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\textbf{Mathematics 262} \\
\textbf{Solutions to practice problems}
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Assignment: 
\begin{itemize}
\item Section 8.4: 12 
\item Section 8.5: 1, 3, 4, 10, 12
\end{itemize}

\textbf{\S 8.4, \#12:} 
It's probably worthwhile thinking about how to do this problem without
the material from Section 8.5.  Using that material, though, makes
things easier.  The problem asks you to show that if $V$ is a
finite-dimensional inner product space, then for every linear operator
on $T$ which is both self-adjoint and unitary, there is a subspace $W$
of $V$ so that $T$ acts as the identity $I$ on $W$ and $T$ acts as
$-I$ on the orthogonal complement $W^{\perp}$.  In other words, I want
to find a basis for $V$ with respect to which the matrix for $T$ is of
the form 
\[
A = 
\begin{bmatrix}
I & 0 \\
0 & -I \end{bmatrix}
= 
\begin{bmatrix}
1 & 0 & \dots & 0 & 0 & \dots & 0 \\
0 & 1 & \dots & 0 & 0 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots & & \vdots \\
0 & 0 & \dots & 1 & 0 & \dots & 0 \\
0 & 0 & \dots & 0 & -1 & \dots & 0 \\
\vdots & \vdots & & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & 0 & 0 & \dots & -1
\end{bmatrix}.
\]
If $T$ is both self-adjoint and unitary, then the main theorems of
Section 8.5 apply: there is a basis for $V$ consisting of eigenvectors
for $T$.  What are the eigenvalues for $V$?  Since $T$ is
self-adjoint, they're real.  Since $T$ is unitary, they have norm
$1$.  Hence $1$ and $-1$ are the only eigenvalues.  So I take my basis
of eigenvectors and order it by putting the eigenvectors corresponding
to $1$ first, and then the eigenvectors corresponding to $-1$.  This
does the trick.

\textbf{\S 8.5, \#1:} 
All you have to do is find an orthonormal set of eigenvectors and make
them the columns of the matrix $P$.  (It wouldn't hurt to check that
$P^{t}AP$ is actually diagonal, too.)  Here are the matrices $P$ for
the three matrices in the problem:
\[
\begin{bmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{bmatrix}, 
\begin{bmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{bmatrix}, 
\begin{bmatrix}
\cos \frac{\theta}{2} & -\sin \frac{\theta}{2} \\
\sin \frac{\theta}{2} & \cos \frac{\theta}{2}
\end{bmatrix}.
\]
For the third one, you might be able to recognize the matrix $A$ as a
reflection, and in particular, reflection across the line through the
origin at angle $\frac{\theta}{2}$ from the horizontal.  So any vector
parallel to this line is an eigenvector with eigenvalue $1$; any
vector perpendicular to this is an eigenvector with eigenvalue $-1$.
If you don't see this, you have to use a few trig identities to get
the right answer.

\textbf{\S 8.5, \#3:} 
The entries of $D$ are the eigenvalues of $A$.  The characteristic
polynomial is $x^{3} - 9x^{2} - 6x$.  This factors as $x
(x^{2}-9x-6)$, and so has roots $0$, $\frac{9 \pm \sqrt{105}}{2}$.
There are several choices for $D$, depending on the order in which we
put the eigenvalues; one choice would be
\[
D = \begin{bmatrix}
0 & 0 & 0 \\
0 & \frac{9 + \sqrt{105}}{2} & 0 \\
0 & 0 & \frac{9 - \sqrt{105}}{2}
\end{bmatrix}.
\]

\textbf{\S 8.5, \#4:} 
To show that $T$ is normal, you have to check that $AA^{*} = A^{*}A$.
Then you find the eigenvalues for $A$, which are $1+i$ and $1-i$.
Then you find corresponding eigenvectors, which are $(1,1)$ and
$(-1,1)$.  Then you normalize to get an orthonormal basis:
\[
(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), 
(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}).
\]

\textbf{\S 8.5, \#10:} 
We did this in class using the spectral theorem.  $A$ is a
\emph{positive} matrix if $A$ is Hermitian and $X^{*}AX > 0$ for every
nonzero column vector $X$.  Since $A$ is Hermitian, then $V$ has an
orthonormal basis consisting of eigenvectors for $A$; with respect to
that basis, $A$ becomes a diagonal matrix $D$.  Since $X^{*}AX$ is
always positive, then $\varepsilon_{i}^{*}D \varepsilon_{i}$ is
positive for each $i$, where $\varepsilon_{i}$ is the $i$th standard
basis vector.  On the other hand $\varepsilon_{i}^{*}D
\varepsilon_{i}$ is the $i$th entry on the diagonal of $D$, so we can
conclude that all of the entries of $D$ are positive.  Therefore $D$
has a square root.  Changing back to the original basis gives a square
root for $A$.  (I've left out lots of details here, and you might want
to fill them in.)

\textbf{\S 8.5, \#12:} 
Let $T$ be normal.  If I view $T$ as operating on a complex vector
space $V$, then by the main theorem (Theorem 22 in the book), there is
an orthonormal basis of $V$ consisting of eigenvectors of $T$.  Note
the word ``orthonormal.''

(This isn't quite enough.  I should really start with a normal
operator $T$ acting on any vector space $V$.  If $V$ happens to be
complex, the previous paragraph solves the problem.  Otherwise, if $V$
is real and if $\alpha$ and $\beta$ are two eigenvectors corresponding
to distinct eigenvalues $c$ and $d$, then I have to extend the action
of $T$ on $V$ to an action of $T$ on a complex vector space $W$, note
that $\alpha$ and $\beta$ are still eigenvectors when viewed as
elements of $W$, and then apply the argument of the previous
paragraph.)

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