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\begin{document}
\centerline{Math 322 lecture for Monday, Week 3}
\bigskip

Let~$F=\R$ or~$\C$, and let~$M_n(F)$ denote~$n\times n$ matrices with
coefficients in~$F$.  The derivative of a curve~$x(t)=(x_1(t),\dots,x_n(t))$
in~$F^n$ with respect to~$t$ gives the curve's tangent direction or velocity at
time~$t$:
\[
  \dot{x}:=x'(t):=\left( \frac{dx_1}{dt},\dots,\frac{dx_n}{dt} \right).
\]
We are interested in finding~$x$ such that
\[
  x'=Ax
\]
and satisfying some initial condition~$x(0)=x_0\in F^n$.  If~$n=1$, then~$A=a\in
F$, and we have already seen the solution~$x=x_0e^{at}=e^{at}x_{0}$.  It turns
out that the solution in the case~$n=1$ is just a space case of the solution
for~$n\geq1$:
\begin{equation}\label{eqn1:03mon}
  x=e^{At}x_0.
\end{equation}
Our first goal is to make sense of equation \eqref{eqn1:03mon} (e.g., what does it mean to
exponentiate a matrix?) and then prove that it is the
unique solution.  
\medskip

\textbf{Definition.} A \emph{norm} on a vector space~$V$ over~$F$ is a mapping
\[
  \|\ \|\colon V\to\R
\]
satisfying
\begin{enumerate}
  \item (positive definite) $\|v\|\geq0$ for all~$v\in V$, and~$\|v\|=0$ if and only
    if~$v=0$.
  \item (absolute homogeneity) $\|\alpha v\|=|\alpha|\|v\|$ for all~$v\in V$
    and~$\alpha\in F$.
  \item (triangle inequality) $\|v+w\|\leq \|v\|+\|w\|$ for all~$v,w\in V$.
\end{enumerate}
\medskip

{\bf Examples.} The usual absolute value on~$F^n$ is a norm.  If~$F=\R$, we have
\[
  \|x\|:=|x|:=\sqrt{x\cdot x}=\sqrt{\textstyle\sum_{j}x_j^2}
\]
and if~$F=\C$, we have
\[
  \|x\|:=|x|=\sqrt{x\cdot\bar{x}}=\sqrt{\textstyle\sum_{j}|x_j|^2}.
\]
Note: if~$x_j=a_j+b_ji$ with~$a_j,b_j\in\R$, then
\[
  \|x\|=|x|=\sqrt{\textstyle\sum_j(a_j^2+b_j^2)},
\]
which is the length of~$x\in\C^n$ thought of as a vector
in~$\R^{2n}$.  As indicated above, we use the usual absolute value notation,
$|x|$ for this norm.

The case~$n=1$ says the usual absolute value on~$F$ is a norm on~$F$.
\medskip

Given a norm~$\|\ \|$ on a vector space~$V$, we can define a
\emph{metric} on~$V$ (i.e., a distance function) by
\[
  d(v,w):=\|v-w\|.
\]
The following properties of this distance function are easy to verify:
\begin{enumerate}
  \item (positive definite) $d(v,w)\geq0$ for all~$v,w\in V$, and~$d(v,w)=0$ if and only
    if~$v=w$.
  \item (symmetry) $d(v,w)=d(w,v)$ for all~$v,w\in V$.
  \item (triangle inequality) $d(u,w)\leq d(u,v)+d(v,w)$ for all~$u,v,w\in V$.
\end{enumerate}
\medskip

The following proposition implies that two norms on a vector space will define
the same topology (``sense of closeness'') on that space:
\medskip

\textbf{Proposition.} Let~$\|\ \|_1$ and~$\|\ \|_2$ be two norms on a
finite-dimensional vector
space~$V$ over~$F$.  Then these norms are {\em equivalent} in the following
sense: there exist positive real numbers~$a,b$ such that
\[
  a\|v\|_2\leq\|v\|_1\leq b\|v\|_2
\]
for all~$v\in V$.
\begin{proof}[Sketch of proof]\ 
\smallskip

\textsc{Step 1.}  If the displayed set of inequalities holds, say~$\|\ \|_1\sim\|\ \|_2$.
Prove that~$\sim$ is an equivalence relation.
\medskip

\textsc{Step 2.} By Step 1, it suffices to prove the result when $\|\ \|_2=|\
|$, the usual absolute value norm, discussed above, and $\|\ \|_{1}$ is
arbitrary. There is nothing to prove if~$v=0$, since any positive constants~$a$
and~$b$ work in that case.  Assume from now an that $v\neq0$. Then, dividing
through by~$|v|$ and using properties of the norm, we see that
$a|v|\leq\|v\|_1\leq b|v|$ is equivalent to $a\leq \|u\|_{1}\leq b$
where~$u=v/|v|$ has (usual) norm~$|u|=1$.
\medskip

\textsc{Step 3.} Show that $v\to\|v\|_1$ is a continuous function with respect
to~$|\ |$.  That is, given~$v\in V$ and~$\varepsilon>0$, show there
exists~$\delta>0$ such that if~$w\in V$ and $|v-w|<\delta$, then
\[
  |\, \|v\|_1-\|w\|_1\,|<\varepsilon.
\]
\medskip

