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

From now on, page references are to our text.  Recall that we will always be
working over the field~$F=\R$ or~$\C$.
\medskip

\textbf{Definition.} A sequence~$(v_k)$ in a normed vector
space~$(V,\|\ \|)$ is a {\em Cauchy sequence} if for all~$\ve>0$ there
exists~$N\in\R$ such that for all~$m,n>N$, we have
\[
  \|v_n-v_m\|<\ve.
\]
A theorem from analysis says that if~$V$ is finite-dimensional then it is {\em
complete}: a sequence~$(v_k)$ converges if and only if it is a Cauchy
sequence.
\medskip

\textbf{Lemma.} (Weierstrass~$M$-test) 
Let~$V$ and~$W$ be normed vector spaces with~$V$ finite-dimensional.  For each $k\geq0$,
let~$f_k\colon W \to V$ be a function.
Let $C\subseteq W$, and suppose there exists a sequence~$(M_k)_k$ of positive numbers such
that
\[
  \|f_k(x)\|\leq M_k
\]
for all~$x\in C$ and for all~$k$.  Suppose further that~$\sum_k M_k$ converges.
Then~$\sum_kf_k$ is absolutely and uniformly convergent on~$C$.
\smallskip

{\bf Proof.} A sequence in a normed space over~$F$ converges if and only if it's a Cauchy
sequence.  Let~$\ve>0$.  Since~$\sum_k M_k$ converges, there exists~$N\in\R$
such that for all~$n\geq m>N$, we have
\[
  |\textstyle\sum_{k=0}^n M_k-\sum_{k=0}^m M_k|=|\sum_{k=m+1}^n M_k|<\ve.
\]
But then for~$n\geq m>N$ is follows that for all~$x\in C$
\[
  \textstyle \|\sum_{k=m+1}^nf_k(x)\|\leq
  \sum_{k=m+1}^n\|f_k(x)\|\leq\sum_{k=m+1}^nM_k<\ve.
\]
Thus~$\sum_k f_k$ is uniformly Cauchy.
\hfill$\Box$

\medskip

We are now ready to prove that it makes sense to exponentiate a matrix:
\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~$t\in[-t_0,t_0]$.
\smallskip

\begin{proof}
  Let~$a:=\|A\|$ and suppose that~$|t|\leq t_0$.  Then from Lemma~1
  in the previous lecture,
  \[
    \left\|\frac{A^kt^k}{k!}\right\|
    \leq \frac{\|A\|^k|t|^k}{k!}\leq\frac{\|A\|^kt_0^k}{k!}=\frac{a^kt_0^k}{k!}=:M_k.
  \]
  It follows that
  \[
    \sum_{k\geq0}M_k=e^{at_0},
  \]
  the usual exponential function.  The result follows by the
  Weierstrass~$M$-test.
\end{proof}

\textbf{Definition.} Let~$A\in M_n(F)$ and~$t\in\R$.  Then
\[
  e^{At}:=\sum_{k\geq 0}\frac{A^kt^k}{k!}.
\]
\medskip

{\bf Note:} The proof of the previous theorem shows that~$e^{At}$ is absolutely
convergent and uniformly convergent on any closed interval for~$t$.  Further, 
\[
  \|e^{At}\|\leq e^{\|A\||t|}.  
\]
To rigorously prove this last statement, note that
\[
  \left\|\sum_{k=0}^{n} \frac{A^{k}t^{k}}{k!}\right\|
  \leq \sum_{k=0}^{n}\left\|\frac{A^{k}t^{k}}{k!}\right\| 
  = \sum_{k=0}^{n} \frac{\|A\|^k|t|^k}{k!}
\]
The norm is a continuous function and hence commutes with limits, and limits
preserve inequalities.  It therefore follows that
\[
  \| e^{At}\|
  = \left\|\lim_{n \to \infty} \sum_{k=0}^{n} \frac{A^{k}t^{k}}{k!}\right\|
  =\lim_{n \to \infty}  \left\|\sum_{k=0}^{n} \frac{A^{k}t^{k}}{k!}\right\|
  = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{\|A\|^k|t|^k}{k!}
  =e^{\|A\|t||}.
\]
\medskip

