c
        \end{array}\right]
    =
    \left[\begin{array}{r}
          0  \\
          0  \\
          0
        \end{array}\right]
    \]
    We can solve this using row reduction, or just observe that the
    third column is just $-1$ times the second column, and so from the
    column definition of matrix multiplication we know immediately that
    the eigenvector must be,
    \[
    \left[\begin{array}{r}
          0  \\
          1  \\
          1
    \end{array}\right]
    \]
    (Note: if you know about vector calculus, another neat way to
    find the eigenvector is to take the cross-product of any two
    rows of $A - \lambda I$.)

\item Suppose that $A$ is an $n\times n$ matrix, and that $A^2=A$.
What can you say, then, about the eigenvalues of $A$? \vskip 5pt
    \noindent {\bf  ANSWER:}
If $\lambda$ is an eigenvalue of $A$. Then
\[
A \overrightarrow{v} = \lambda \overrightarrow{v}
\]
Where $\overrightarrow{v}$ is an eigenvector corresponding to
$\lambda$. Therefore
\[
A^2 \overrightarrow{v} = A A \overrightarrow{v} = A \lambda
\overrightarrow{v} = \lambda A \overrightarrow{v} = \lambda^2
\overrightarrow{v}
\]
And so $A^2 = A$ implies $A^2 \overrightarrow{v} = A
\overrightarrow{v}$. So it must be that
\[
\lambda = \lambda^2
\]
And $\lambda$ must be $0$ or $1$.

\item Suppose $A$ is a $3\times 3$ matrix with eigenvalues $1$,
$2$ and $3$.  If $v_1$ is an eigenvector for the eigenvalue $1$,
$v_2$ for $2$, and $v_3$ for $3$, then what is $A (v_1+v_2-v_3)$?
\vskip 5pt
    \noindent {\bf  ANSWER:}
\[
A (\overrightarrow{v_1}+\overrightarrow{v_2}-\overrightarrow{v_3})
= A  \overrightarrow{v_1} + A \overrightarrow{v_2} - A
\overrightarrow{v_3} = 1 \overrightarrow{v_1} + 2
\overrightarrow{v_2} - 3 \overrightarrow{v_3}
\]

\end{enumerate}


\end{document}