\textsc{Step 4.} Apply the \emph{extreme value theorem}, a continuous function
on a compact set (closed and bounded) achieves a minimum and a maximum value.
In our case, the compact set is $\{u\in V: \|u\|_1=1\}$ and the minimum and
maximum values are the desired constants~$a$ and~$b$, respectively.
\end{proof}

\textbf{Definition.} The \emph{operator norm} on the vector space $M_{n}(F)$ of~$n\times n$
matrices with coefficients in~$F$ is given by
\[
  \|A\|:=\max_{|x|\leq 1}|Ax|.
\]
for each~$A\in M_n(F)$ where~$|\ |$ is the usual norm on~$F$.
\bigskip

\textbf{Remarks.}
\begin{enumerate}[topsep=2pt]
  \item For the identity matrix, we have~$\|I_n\|=1$.
  \item The real number $\|A\|$ is the most that~$A$ scales any vector:
    \[
      \|A\|=\max_{x\neq 0}A\left( \frac{x}{|x|} \right)=\max_{x\neq 0}\frac{|Ax|}{|x|}.
    \]
    Thus,~$|Ax|\leq\|A\||x|$ for all~$x\in F^n$.  A detailed proof will be given
    below.
  \item When trying to define a norm on~$M_n(F)$, it might seem more natural to
    just think of an~$n\times n$ matrix as an element of~$F^{n^2}$ and use the
    usual norm on~$F^{n^2}$.  However, the norm we have just described is easier to work
    with and, according to the proposition given above, it is equivalent to 
    any other norm on~$M_n(F)$.
\end{enumerate}
\medskip

{\bf Lemma 1.} For all~$A,B\in M_n(F)$ and~$x\in F^n$,
\begin{enumerate}
  \item $|Ax|\leq\|A\||x|$.
  \item $\|AB\|\leq\|A\|\|B\|$.
  \item $\|A^k\|\leq\|A\|^k$.
\end{enumerate}

\begin{proof}
For part~1, first note that the inequality holds when~$x=0$.  So suppose
that~$x\neq 0$, and let~$u=\frac{x}{|x|}$.  We have that~$|u|=1$, and hence,
\[
 \frac{|Ax|}{|x|} = \left|A \frac{x}{|x|}\right| = |Au| \leq \max_{|y|\leq 1}|Ay|=\|A\|.
\]
Multiplying through by~$|x|$ gives~$|Ax|\leq \|A\|x|$, as desired.

 For part~2, note that for all $x\in
 F^n$ with~$|x|\leq 1$, we have from part~1,
 \[
   |(AB)(x)|=|A(Bx)|\leq\|A\||Bx|\leq\|A\|\|B\||x|\leq\|A\|\|B\|.
 \]
 Therefore,
 \[
   \|AB\|:=\max_{|x|\leq 1}|(AB)(x)|\leq\|A\|\|B\|.
 \]
 Part~3 follows from part~2.
\end{proof}

\textbf{Definition.} Let~$(v_k)_{k=0,1,\dots}$ be a sequence in a normed
vector space~$(V,\|\ \|)$.  We say 
\[
\lim_kv_k=v
\]
for some vector~$v\in V$ if for all~$\ve>0$ there exists~$N\in\R$ such
that
\[
\|v-v_k\|<\ve
\]
whenever~$k\geq N$.  A series~$\sum_{k=0}^{\infty} v_k$ converges to~$v$ if its
sequence of partial sums~$v_0$, $v_0+v_{1}$, $v_{0}+v_{1}+v_{2}$,\ldots
converges to~$v$.
\medskip

\textbf{Theorem.} For all~$A\in M_n(F)$ and~$t_0>0$, the function~$\R\to M_n(F)$ given
by
\[
  t\mapsto\sum_{k\geq 0}\frac{A^kt^k}{k!}
\]
converges absolutely and uniformly for~$t\in[-t_0,t_0]$.
\medskip

Before proving this theorem, let's review the notions of absolute and uniform
convergence of series of functions.  First, a series~$\sum_k v_k$ in a normed
vector space~$(V,\|\ \|)$ is {\em absolutely convergent} if~$\sum_k\|v_k||$ converges.
If a series is absolutely convergent then every rearrangement of the series will
converge.


Let~$V$ and~$W$ be normed vector spaces, and let $C\subseteq W$.  (For instance,
we could take~$W=\R$ and~$C=[-t_{0},t_{0}]$.) For each~$n\geq0$, let~$f_n\colon
W \to V$ be a function.  The sequence~$(f_n)$ {\em converges uniformly}
to~$f\colon W\to V$ on~$C$ if for all~$\ve>0$, there exists an~$N(\ve)\in\R$
such that for all~$x\in C$,
\[
  \|f(x)-f_n(x)\|<\ve
\]
whenever~$n>N(\ve)$. {\bf Note:} the word ``uniform'' refers to the fact
that~$N(\ve)$ is independent of~$x$.
\medskip

The notion of uniform convergence makes sense for a series~$\sum_kf_k$ since a
series is just a sequence of partial sums.  
\end{document}

