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\begin{center}
\textbf{Mathematics 262} \\
\textbf{Homework 10 solutions}
\end{center}
\vspace{-0.05in}

Assignment: 
\begin{itemize}
\item Section 8.4: 4, 5, 6, 14
\end{itemize}

\textbf{\S 8.4, \#4:} 
If $U$ is a unitary operator on $V=\mathbf{R}^{2}$, with matrix
$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (with respect to the
standard basis), then since $U$ is unitary and we're working over
$\mathbf{R}$, then we have $U^{-1}=U^{t}$.  Multiplying $U$ by $U^{t}$
and setting the result equal to $I$ gives three equations:
\begin{align*}
a^{2} + b^{2} = 1, \\
ac+bd = 0, \\
c^{2} + d^{2} = 1.
\end{align*}
Because of the first equation, I can let $a=\cos \theta$ and $b=\sin
\theta$ for some angle $\theta$ between $0$ and $2\pi$.  Similarly, I
can let $c=\sin \varphi$ and $d = \cos \varphi$ for some angle
$\varphi$.  The second equation then says that
\[
\cos \theta \sin \varphi + \sin \theta \cos \varphi = 0.
\]
Using a trig identity, this becomes $\sin (\theta + \varphi) = 0$.
Therefore, $\varphi$ is either $2\pi-\theta$ or $\pi-\theta$.  If the
first of these holds, then the matrix for $U$ is $\begin{bmatrix} \cos
\theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.  If
the second holds, then the matrix is $\begin{bmatrix} \cos \theta &
\sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix}$.

(a) $U_{\theta} U_{\varphi}$ is rotation by $\varphi$ followed by
rotation by $\theta$, which is the same as $U_{\theta + \varphi}$,
rotation by the angle $\theta + \varphi$.  (One can also see this by
multiplying out the matrices and using trig identities, but it's not
as pleasant.)

(b) Well, $U^{*} = U^{t}$ since we're working over $\mathbf{R}$, and
$U^{t} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta &
\cos \theta \end{bmatrix}$.  Since $\sin (-\theta) = -\sin \theta$ and
$\cos (-\theta) = \cos \theta$, this matrix is equal to $U_{-\theta}$.
Alternatively, by part (a), $U_{\theta} U_{-\theta} = U_{0} = I$, so
$U_{-\theta} = U_{\theta}^{-1} = U_{\theta}^{*}$.  (Geometrically, if
you rotate by $\theta$ and then by $-\theta$, you have performed the
identity operation.)

(c) The change-of-basis matrix is $U_{\varphi}$, so I want to compute
$U_{\varphi}^{-1} U_{\theta} U_{\varphi}$.  I could compute this by
multiplying out the matrices.  I could also compute this using the
previous parts:
\[
U_{\varphi}^{-1} U_{\theta} U_{\varphi} = U_{-\varphi} U_{\theta}
U_{\varphi} = U_{-\varphi + \theta + \varphi} = U_{\theta}.
\]
(I could also try to argue that if a linear operator is a rotation, it
should treat the rotated basis $\{\alpha_{1}, \alpha_{2}\}$ just like
standard orthonormal basis: $\alpha_{1}$ should get sent to $(\cos
\theta) \alpha_{1} + (\sin \theta) \alpha_{2}$, etc.  So the matrix
for $U_{\theta}$ should be the same in this new basis.)

\textbf{\S 8.4, \#5:} 
Following the suggestion in the book, I let $\alpha = (1,1,1)$ and
$\beta = (1,1,-2)$, and I use Gram-Schmidt to turn this into an
orthonormal basis: $\alpha_{1} = (\frac{1}{\sqrt{3}},
\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ and $\alpha_{2} =
(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$.  I
find a vector orthogonal to both $\alpha$ and $\beta$ by using the
cross product, and I divide by its norm to get $\alpha_{3} =
(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$.

Now, $U$ sends $\alpha_{3}$ to itself, and it rotates the subspace
spanned by $\alpha_{1}$ and $\alpha_{2}$ by the angle $\theta$, so
with respect to the orthonormal basis $\{\alpha_{1}, \alpha_{2},
\alpha_{3}\}$, $U$ is represented by the matrix
\[
A = \begin{bmatrix}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & cos \theta & 0 \\
0 & 0 & 1
\end{bmatrix}.
\]
Let $P$ be the change-of-basis matrix:
\[
P = \begin{bmatrix}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0
\end{bmatrix}.
\]
This is supposed to be unitary (i.e., orthogonal, since we're working
over $\mathbf{R}$), and you should check this (e.g., check that the
columns are an orthonormal set, or that $P^{-1} = P^{t} = P^{*}$).
With respect to the standard basis, then, $U$ is represented by the
matrix
\[
PAP^{-1} = PAP^{t} = \begin{bmatrix}
\frac{1}{2} \cos \theta + \frac{1}{2} & \frac{1}{2} \cos \theta -
\frac{1}{2} &  \frac{1}{\sqrt{2}} \sin \theta \\
\frac{1}{2} \cos \theta - \frac{1}{2} & \frac{1}{2} \cos \theta +
\frac{1}{2} &  \frac{1}{\sqrt{2}} \sin \theta \\
-\frac{1}{\sqrt{2}} \sin \theta & -\frac{1}{\sqrt{2}} \sin \theta &
\cos \theta
\end{bmatrix}.
\]
(Since I'm changing from the basis $\{\alpha_{1}, \alpha_{2},
\alpha_{3} \}$ to the standard basis, I want $PAP^{-1}$.  If I were
changing from the standard basis to $\{\alpha_{1}, \alpha_{2},
\alpha_{3} \}$, I would want $P^{-1}AP$.)  You could check this by
seeing if $U\alpha_{3} = \alpha_{3}$, and seeing if $U\alpha_{i}$
looks reasonable for $i=1,2$.  (For instance, it's not so hard to see
that $U\alpha_{1}$ is a linear combination of $\alpha_{1}$ and
$\alpha_{2}$, so at least it's in the right general area.)