\textbf{Proposition.} (p.~13) Let~$A,P\in M_n(F)$ with~$P$ invertible.  Then
\[
  e^{P^{-1}AP}=P^{-1}e^{A}P.
\]
\smallskip

\begin{proof}
 Recall the trick from linear algebra:
 \begin{align*}
   (P^{-1}AP)^k
   &= (P^{-1}AP)(P^{-1}AP)(P^{-1}AP)\cdots(P^{-1}AP)\\
   &= P^{-1}A(PP^{-1})A(PP^{-1})A(P\cdots P^{-1})AP\\
   &= P^{-1}A^kP.
 \end{align*}
 Therefore,
 \begin{align*}
   e^{P^{-1}AP}&:=\sum_{k\geq0}\frac{(P^{-1}AP)^k}{k!}\\[8pt]
 	      &= \sum_{k\geq0}\left(P^{-1}\frac{A^k}{k!}\,P\right)\\[8pt]
 	      &= P^{-1}\left( \sum_{k\geq0}\frac{A^k}{k!}\right)P\\[8pt]
 	      &= P^{-1}e^{A}P.
 \end{align*}
 The matrices~$P^{-1}$ and~$P$ can be pulled out of the sum since multiplication
 by these represent linear transformations, which are continuous, and the sum is
 a limit---limits commute with continuous functions (by definition of
 continuity).
\end{proof}

\textbf{Proposition.} (p.~13) Let~$A,B\in M_n(F)$.  If~$A$ and~$B$ commute,
then~$e^{(A+B)}=e^Ae^B$.

\begin{proof}
 \begin{align*}
   e^{(A+B)}
   &= \sum_{n\geq0}\frac{1}{n!}(A+B)^n\\[8pt]
   &=\sum_{n\geq0}\frac{1}{n!}\left(\sum_{i+j=n}\frac{n!}{i!j!}A^iB^j\right)\\[8pt]
   &=\sum_{i\geq0}\frac{1}{i!}A^i\left(\sum_{j\geq0}\frac{1}{j!}B^j\right)\\[8pt]
   &= e^Ae^B.
 \end{align*}
\end{proof}

\textbf{Corollary.} (p.~13) If~$A\in M_n(F)$, then
\[
  e^{-A}=\left( e^A \right)^{-1}.
\]

\begin{proof}
 Since~$A$ and~$-A$ commute, 
 \[
   I_n=e^{0}=e^{(A+(-A))}=e^{A}e^{-A}.
 \]
\end{proof}

\textbf{Example.} The above proposition only holds, in general, if the
matrices~$A$ and~$B$ commute.  Consider,
\[
  A =  \left(\begin{array}{rr}
      0 & 1 \\
      0 & 0
  \end{array}\right)
  \quad\text{and}\quad
  B = 
  \left(\begin{array}{rr}
      1 & 0 \\
      0 & 2
  \end{array}\right)
\]
It is easy to check that~$AB\neq BA$.

Since~$A^k=0$ for~$k>1$,
\[
  e^A=I+A = 
  \left(\begin{array}{rr}
      1 & 1 \\
      0 & 1
  \end{array}\right),
\]
and
\[
  e^B
  =\sum_{k\geq 0}\frac{1}{k!}
  \left(\begin{array}{rr}
      1 & 0 \\
      0 & 2
  \end{array}\right)^k
  =\sum_{k\geq 0}\frac{1}{k!}
  \left(\begin{array}{rr}
      1^k& 0 \\
      0 & 2^k
  \end{array}\right)
  =\sum_{k\geq 0}
  \left(\begin{array}{rr}
      1/k!& 0 \\
      0 & 2^k/k!
  \end{array}\right)
  =
  \left(\begin{array}{rr}
      e& 0 \\
      0 & e^2
  \end{array}\right).
\]
Thus,
\[
  e^Ae^B=
  \left(\begin{array}{rr}
      e& e^2 \\
      0 & e^2
  \end{array}\right).
\]
On the other hand, you can check by induction that
\[
  (A+B)^k=
  \left(\begin{array}{cc}
      1& 2^k-1 \\
      0 & 2^k
  \end{array}\right).
\]
Hence,
\[
  e^{A+B}=
  \sum_{k\geq0} \frac{1}{k!}
  \left(\begin{array}{cc}
      1& 2^k-1 \\
      0 & 2^k
  \end{array}\right)
  =
  \left(\begin{array}{cc}
      e & e^{2} - e \\
      0 & e^{2}
  \end{array}\right)
  \neq e^{A}e^{B}.
\]
\end{document}


