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\begin{document}
\begin{center}
{\bf Homework 3}\\
(due Wednesday, February 8)
\end{center}
\vspace*{.3in}

\noindent In computational problems, you're free to use a calculator or computer to compute eigenvalues and eigenvectors, and handle matrix multiplications/inversions.  Just make sure you present the results in sufficient detail for Josh to know what methods you're using.  

In 3.4 and 3.5 where you're asked to find singular value decompositions of a matrix $A$, give your answer in the form $A = W\Sigma V^*$ where $\Sigma$ is a diagonal matrix whose non-zero entries are the non-zero singular values of $A$.  Note that my matrix version of SVD differs slightly from the book's.  The matrices I use are smaller than those used in Treil's book since I only use data corresponding to non-zero singular values (i.e. I throw out eigenvectors corresponding to zero singular values, and I don't complete the vectors $\vw_1,\dots,\vw_k$ in the range to get an orthonormal basis for the codomain.  You can use the book's format or my own, but mine is easier.

\vspace*{.2in}
\noindent{\bf From Treil}
\begin{description}
 \item[Section 6.3, Pages 171-172] 3.5, 3.6, 3.9, 3.10
\item[Section 6.4, page 178] 4.3
\end{description}

\begin{prob}
Find polar decompositions of the matrices 
$
\left[\begin{matrix}
1 & 1 \\ 2 & -2
\end{matrix}\right]
$
and
$
\left[\begin{matrix}
20 & 4 & 0 \\ 0 & 0 & 1 \\ 4 & 20 & 0
\end{matrix}\right].
$
Say whether and why in each case your answer is the only one possible, i.e. whether the polar decomposition is unique.  For the $2\times 2$ matrix, use your answer (or the computations involved in reaching it) to illustrate what the associated linear transformation $T:\R^2\to\R^2$ does to the unit circle.
\end{prob}

\begin{prob}
In class I stated and mostly proved the following assertion: \emph{if $T:V\to V$ is a non-negative, self-adjoint operator on a finite dimensional inner product space, then $T$ has a unique non-negative self-adjoint square root $S:V \to V$.}  The uniqueness assertion turned on the fact that if $S:V\to V$ is any non-negative self-adjoint square root of $T$, then each eigenvector $\vv\in V$ for $T$ is also an eigenvector for $S$.  I proved this fact in the case when the corresponding eigenvalue is non-zero.  So you prove it when the eigenvalue is zero.
\end{prob}

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