\textbf{\S 8.4, \#6:} 
Choose orthonormal bases for $W$ and $W^{\perp}$.  Together, they form
an orthonormal basis for $V$.  With respect to this basis, $U$ is
represented by the matrix $A = \begin{bmatrix} I & 0 \\ 0 &
-I\end{bmatrix}$.  (The $I$ in the upper left is the identity matrix
on $W$.  The $-I$ in the lower right is negative the identity matrix
on $W^{\perp}$.)

(a) From the matrix description, $A = A^{t} = A^{*}$, so $U$ is
self-adjoint.  Also, $A A^{*} = A A^{t} = A^{2} = I$, so $U$ is
unitary (and also orthogonal).

(b) I guess I should find orthonormal bases for $W$ and $W^{\perp}$
and do a change of basis: $W$ is spanned by $(\frac{1}{\sqrt{2}}, 0,
\frac{1}{\sqrt{2}})$.  $W^{\perp}$ is spanned by any two vectors
orthogonal to this one, say $(1, 0, -1)$ and $(0,1,0)$.  These happen
to be orthogonal (if they weren't, I'd use Gram-Schmidt), so an
orthonormal basis for $W^{\perp}$ is $\{(\frac{1}{\sqrt{2}}, 0,
-\frac{1}{\sqrt{2}}), (0, 1, 0)\}$.  Let $P$ be the change-of-basis
matrix
\[
P = \begin{bmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
0 & 0 & 1 \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0
\end{bmatrix}.
\]
This is a unitary matrix (because the columns are an orthonormal
basis), so $P^{-1} = P^{*} = P^{t}$.  With respect to this basis, $U$
is represented by the matrix 
\[
A = \begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{bmatrix}.
\]
So with respect to the standard basis, $U$ is represented by 
\[
PAP^{-1} = PAP^{t} = \begin{bmatrix}
0 & 0 & 1 \\
0 & -1 & 0 \\
1 & 0 & 0
\end{bmatrix}.
\]
(Again, since I'm changing from some other basis to the standard
basis, I look at $PAP^{-1}$, not $P^{-1}AP$.)  Does this answer make
sense?  Well, $\varepsilon_{2}$ is in $W^{\perp}$, so
$U\varepsilon_{2} = -\varepsilon_{2}$.  To compute $U\varepsilon_{1}$,
I write $\varepsilon_{1} = \frac{1}{2}(1, 0, 1) + \frac{1}{2}(1, 0,
-1)$; the first of these vectors is in $W$, the second in $W^{\perp}$.
Thus $U\varepsilon_{1} = \frac{1}{2}(1,0,1) - \frac{1}{2}(1,0,-1) =
(0,0,1) = \varepsilon_{3}$.  Similarly, $U\varepsilon_{3} =
\varepsilon_{1}$.  So this is indeed the correct matrix.

\pagebreak

\textbf{\S 8.4, \#14:} 

(a) Let's see.  If $T(0) = 0$, then 
\[
\norm{T\alpha} = \norm{T\alpha - 0} = \norm{T\alpha - T0} =
\norm{\alpha - 0} = \norm{\alpha},
\]
so if I knew that $T$ were linear, I could conclude that $T$ is
unitary.  So I have to prove linearity.

