\documentclass[minion]{homework}
\usepackage{cmacros}
\def\calB{\mathcal{B}}
\DeclareMathOperator{\Ker}{{\rm Ker}}
\DeclareMathOperator{\Image}{{\rm Image}}
\begin{document}
\doclabel{Math 617: Homework 11}
\docdate{Due: April 11, 2012}
\begin{aproblems}
\hproblem 


\end{aproblems}
\end{document}
\hproblem \solver{David Maxwell}
Suppose $A:X\ra X$ is a compact normal operator.
Let $B=A^*A$.  
\begin{subproblems}
\item Suppose $\lambda\neq 0$ is an eigenvalue of $B$ with eigenvector $x$.
Show that $Ax$ is also an eigenvector of $B$ with eigenvalue $\lambda$.
\item Show that $\Ker A=\Ker B$.
\item It is a result from linear algebra that every normal operator
on a finite-dimensional inner-product space has an orthonormal basis of eigenvectors.  Conclude that every compact normal operator on
a Hilbert space has an orthonormal basis of eigenvectors.
\end{subproblems}

\hproblem\solver{Will Mitchell}
Let $X$ be a Hilbert space and $K:X\ra X$ be a compact operator.
Consider
$$
A=I+K,
$$
i.e. a compact perturbation of the identity.
Show that exactly one of the following two possibilities holds.
\begin{itemize}
\item[a)]  The kernel of $A$ is trivial and for all $y\in X$ 
there is a unique solution of $Ax=y$
\item[b)]  The kernel of $A$ has positive but finite dimension.
In this case $Ax=y$ has a solution if and only if $y\in (\Ker A)^\perp$.
Moreover, when a solution exists, there is finite dimensional subspace of solutions (the dimension being the same as the dimension of the kernel
of A).
\end{itemize}

\hproblem D \& M 61 \solver{Lyman Gilispie} 

\hproblem D \& M 65 \solver{TJ Barry}

\end{aproblems}
\end{document}