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{\bf Math 366, Assignment 1 \\ Profs personal problems (assigned 1/16/03)}
\end{center}
\vspace*{.3in}

\begin{enumerate}
\item Find the operator norm of the linear transformations
$L:\R^2\to\R^2$ with matrices
$$
\left(\begin{matrix} 4 & 0 \\ 0 & -4\end{matrix}\right)\text{ and }
\left(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}\right).
$$
\vspace*{.2in}

\item Let $V$ be a vector space over the field $\R$ (or $\C$).  A 
\emph{norm} on $V$ is a function $\norm{\cdot}:V\to \R$ such that
for all $\lambda\in\R$ and $\mathbf{v},\mathbf{w}\in V$,
\begin{itemize}
\item $\norm{\mathbf{v}} \geq 0$ with equality if and only if 
$\mathbf{v} = 0$.
\item $\norm{\mathbf{\lambda v}} = |\lambda|\norm{\mathbf{v}}$
\item $\norm{\mathbf{v} + \mathbf{w}} \leq 
  \norm{\mathbf{v}} + \norm{\mathbf{w}}$.
\end{itemize}
Given a norm $\norm{\cdot}$ on $V$, show that 
$$
d(\mathbf{v},\mathbf{w}) = \norm{\mathbf{v} - \mathbf{w}}
$$
defines a metric on $V$.  A set $U$ is said to be \emph{open with respect to
$\norm{\cdot}$} if it is open with respect to the associated metric $d$.
\vspace*{.2in}

\item Different norms $\norm{\cdot}$ and $\norm{\cdot}'$ on the same
vector space are called \emph{comparable} if there are constants $C_1,C_2>0$
such that
$$
C_1\norm{\mathbf{v}} \leq \norm{\mathbf{v}}' \leq C_2\norm{\mathbf{v}}
$$
for all $\mathbf{v}\in V$.  

Supposing that $\norm{\cdot},\norm{\cdot}'$ are comparable, show that
a set $U\subset V$ is open with respect to $\norm{\cdot}$ if and only
if it is open with respect to $\norm{\cdot}'$.  Does the same conclusion
hold if you replace `open' with `closed'? `compact'? `connected'?
Explain.

\vspace*{.2in}
\item Let $n,m\in\Z^+$ be given and $V = L(\R^n,\R^m)$ be the vector space
of linear transformations from $\R^n$ to $\R^m$.  Let $T = (a_{ij})\in V$ be
an arbitrary element.  Show that the following norms on $V$ are all comparable
to the operator norm on $V$.
\begin{itemize}
\item $\norm[\infty]{T} = \max_{i,j}|a_{ij}|$
\item $\norm[1]{T} = \sum_{i,j} |a_{ij}|$
\item $\norm[2]{T} = \sqrt{\sum_{i,j} |a_{ij}|^2}$
\end{itemize}
In fact, it can be shown that pretty much any two norms on a finite dimensional
vector space are comparable (Prove this and you take care of all the
above items at once.  And I'll give you five extra credit points).  

\vspace*{.2in}
\item
Show that the norms
\begin{itemize}
\item $\norm[\infty]{f} = \max_{x\in[0,1]} |f(x)|$
\item $\norm[1]{f} = \int_0^1 |f(x)|\,dx$
\end{itemize}
on the (infinite dimensional) vector space $C([0,1],\R)$ are not
comparable.  

\end{enumerate}
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