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\begin{center}
\textbf{Mathematics 262} \\
\textbf{Homework 9 solutions}
\end{center}
\vspace{-0.05in}

Assignment: 
\begin{itemize}
\item Section 8.3: 2, 5, 9, 12
\end{itemize}

\textbf{\S 8.3, \#2:} 
Since I know what $T$ does to the standard basis vectors, I can
write down its matrix: $A=\begin{bmatrix} 1+i & i \\ 2 & i
\end{bmatrix}$.  Since the standard basis is orthonormal, then the
matrix for $T^{*}$ is the conjugate transpose of this:
$A^{*}=\begin{bmatrix} 1-i & 2 \\ -i & -i \end{bmatrix}$.
To check whether $T$ commutes with $T^{*}$, just compute $AA^{*}$ and
$A^{*}A$, and see if they're the same.  I think the results are:
\[
AA^{*} = \begin{bmatrix} 3 & 3+2i \\ 3-2i & 5 \end{bmatrix},
\hspace{0.5in}
A^{*}A = \begin{bmatrix} 6 & 1+3i \\ 1-3i & 2 \end{bmatrix}.
\]
So they don't commute.  (If you'd rather work with linear operators,
you can evaluate $TT^{*}(\varepsilon_{j})$ and
$T^{*}T(\varepsilon_{j})$ for $j=1,2$, and see if you get the same
thing.)

\textbf{\S 8.3, \#5:}
The easiest thing to do is to use the properties of the adjoint: if
$T$ is invertible, then $I = TT^{-1}$.  Take the adjoint of both
sides, and use the fact that $I^{*} = I$:
\[
I =I^{*} = (TT^{-1})^{*} = (T^{-1})^{*} T^{*}.
\]
Similarly, the equation $I = T^{-1}T$ yields $I=T^{*} (T^{-1})^{*}$.
Therefore the operator $T^{*}$ is invertible, because I have found its
inverse: $(T^{-1})^{*}$.  (A linear transformation $U$ is invertible
if and only if there is another linear transformation $S$ so that $SU
= I$ and $US=I$.)

\textbf{\S 8.3, \#9:}
I know that if I can find the matrix for $D$ with respect to an
\emph{orthonormal} basis, then it's easy to get the matrix for
$D^{*}$.  Fortunately, as part of a previous assignment, I know such
a basis.  Actually, the previous assignment gives an orthogonal basis,
so I find the norm of each basis vector and divide by it to get the
following orthonormal basis: 
%\[
%(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}) = \left(1,
%2 \sqrt{3}(x-\frac{1}{2}), 6 \sqrt{5}(x^{2}-x+\frac{1}{6}),
%20 \sqrt{7}(x^{3}-\frac{3}{2}x^{2} + \frac{3}{5}x - \frac{1}{20})
%\right).
%\]
\begin{align*}
(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}) 
&= \left(1,
2 \sqrt{3}(x-\frac{1}{2}), 6 \sqrt{5}(x^{2}-x+\frac{1}{6}),
20 \sqrt{7}(x^{3}-\frac{3}{2}x^{2} + \frac{3}{5}x - \frac{1}{20})
\right) \\
&= \left(1,
\sqrt{3}(2x-1), \sqrt{5}(6x^{2}-6x+1), \sqrt{7}(20x^{3}-30x^{2} + 12x
- 1)\right).
\end{align*}
To compute the matrix for $D$, I apply $D$ to each basis element and
expand the result in terms of this basis; in other words, as we saw in
class, the $(i,j)$-entry of the matrix for $D$ is $(D\alpha_{j} |
\alpha_{i}) = \int_{0}^{1} \frac{d\alpha_{j}}{dt} \alpha_{i} dt$.
\begin{align*}
D(\alpha_{1}) & = D(1) = 0.\\
D(\alpha_{2}) & = 2 \sqrt{3} D(x-\frac{1}{2}) = 2 \sqrt{3} = 2 \sqrt{3}
\alpha_{1}. \\
D(\alpha_{3}) & = 6 \sqrt{5} D(x^{2}-x+\frac{1}{6}) = 6 \sqrt{5}(2x-1) =
2 \sqrt{15} (\alpha_{2}). \\
D(\alpha_{4}) & = 20 \sqrt{7}
D(x^{3}-\frac{3}{2}x^{2}+\frac{3}{5}x-\frac{1}{20}) \\
 &= 20 \sqrt{7} (3x^{2} - 3x + \frac{3}{5}) \\
 & = 20 \sqrt{7} \left( 3 (x^{2}-x+\frac{1}{6}) - \frac{1}{2} +
\frac{3}{5}\right) \\
 & = 20 \sqrt{7}(3\alpha_{3} + \frac{1}{10} \alpha_{1}) \\
 & = 2 \sqrt{35} \alpha_{3} + 2\sqrt{7} \alpha_{1}.
\end{align*}
So the matrix for $D$ is 
\[
\begin{bmatrix}
0 & 2\sqrt{3} & 0 & 2\sqrt{7} \\
0 & 0 & 2 \sqrt{15} & 0 \\
0 & 0 & 0 & 2 \sqrt{35} \\
0 & 0 & 0 & 0
\end{bmatrix}.
\]
Therefore the matrix for $D^{*}$ is the conjugate transpose of this,
or actually just the transpose, since we are working with real numbers:
\[
\begin{bmatrix}
0 & 0 & 0 & 0 \\
2\sqrt{3} & 0 & 0 & 0 \\
0 & 2 \sqrt{15} & 0 & 0 \\
2\sqrt{7} & 0 & 2\sqrt{35} & 0
\end{bmatrix}.
\]
For example, 
\[
D^{*}(1) =  D^{*}(\alpha_{1}) = 2\sqrt{3} \alpha_{2} +
2\sqrt{7}\alpha_{4} = 280x^{3} - 420x^{2} + 180x - 20.
\]

