Section 4.1 Definition of
Definition 4.1.1.
For any
Note that the sum above always converges.
Theorem 4.1.2.
Suppose
Furthermore, if
Proof.
We have that
\begin{align*}
\exp(tA) \amp = I_n + \sum_{k=1}^\infty \frac{A^kt^k}{k!},\\
\amp = I_n + \sum_{k=1}^\infty \frac{V D^k V^{-1} t^k}{k!}, \\
\amp = V \biggl( I_n + \sum_{k=1}^\infty \frac{D^k t^k}{k!} \biggr) V^{-1}, \\
\amp = V e^{tD} V^{-1}.
\end{align*}
For the second part, note that if \(D = \bmat a00b\text{,}\) then
\begin{gather*}
D^k = \bmat{a^k}00{b^k},
\end{gather*}
so,
\begin{align*}
e^{tD} \amp = I_n + \sum_{k=1}^\infty \frac{D^k t^k}{k!} = \sum_{k=0}^\infty \frac{t^k}{k!} \bmat{a^k}00{b^k} = \bmat{e^{at}}00{e^{bt}}.
\end{align*}
Example 4.1.3. Calculating
Note that
\begin{align*}
A^2 \amp = \bmat0100^2 = \bmat0000,
\end{align*}
so \(A^k = \bmat0000\) for all \(k \geq 2\text{.}\) In calculating \(\exp(tA)\text{,}\) we get that
\begin{align*}
\exp(tA) \amp = I_2 + \sum_{k=1}^\infty \frac{A^kt^k}{n!} = \bmat1001 + t\bmat0100 = \bmat1t01.
\end{align*}
Question 4.1.4. Why does this occur?
Take a look at the series representation of \(e^{tA}\) and differentiate it with respect to \(t\text{:}\)
\begin{align*}
\frac d{dt} e^{tA} \amp = \sum_{k=1}^\infty \frac{k A^k t^{k-1}}{k!} = A \biggl( I_n + \sum_{k=1}^\infty \frac{A^k t^k}{k!} \biggr) = Ae^{tA}.
\end{align*}