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\title{Math 23a 2nd Midterm Solutions}
\year1999\month11\day19

\begin{document}
\maketitle

\begin{enumerate}
\item (George)   % 1
a)  $A \subset \R^n$ is closed if and only if for all sequences
$\{x_i\} \subset A$ such that $\{x_i\}$ converges to $x$ in $\R^n$,
$\{x_i\}$ converges in $A$ (that is, $x \in A$).

b)  Suppose $A$ is closed according to the above definition.  Consider $y
\in \R^n$ such that there is no $\eps$ such that $B_{\eps}(y) \subset \R^n
\setminus A$.  Thus for any $i$ there is an $x_i \in A$ with $|x_i - y| <
1/i$.  It is clear that $\{x_i\}$ converges to $y$ in $\R^n$ so by the
definition above $y$ is in $A$.  Therefore for every $z$ in $\R^n
\setminus A$ there is an $\eps$ such that $B_{\eps}(z) \subset \R^n
\setminus A$ so $\R^n \setminus A$ is open.

Suppose $\R^n \setminus A$ is open and $\{x_i\}$  converges to $x$
in $\R^n$ and has terms in $A$.  Given $\eps > 0$ there is $N$ such that
$x_i \in B_{\eps}(x)$ for all $i > N$.  In particular, there is at least
one
element of $A$ in $B_{\eps}(x)$ so no neighborhood of $x$ lies in $\R^n
\setminus A$ and since $\R^n \setminus A$ is open that means that $x \in
A$.

c)  Suppose $f$ is continuous and $A$ is a closed set in $\R$.  Consider a
sequence $\{x_i\}$ that converges to $x$ in $\R^n$ such that each $x_i$ is
in $f^{-1}(A)$.  Since $f$ is continuous we know that $\lim(f(x_i)) =
f(\lim(x_i))$ so $f(x) = \lim(f(x_i))$.  It is clear that the sequence
$\{f(x_i)\}$ is in $A$ and since $A$ is closed so is $\lim(f(x_i)) =
f(x)$.  Therefore $x \in f^{-1}(A)$ so $f^{-1}(A)$ is closed.

Now suppose $f^{-1}(A)$ is open for every open set $A$.  Given $x \in
\R^n$
and $\eps > 0$ consider $f^{-1}(B_{\eps}(f(x))) \subset \R^n$.  Since
$(B_{\eps}(f(x))$ is open in $R$ $f^{-1}(B_{\eps}(f(x)))$ is open in
$\R^n$
so there is a $\delta > 0$ such that $$B_{\delta}(x) \subset
f^{-1}(B_{\eps}(f(x)))$$
This means that $|x-y| < \delta$ (i.e. $y \in
B_{\delta}(x)$) implies that $|f(x) - f(y)| < \eps$ (i.e. $f(y) \in
B_{\eps}(f(x))$).  This is exactly the desired result.

\item (Matt)     % 2
Define $f\colon {\mathbb R}^2 \to {\mathbb R}^3$ by
$f\left(\begin{array}{ccc}x\\y\end{array}\right)=
\left(\begin{array}{ccc}\frac{x+y}{2}\\\frac{x-y}{2}\\{xy}
\end{array}\right)$.
Define $g\colon {\mathbb R} \to {\mathbb R}^2$ by
$g(t)=\left(\begin{array}{ccc}{e^t}\\{e^{-t}}\end{array}\right)$.
Find D$(f \circ g)(t)$.

1) using the Chain Rule

$f$ and $g$ have continuous partial derivatives, so their derivatives
are just their Jacobian matrices:

${\D}f\left(\begin{array}{ccc}x\\y\end{array}\right)=
\left(\begin{array}{ccc}{\D}_1 f_1 \ {\D}_2 f_1 \\
                        {\D}_1 f_2 \ {\D}_2 f_2 \\
                        {\D}_1 f_3 \ {\D}_2 f_3 \\
\end{array}\right)
=\left(\begin{array}{ccc}\frac12 \ \frac12 \\
                         \frac12 \ {-\frac12} \\
                         y \ x \\
\end{array}\right)$

${\D}g(t)=
\left(\begin{array}{ccc}{\D}_1 g \\
{\D}_2 g \\
\end{array}\right)
=\left(\begin{array}{ccc}e^t\ \\
-e^{-t} \\
\end{array}\right)$

So we have
$[\D(f \circ g)(t)]=[{\D}f(g(t))] \circ [{\D}g(t)]=
\left(\begin{array}{ccc}\frac12 \ \frac12 \\
                         \frac12 \ {-\frac12} \\
                         e^{-t} \ e^t\\
\end{array}\right)
\left(\begin{array}{ccc}e^t\ \\
-e^{-t} \\
\end{array}\right)=
{\D}(f \circ g)(t)=\left(\begin{array}{ccc}
\frac{e^t-e^{-t}}2
\\\frac{e^t+e^{-t}}2
\\0\end{array}\right)$.