\textbf{\S 8.3, \#12:}
If $T$ is self-adjoint, then $\inner{T\alpha}{\beta} =
\inner{\alpha}{T\beta}$ for all $\alpha, \beta$.  In particular,
$\inner{T\alpha}{\alpha} = \inner{\alpha}{T\alpha}$.  On the other
hand, by switching factors, $\inner{\alpha}{T\alpha} =
\overline{\inner{T\alpha}{\alpha}}$.  We conclude that
$\inner{T\alpha}{\alpha} = \overline{\inner{T\alpha}{\alpha}}$ for all
$\alpha$, and therefore $\inner{T\alpha}{\alpha}$ is real for all
$\alpha$.

Suppose, conversely, that $\inner{T\alpha}{\alpha}$ is real for all
$\alpha$.  I want to show that for all $\alpha$ and $\beta$,
\[
\inner{T\alpha}{\beta} = \inner{\alpha}{T\beta}.
\]
I'll do this in sort of a roundabout way.  Let $z$ be a complex
number, and consider
\[
\inner{T(\alpha + z\beta)}{\alpha + z\beta} = \inner{T\alpha}{\alpha}
+ \overline{z} \inner{T\alpha}{\beta} + z \inner{T\beta}{\alpha} +
|z|^{2} \inner{T\beta}{\beta}.
\]
This is a real number (by assumption, using $\alpha+z\beta$ instead of
$\alpha$), as are $\inner{T\alpha}{\alpha}$ and $|z|^{2}
\inner{T\beta}{\beta}$.  (Recall that $|z|^{2} = z \overline{z}$ is
always a real number.)  Therefore $\overline{z} \inner{T\alpha}{\beta}
+ z \inner{T\beta}{\alpha}$ is real for every complex number $z$.  I
claim that this implies that $\inner{T\alpha}{\beta} =
\overline{\inner{T\beta}{\alpha}}$.  If I can verify the claim, then
\[
\inner{T\alpha}{\beta} = \overline{\inner{T\beta}{\alpha}} =
\inner{\alpha}{T\beta},
\]
as desired.

I have two complex numbers, $r=\inner{T\alpha}{\beta}$ and
$s=\inner{T\beta}{\alpha}$, and I know that for every complex number
$z$, $\overline{z}r + zs$ is a real number.  I want to show that
$r=\overline{s}$.  (The idea is that $\overline{z} \overline{s} + zs$
is real, and $r=\overline{s}$ should be the only solution that works
for every $z$.)  This is pretty easy: let $r=a+bi$, and let $s=c+di$.
I want to show that $a=c$ and $b=-d$.  When $z=1$, I have
\[
\overline{z}r + zs = r + s = (a+c) + i(b+d).
\]
Since this is a real number, then $b=-d$.  When $z=i$,
\[
\overline{z}r + zs = -ir + is = -ia+b +ic -d = (b-d) + i(c-a).
\]
Since this is a real number, then $a=c$.  This finishes the claim, and
hence the problem.

(Alternatively, if I know that $\overline{z}r + zs$ is real for every
complex number $z$, then it is equal to its complex conjugate.  For
$z=1$ and $z=i$, this gives:
\begin{gather*}
r + s = \overline{r} + \overline{s}, \\
-ir + is = i\overline{r} - i \overline{s}.
\end{gather*}
Multiply the second equation by $i$:
\[
r - s = -\overline{r} + \overline{s}.
\]
Now add it to the first equation:
\[
2r = 2\overline{s}.
\]
Dividing by 2 gives the desired result.)

\end{document}



Stopping at this point gives a good enough answer.  For more
information, though, it would be nice to have a formula for what
$D^{*}$ does to an arbitrary polynomial in $V$.  Given a polynomial
$f(x) = a_{3} x^{3} + a_{2}x^{2} + a_{1}x + a_{0}$, to compute
$D^{*}(f)$, we need to expand $f$ with respect to the above basis.
Since $f$ has already been written in terms of the ``standard basis''
$(1, x, x^{2}, x^{3})$, then we are really asking to change bases
between the standard one and the orthonormal one.  One way to do this
is to expand the standard basis vectors in terms of the orthonormal
ones:
\begin{gather*}
1 = \alpha_{1}. \\
x = \frac{\sqrt{3}}{6} \alpha_{2} + \frac{1}{2} \alpha_{1}.\\
x^{2} = \frac{\sqrt{5}}{30} \alpha_{3} + \frac{\sqrt{3}}{6} \alpha_{2}
+ \frac{1}{3} \alpha_{1}.\\
x^{3} = \frac{\sqrt{7}}{140} \alpha_{4} + \frac{\sqrt{5}}{20}
\alpha_{3} + \frac{3\sqrt{3}}{20} \alpha_{2} + \frac{1}{4} \alpha_{1}.
\end{gather*}
So with respect to this basis, 
\begin{gather*}
D^{*}(1) = D^{*}(\alpha_{1}) = 2\sqrt{3} \alpha_{2} +
2\sqrt{7}\alpha_{4} = \dots \\
D^{*}(x) = D^{*}(\frac{\sqrt{3}}{6} \alpha_{2} +
\frac{1}{2}\alpha_{1}) = \dots \\
D^{*}(x^{2}) = \dots \\
D^{*}(x^{3}) = \dots \\
\end{gather*}