First, I'll show that $T$ preserves inner products.  Since
$\inner{\alpha}{\beta} = \frac{1}{4} (\norm{\alpha + \beta}^{2} -
\norm{\alpha -\beta}^{2})$, and since $\norm{T\alpha - T\beta} =
\norm{\alpha -\beta}$, it's good enough to show that $\norm{T\alpha +
T\beta} = \norm{\alpha + \beta}$.  Well,
\[
\norm{T\alpha + T\beta} = \norm{T\alpha - (- T\beta)}.
\]
If I knew that $-T\beta = T(-\beta)$, then this would equal 
\[
\norm{T\alpha - T(-\beta)} = \norm{\alpha - (-\beta)} = \norm{\alpha +
\beta}.
\]
So it's good enough to show that $T(-\beta) = -T\beta$ for each vector
$\beta$.  Find a vector $\gamma$ perpendicular to $\beta$.  Then
$\norm{T\beta} = \norm{\beta}$, $\norm{T\gamma} = \norm{\gamma}$, and
$\norm{T\gamma - T\beta} = \norm{\gamma - \beta}$.  Since the vectors
$\beta$, $\gamma$, and $\gamma - \beta$ form a right triangle, and
since the vectors $T\beta$, $T\gamma$, and $T\gamma - T\beta$ have the
same lengths, then they must also form a right triangle.  In
particular, $T\beta$ is perpendicular to $T\gamma$.  Similarly,
$T(-\beta)$ is perpendicular to $T\gamma$.  Two vectors in
$\mathbf{R}^{2}$ which are perpendicular to the same vector must be
parallel, so $T\beta$ and $T(-\beta)$ are parallel.  They also have
the same length, so $T(-\beta) = \pm T\beta$.  Lastly, $\norm{T\beta -
T(-\beta)} = \norm{\beta - (-\beta)} = 2 \norm{\beta}$; therefore
$T(-\beta)$ must be $-T\beta$.  Therefore $T$ preserves inner
products.

Let $\varepsilon_{1} = (1,0)$ and $\varepsilon_{2} = (0,1)$.  Since
$T$ preserves norms and inner products, then $T\varepsilon_{1}$ and
$T\varepsilon_{2}$ are an orthonormal basis for $\mathbf{R}^{2}$.  I
claim that $T$ is linear.  Suppose $\alpha = (x_{1},x_{2}) =
x_{1}\varepsilon_{1} + x_{2}\varepsilon_{2}$.  Then
$\inner{T\alpha}{T\varepsilon_{i}} = \inner{\alpha}{\varepsilon_{i}} =
x_{i}$, so $T\alpha = x_{1} T\varepsilon_{1} + x_{2}
T\varepsilon_{2}$.  In other words,
\[
T(x_{1}\varepsilon_{1} + x_{2}\varepsilon_{2}) = x_{1}
T\varepsilon_{1} + x_{2}T\varepsilon_{2}.
\]
This implies that $T$ is linear.  If you want to be explicit: if
$\beta = y_{1}\varepsilon_{1} + y_{2} \varepsilon_{2}$, then
\begin{align*}
T(c\alpha + \beta) &= T((cx_{1} + y_{1}) \varepsilon_{1} + (cx_{2} +
y_{2}) \varepsilon_{2}) \\
 &= (cx_{1} + y_{1}) T\varepsilon_{1} + (cx_{2} + y_{2})
T\varepsilon_{2} \\
 &= (cx_{1} T\varepsilon_{1} + cx_{2} T\varepsilon_{2}) + (y_{1}
T\varepsilon_{1} + y_{2} T\varepsilon_{2}) \\
 &= cT\alpha + T\beta.
\end{align*}

(b) Given a rigid motion $f$ of $\mathbf{R}^{2}$, find a point
$\gamma$ so that $f(\gamma) = 0$ (justification below), and define a
function $U: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}$ by $U(\alpha)
= f(\alpha + \gamma)$.  Then $U(0) = f(\gamma)=0$ and $U$ is a rigid
motion:
\[
\norm{U\alpha - U\beta} = \norm{f(\alpha + \gamma) - f(\beta +
\gamma)} = \norm{(\alpha + \gamma) - (\beta + \gamma)} = \norm{\alpha
-\beta}.
\]
By part (a), then, $U$ is a unitary (linear) operator.  Furthermore,
for all vectors $\alpha$, $f(\alpha + \gamma)= U(\alpha)$.  Letting
$\beta = \alpha - \gamma$, we get 
\[
f(\beta) = U(\beta - \gamma) = U (T_{-\gamma}(\beta)) = U \circ
T_{-\gamma} (\beta),
\]
where $T_{-\gamma}$ is translation by $-\gamma$: $T_{-\gamma}(\beta) =
\beta - \gamma$.

I still owe you the justification that there is a point $\gamma$ so
that $f(\gamma) = 0$.  The idea is that if $f$ is a rigid motion, then
it preserves all distances, angles, and other geometric information.
So if $f$ takes $0$ to some point $\alpha$, then $f$ takes the circle
of radius $\norm{\alpha}$ centered at $0$ to the circle of radius
$\norm{\alpha}$ centered at $\alpha$.  In particular, since $0$ is on
this second circle, then $f$ takes some point on the first circle to
$0$.  One can do this more carefully (just from the definition of
rigid motion, without hand-waving about ``preserving other geometric
information''), but I don't feel like writing down the details.

(c) This follows from problem 4.  Every unitary operator on
$\mathbf{R}^{2}$ is either a rotation or a reflection followed by a
rotation, and every rigid motion is a translation followed by one of
these.


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