2) by directly differentiating $(f \circ g)$.
First we compute $(f \circ g)$:

$(f \circ g)(t)=\left(\begin{array}{ccc} \frac{e^t+e^{-t}}2
\\\frac{e^t-e^{-t}}2
\\1\end{array}\right)$.

Then we determine the partial derivatives of $(f \circ g)$, find that they are all
continuous, and use them to write the Jacobian matrix:

${\D}(f \circ g)(t)=\left(\begin{array}{ccc}
\frac{e^t-e^{-t}}2
\\\frac{e^t+e^{-t}}2
\\0\end{array}\right)$.

We see that the Chain Rule yields the same answer as does direct differentiation.

\item (David)    % 3
Let $F:M_n \rightarrow M_n$ be the map given by $F(A)=AA^T$.

a)Prove that $F$ is continuous.

To prove that $F$ is continuous at $A\in M_n$ we have to show that 
$|F(A+H)-F(A)|$ is small if $|H|$ is small. By the definition 
$$F(A+H)-F(A)=(A+H)(A^T +H^T)-AA^T=AH^T +A^T H+HH^T$$ Therefore 
\begin{multline}
|F(A+H)-F(A)|=|AH^T +A^T H+HH^T|\leq |AH^T| +|A^T H|+|HH^T|\\
\leq |A||H^T| +|A^T||H|+|H||H^T|\leq 2|A||H|+|H|^2
\end{multline}
So  $|F(A+H)-F(A)|$ is small if $|H|$ is small.

b)Prove that $F$ is differentiable and find $D(F)(A):M_n\rightarrow M_n$
for $A\in M_n$.

To prove that  $F$ is differentiable at  $A\in M_n$ we have to show a
linear map $C:M_n\rightarrow M_n$ such that for small $H\in M_n$ we have
$$|F(A+H)-F(A)-C(H)|/|H|\rightarrow 0 {}for H\rightarrow 0$$. In this
case  $D(F)(A)=C$. Define a linear map $C:M_n\rightarrow M_n$ by 
$C(H)=AH^T +A^T H$. Then $F(A+H)-F(A)-C(H)=HH^T$ and therefore 
$|F(A+H)-F(A)-C(H)|/|H|=|HH^T|/|H|\leq |H|$. So  $F$ is differentiable
at  $A$ and  the linear map  $D(F)(A):M_n\rightarrow M_n$ is given by 
$D(F)(A)(H)=AH^T +A^T H$

\item (Nils)     % 4
Recall from problem set 5/solution set 5 that any choice of basis on
$V=\bR^n$ gives an isomorphism of bilinear forms on $\bR^n$ with
$n\times n$ matrices, \ie $\cB(\bR^n)\cong \Mat(n,n)$.
Since a bilinear form is antisymmetric iff the matrix in one/any basis
is antisymmetric, this means that we also get
$\cB_{an}(\bR^n)\cong \Mat_{an}(n,n)$, that is, the space of antisymmetric
bilinear forms on $\bR^n$ is isomorphic (given a choice of basis) with
the space of antisymmetric $n\times n$ matrices.
To compute the dimension of this latter space, note that every
antisymmetric $n\times n$ matrix is of the form:
 \begin{equation}
  \begin{pmatrix}
  0 & a_1 & a_2 & \hdotsfor{2}\\
  -a_1 & 0 & a_n & \hdotsfor{2}\\
  \hdotsfor{1}&\ddots&\ddots&\ddots&\hdotsfor{1}\\
  \hdotsfor{2} &\ddots& 0 & a_N\\
  \hdotsfor{3} &-a_N & 0
  \end{pmatrix},
 \end{equation} 
 where\footnote{Recall that the binomial coefficient
$\binom{n}{k}=\frac{n!}{(n-k)!k!}$.}
 $N=\sum_{i=1}^{n-1} i=\frac{(n-1)n}{2}=\binom{n}{2}$,
 so $\dim \cB_{an}(\bR^n)=\binom{n}{2}$.
 In particular,
\begin{enumerate}
 \item For $n=1$, $N=0$.
 \item For $n=2$, $N=1$.
 \item For $n=3$, $N=3$.
 \item For $n=10$, $N=45$.
\end{enumerate}

Note in fact that one can ``pull'' the direct sum decomposition
$\cB(\bR^n)=\cB_s(\bR^n)\oplus \cB_{an}(\bR^n)$ over to matrices, so
$\Mat(n,n)=\Mat_s(n,n)\oplus \Mat_{an}(n,n)$. This gives $\dim
\cB_s(\bR^n)=n^2-\binom{n}{2}=\binom{n+1}{2}$.

