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\begin{center}
\textbf{Mathematics 262} \\
\textbf{Homework 8 solutions}
\end{center}
\vspace{-0.05in}

Assignment: 
\begin{itemize}
\item Section 8.1: 1, 3
\item Section 8.2: 2, 3, 6, 9, 13, 17
\end{itemize}

\textbf{\S 8.1, \#1:}
(a) We have 
\[
\inner{\vec{0}}{\beta} = \inner{\vec{0} - \vec{0}}{\beta} 
= \inner{\vec{0}}{\beta} - \inner{\vec{0}}{\beta} = 0
\]

(b) If $\inner{\alpha}{\beta} = 0$ for all $\beta \in V$, then in
particular, $\inner{\alpha}{\alpha} = 0$.  Therefore $\alpha$ must be
$\vec{0}$.

\textbf{\S 8.1, \#3:}
I claim that any inner product on $\mathbf{R}$ (or on $\mathbf{C}$) is
determined by the value of $\inner{1}{1}$.  After all, the inner
product of two arbitrary numbers $x$ and $y$ must be $\inner{x}{y} = x
\overline{y} \inner{1}{1}$ by the linearity properties of the inner
product.  Because $\inner{x}{x}$ must be positive whenever $x$ is
nonzero, then $\inner{1}{1}$ must be a positive real number.  Hence
every inner product on $\mathbf{R}$ (or on $\mathbf{C}$) is of the
form
\[
\inner{x}{y} = x \overline{y} c
\]
for some positive real number $c$.  Conversely, every positive $c$
determines an inner product via this formula.

\textbf{\S 8.2, \#2:}
Gram-Schmidt gives $\{(1,0,1), (1,0,-1), (0,3,0) \}$, so normalizing
(dividing each vector by its norm) gives the following orthonormal
basis: $\{\frac{1}{\sqrt{2}} (1,0,1), \frac{1}{\sqrt{2}} (1,0,-1),
(0,1,0) \}$.

\textbf{\S 8.2, \#3:}
Gram-Schmidt gives $\{(1,0,i), \frac{1}{2} (1+i,2,1-i) \}$, so
normalizing gives the following orthonormal basis:
$\{\frac{1}{\sqrt{2}} (1,0,i), \frac{1}{2\sqrt{2}} (1+i,2,1-i)\}$.

\textbf{\S 8.2, \#6:}
First, I'll find an orthonormal basis for $W$.  Since $W$ is
one-dimensional, this just corresponds to normalizing the single basis
vector: $\alpha = \frac{1}{5}(3,4)$.

(a) The formula for orthogonal projection onto $W$ is 
\begin{align*}
E(x_{1}, x_{2}) &= \left( (x_{1},x_{2}) \cdot \alpha \right) \alpha \\
 &= \frac{1}{5} (3x_{1} + 4x_{2}) \alpha \\
 &= \frac{1}{25} (9x_{1} + 12x_{2}, 12x_{1} + 16x_{2}).
\end{align*}

(b) So the matrix for $E$ is $\frac{1}{25} \begin{bmatrix} 9 & 12 \\
12 & 16 \end{bmatrix}$.

(c) $W^{\perp}$ is one-dimensional, so I just have to find one vector
orthogonal to $(3,4)$.  This is easy: $(4,-3)$ is an example, so
$W^{\perp}$ is the subspace spanned by $(4,-3)$.

(d) Let $\beta = \frac{1}{5}(4,-3)$.  Then with respect to the
orthonormal basis $(\alpha, \beta)$, the matrix for $E$ is as desired.

\textbf{\S 8.2, \#9:}
(a) Suppose that $g(t)=a_{0} + a_{1}t + a_{2}t^{2} + a_{3}t^{3}$ is in
the orthogonal complement of the subspace of scalar polynomials.  Then 
\[
\int_{0}^{1} cg(t) dt = c \int_{0}^{1} g(t) dt = 0
\]
for every scalar $c$.  In particular, $\int_{0}^{1} g(t) dt = 0$.  We
can evaluate this integral: it is $a_{0} + \frac{a_{1}}{2} +
\frac{a_{2}}{3} + \frac{a_{3}}{4}$.  So the orthogonal complement is
the subspace of all polynomials $g(t)=a_{0} + a_{1}t + a_{2}t^{2} +
a_{3}t^{3}$ satisfying
\[
a_{0} + \frac{a_{1}}{2} + \frac{a_{2}}{3} + \frac{a_{3}}{4} = 0.
\]

