\documentclass{article}
\usepackage{amsmath}
\newcommand{\myline}{\vspace{-0.2in}\begin{center}\underline{\hspace{3in}}\end{center}}
\setlength{\parindent}{0in}
\setlength{\parskip}{0.08in}
\addtolength{\topmargin}{-0.5in}
\addtolength{\textheight}{1in}
\addtolength{\textwidth}{1.5in}
\addtolength{\oddsidemargin}{-0.75in}
\pagestyle{empty}
\begin{document}

\begin{center}
\textbf{Mathematics 262} \\
\textbf{Homework 5 solutions}
\end{center}
\vspace{-0.05in}

Assignment: Section 6.2, problems 11, 12, 13; Section 6.3, problems 1,
5, 6, 8.

\S 6.2, \#11:
If $N$ is not the zero matrix, then there is a (nonzero) vector
$\alpha$ so that $N \alpha \neq \vec{0}$.  Let $\beta = N \alpha$;
then
\[
N \beta = N (N \alpha) = N^{2} \alpha = \vec{0} = 0 \beta.
\]
This means that 0 is an eigenvalue, with eigenvector $\beta$.  Since
$\beta$ is in the nullspace of $N$ and $\alpha$ isn't, then $\alpha$
and $\beta$ are linearly independent, and hence form a basis.  With
respect to this basis, the matrix for $N$ looks like $\begin{bmatrix}
0 & 0 \\ 1 & 0 \end{bmatrix}$.

\S 6.2, \#12:
[partial solution]
Since we are working over the complex numbers, then the characteristic
polynomial for $A$ factors into linear terms, say as $(x-a)(x-b)$.  If
$a$ and $b$ are distinct, then we know (by problem 7, for instance,
which was one of the practice problems last week) that $A$ is
diagonalizable, so $A$ is similar to $\displaystyle
\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$.  

If $a=b$, then the eigenspace for $a$ has dimension 1 or 2.  If it has
dimension 2, then by the theorem from class, $A$ is diagonalizable,
and hence is similar to $\displaystyle \begin{bmatrix} a & 0 \\ 0 & a
\end{bmatrix}$ (i.e., to $\displaystyle \begin{bmatrix} a & 0 \\ 0 & b
\end{bmatrix}$, where $a=b$).  If the eigenspace for $a$ has dimension
1, then $A$ is not diagonalizable, so we'd better show that it's
similar to $\displaystyle \begin{bmatrix} a & 0 \\ 1 & a
\end{bmatrix}$.  You should keep working on this, and turn it in next
week.

\S 6.2, \#13:
Suppose that for some function $f$, we have $T(f) = cf$.  Then
\[
c f(x) = \int_{0}^{x} f(t) \ dt
\]
for every $x$.  (In particular, $f(0) = 0$.)  Differentiating both
sides and using the fundamental theorem of calculus gives
\[
c f'(x) = f(x).
\]
If $c=0$, then obviously $f(x)=0$ for all $x$.  Otherwise, this is a
separable differential equation; its only solution satisfying the
initial condition $f(0)=0$ is the zero function $f(x) = 0$.  Since the
zero function (i.e., vector) can never be an eigenvector, there are no
eigenvectors, so there are no eigenvalues.

\S 6.3, \#1: 
The minimal polynomial for $I$ is $p(x)=x-1$, since for a matrix $A$
this becomes $p(A) = A-I$.  If $A=I$, then $I-I$ is certainly 0.  The
minimal polynomial for the zero operator is $p(x)=x$.

\S 6.3, \#5:
Since $T$ satisfies $g(x)=x^{k}$, then $p(x)$ must divide evenly into
$x^{k}$.  Therefore $p(x)=x^{j}$ for some $j$.  Since the degree of
$p(x)$ is at most $n$, then $j\leq n$.  Well, if $T^{j}=0$ for some
$j$ with $j\leq n$, then also $T^{n}=0$.

\S 6.3, \#6:
$\displaystyle A= \begin{bmatrix}
0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 
\end{bmatrix}$.

\S 6.3, \#8: 
Given a vector $\alpha$, write it in coordinates as $\alpha
=(\alpha_{1}, \alpha_{2})$.  Then $P$ is defined by $P\alpha =
\alpha_{1}$.  It's easy to check that $P(\alpha + \beta) = \alpha_{1}
+ \beta_{1} = P(\alpha) + P(\beta)$, and that $P(c \alpha) =
c\alpha_{1} = c P (\alpha)$.  (For one thing, you can write $P$ as the
matrix $\displaystyle \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$.)
The minimal polynomial is $p(x)=x^{2}-x$.



\end{document}