\item (Li-chung) % 5
\begin{enumerate}
\item % (a)
We claim that the partial derivatives of $f$
are continous.
Let $F_1, F_2,\dots,F_n$ be the component functions of $F$, i.e.
$F(u) = (F_1(u), F_2(u),\dots,F_n(u))$.  Then all the partial derivatives
of $F_1,\dots,F_n$ are continuous because $F$ is continuously differentiable.
Let $v = (v_1,v_2,\dots,v_n)$.  Then $f(u)$ is simply
$$(F_1(u)-v_1)^2 + (F_2(u)-v_2)^2 + \dots + (F_n(u)-v_n)^2.$$
So for each $j = 1,\dots,n$,
$$D_jf(u) 
= 2(F_1(u)-v_1)\cdot D_jF_1(u) + \dots + 2(F_n(u)-v_n)\cdot D_jF_n(u).
$$
This is a continuous function because $D_jF_i$ is continous for each $i$
(as a function of $u$),
$F_i$ is continuous for each $i$ (since $F$ being differentiable implies 
each $F_i$ is continuous), and $D_jF$ is just a polynomial expression of
continuous functions.  (Use theorem 1.5.27 repeatedly, and also note that
constant functions are continuous.)

Because $f$ has continuous partial derivatives everywhere, $f$ is 
differentiable everywhere.
\item % (b)
We will use the formula of $D_jf$ derived above.
\begin{align*}
& Df(u) = 0 \\
\iff & D_1f(u) = D_2f(u) = \dots = D_nf(u) = 0 \\
\iff &
\begin{bmatrix}
0 \\ 0 \\ \vdots \\ 0
\end{bmatrix}
=
\begin{bmatrix}
D_1f(u) \\ D_2f(u) \\ \vdots \\ D_nf(u)
\end{bmatrix}
= 
\begin{bmatrix}
2(F_1(u)-v_1)\cdot D_1F_1(u) + \dots + 2(F_n(u)-v_n)\cdot D_1F_n(u) \\
2(F_1(u)-v_1)\cdot D_2F_1(u) + \dots + 2(F_n(u)-v_n)\cdot D_2F_n(u) \\
\vdots \\
2(F_1(u)-v_1)\cdot D_nF_1(u) + \dots + 2(F_n(u)-v_n)\cdot D_nF_n(u) \\
\end{bmatrix}
\\
= &
\begin{bmatrix}
D_1F_1(u) & D_1F_2(u) & \dots & D_1F_n(u) \\
D_2F_1(u) & D_2F_2(u) & \dots & D_2F_n(u) \\
\vdots & \vdots & & \vdots \\
D_nF_1(u) & D_nF_2(u) & \dots & D_nF_n(u)
\end{bmatrix}
\begin{bmatrix}
2(F_1(u)-v_1) \\ 2(F_2(u)-v_2) \\ \vdots \\ 2(F_n(u)-v_n)
\end{bmatrix}
\\
= &
{DF(u)}^T
\begin{bmatrix}
2(F_1(u)-v_1) \\ 2(F_2(u)-v_2) \\ \vdots \\ 2(F_n(u)-v_n)
\end{bmatrix}.
\end{align*}

Now $DF(u)$ is invertible, so ${DF(u)}^T$ is too; specifically,
$(DF(u)^{-1})^T$ is the inverse because
${DF(u)}^T(DF(u)^{-1})^T = \left(DF(u)^{-1}DF(u)\right)^T = I^T = I$ and
$(DF(u)^{-1})^TDF(u)^T = \left(DF(u)DF(u)^{-1}\right)^T = I^T = I$.  Hence
${DF(u)}T$ as a linear transformation $\R^n \to \R^n$ is an isomorphism.
This implies that
$\begin{bmatrix}
0 \\ 0 \\ \vdots \\ 0
\end{bmatrix}
= {DF(u)}^T
\begin{bmatrix}
2(F_1(u)-v_1) \\ 2(F_2(u)-v_2) \\ \vdots \\ 2(F_n(u)-v_n)
\end{bmatrix}$ occurs if and only if
$
\begin{bmatrix}
2(F_1(u)-v_1) \\ 2(F_2(u)-v_2) \\ \vdots \\ 2(F_n(u)-v_n)
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ \vdots \\ 0
\end{bmatrix}$, which occurs if and only if
$F_i(u)=v_i$ for all $i$.  This is the same as saying
$F(u)=v$.
\end{enumerate}
\end{enumerate}
\end{document}