(b) Start with $\alpha_{0} = 1$.  This has norm 1.  Then let
\begin{align*}
\alpha_{1} &= x - \inner{x}{\alpha_{0}} \alpha_{0} \\
   &= x - (\int_{0}^{1} t dt) 1 \\
   &= x - \frac{1}{2}.
\end{align*} 
This has norm $(\int_{0}^{1}(t-\frac{1}{2})^{2}dt)^{\frac{1}{2}} =
\frac{1}{\sqrt{12}}$.  Then let
\begin{align*}
\alpha_{2} &= x^{2} -
\frac{\inner{x^{2}}{\alpha_{1}}}{\norm{\alpha_{1}}^{2}} \alpha_{1} -
\inner{x^{2}}{\alpha_{0}} \alpha_{0} \\
   &= x^{2} - x + \frac{1}{6}.
\end{align*}
This has norm $\frac{1}{\sqrt{180}}$.  Finally,
\[
\alpha_{3} = x^{3} - \frac{3}{2} x^{2} + \frac{3}{5} x - \frac{1}{20}.
\]

\textbf{\S 8.2, \#13:}
It suffices to show that each element $\alpha$ in $S$ is in
$(S^{\perp})^{\perp}$.  By definition of $S^{\perp}$, $\alpha$ is
orthogonal to every element in $S^{\perp}$, and therefore $\alpha$ is
in $(S^{\perp})^{\perp}$.

Suppose that $V$ is finite-dimensional, and let $W$ be the subspace
spanned by $S$.  Then by one of our theorems, $V = W \oplus
W^{\perp}$, and $W^{\perp}$ is equal to $S^{\perp}$, so $V=W \oplus
S^{\perp}$.  If we apply the theorem to $S^{\perp}$, then we find that
$V=S^{\perp} \oplus (S^{\perp})^{\perp}$.  Counting dimensions, we
find that $\dim W = \dim (S^{\perp})^{\perp}$; since $W \subseteq
(S^{\perp})^{\perp}$, then these two subspaces must be equal.

\textbf{\S 8.2, \#17:}
I would guess that the even functions (functions $g(t)$ satisfying
$g(t)=g(-t)$ for all $t$) would be the orthogonal complement.  Let's
see what happens.

If $f(t)$ is odd and $g(t)$ is even, then $f(t)g(t)$ is odd, so
\[
\inner{f}{g} = \int_{-1}^{1} f(t) g(t) dt = 0.
\]
Therefore the subspace of even functions is contained in $W^{\perp}$.

Let $h(t)$ be any function.  Then the formula
\[
h(t)= \frac{1}{2}(h(t)+h(-t)) + \frac{1}{2}(h(t)-h(-t))
\]
displays $h(t)$ as being a sum of an even function and an odd
function.  For shorthand, write $h(t)=g(t) + f(t)$, where $g(t)$ is
even and $f(t)$ is odd.  If $h(t) \in W^{\perp}$, then $\inner{h}{f} =
0$.  But $\inner{h}{f} = \inner{g}{f} + \inner{f}{f}$.
$\inner{g}{f}=0$ because $g$ is even and $f$ is odd.  So if $h\in
W^{\perp}$, with $h=g+f$, then $\inner{f}{f}=0$, hence $f(t)=0$, and
hence $h=g$.  In other words, if $h(t)$ is in $W^{\perp}$, then $h(t)$
is even.  Therefore, $W^{\perp}$ is contained in the subspace of even
functions.

Therefore $W^{\perp}$ is equal to the subspace of even functions.

(Alternatively, the formula $h=g+f$ shows that $V$ is the sum of the
subspaces of even and odd functions.  The only function that is both
even and odd is the zero function, and therefore this is actually a
direct sum: $V = W \oplus W'$, where $W'$ is the subspace of even
functions.  We showed that $W^{\perp} \supseteq W'$; this direct sum
decomposition shows that $W^{\perp} \subseteq W'$.  Therefore
$W^{\perp} = W'$.)

\end{document}
