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\begin{document}
\noindent
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\begin{center}
{\bf{\Large Section 2.2:}\bf 1 (optional), 3 (optional), 4, A, B}
\end{center}
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\begin{enumerate}
\item[1.]  Let $X=\RR^{n\times n}$ be the space of all $n\times n$
  matrices with real entries, equipped with the Frobenius norm, $\displaystyle\|A\|=\left(\sum_{i,j=1}^n|a_{ij}|^2\right)^{1/2}$.
\begin{enumerate}
\item Show that the set of all invertible matrices in $\RR^{n\times
    n}$ (i.e. the general linear group $GL_n(\RR)$) is open.
\item Show that the set of all symmetric matrices is closed.
\end{enumerate}
\begin{quote}
{\bf Hint:}  You may use properties of the Frobenius norm such as
$\|AB\|\leq \|A\|\,\|B\|$ and $\|A\mb{x}\|_2\leq \|A\|\,\|\mb{x}\|_2$,
where $\|\mb{y}\|_2=\left(|y_1|^2+\cdots+|y_n|^2\right)^{1/2}$ is the
Euclidean norm for $\mb{y}=(y_1,\ldots,y_n)\in\RR^n$.  For the first
problem, aim to show that $B(A,1/\|A^{-1}\|)\subset GL_n(\RR)$ when $A\in GL_n(\RR)$.
\end{quote}

\item[3.]  Let $\|\cdot\|$ and $\enorm{\cdot}$ be two norms on $X$,
  and let $(x_n)_{n=1}^\infty$ be a sequence in $X$.  Suppose that
  $\|x_n-x\|\to 0$ for some $x\in X$ and $\enorm{x_n-y}\to 0$ for some
  $y\in X$.  Must it be the case that $y=x$?
  \begin{quote}
{\bf Hint:} If $\|\cdot\|$ and $\enorm{\cdot}$ are equivalent norms,
then $y=x$, so if you are looking for counter-examples, the norms
cannot be equivalent, which implies that you must consider infinite
dimensional vector spaces.  One possibility is to consider the space
$\cP$ of polynomials of arbitrary (but finite) order, and construct
norms in terms of coefficients with respect to different bases.
  \end{quote}

\item[4.] Let $K$ be a (non-empty) compact subset of the normed vector space
  $(X,\|\cdot\|)$.
  \begin{enumerate}
  \item Show that, for each $x\in X$ there is a $y\in K$ such that
    $\|x-y\|=\inf_{z\in K}\|x-z\|$~.
  \item Show that, if for each $x\in X$ there is a unique(!) $y\in K$ such that
    $\|x-y\|=\inf_{z\in K}\|x-z\|$, then the mapping defined by $P(x)=y$, is continuous.
  \end{enumerate}
  \begin{quote}
    {\bf Note:} You should not assume that $P$ is a linear mapping. \\[3pt]
    {\bf Hint:} For the first part, you should first argue
    that, for a fixed $x$, the function $f:X\to\RR$ defined by
    $f(z)=\|x-z\|$ is continuous.  For the second part, you should
    first argue that the function $d_K:X\to\RR$ defined by
    $d_K(x)=\inf_{z\in K}\|x-z\|$ is continuous.
  \end{quote}

   \item[A.] Suppose that $Z$ is a subspace of the vector space $X$ and
    that $\dim(X/Z)=1$.  Show that, for each $y\in X\setminus Z$, we
    have $X=Z\oplus\spn\{y\}$.  Note: When $Z$ is a subspace of $X$,
    $\dim(X/Z)$ is typically called the \textit{codimension} of $Z$ in
    $X$.
  
  \item[B.] (Complexifying a real normed vector space).  Let
    $(X,\|\cdot\|)$ be a normed vector space over $\RR$.  We define
    the complexification of $X$ by
    \begin{align*}
      X_\CC=\{x+\ii y:\, x,y\in X\}=X\oplus \ii X~,
    \end{align*}
    together with the operations
    \begin{itemize}
    \item[(VA)] $(x+\ii y)+(u+\ii v)\doteq (x+u)+\ii (y+v)$\hfill $x,y,u,v\in X$
    \item[(SM)] $(a+\ii b)(x+\ii y)\doteq (ax-by)+\ii (ay+bx)$\hfill
      $a,b\in\RR$, $x,y\in X$
    \end{itemize}
    This defines a complex vector space---you can verify the vector
    space axioms if you wish, but I am not requesting it.  The most
    common (and natural)
    extension of the norm $\|\cdot\|$ to $X_\CC$ is as follows:
    \begin{align*}
      \|x+\ii y\|_\CC\doteq \sup_{\theta\in\RR}\|(\cos\theta)x+(\sin\theta)y\|
    \end{align*}
    \begin{enumerate}
    \item Show that $\|\cdot\|_\CC=\|\cdot\|$ on $X$.  (This is not
      meant to be tricky).
    \item Show that $\|\cdot\|_\CC$ satisfies the properties of a norm on
        $X_\CC$. Note: Homogeneity is the trickiest property to establish.
  \end{enumerate}
\end{enumerate}

\end{document}