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Section 4.1 Definition of \(\exp(A)\)

Definition 4.1.1.

For any \(n \times n\) matrix \(A\text{,}\) the matrix \(\exp(A)\) or \(e^A\) is called the matrix exponential of \(A\) and is calculated by

\begin{align*} \exp(A) \amp = I_n + \sum_{k=1}^\infty \frac{A^k}{k!}. \end{align*}

Note that the sum above always converges.

We'll be interested in \(2 \times 2\) matrices \(A\) and the exponential of the form \(\exp(tA)\text{.}\)

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*}

We don't need \(A\) to be diagonalizable in order for \(\exp(tA)\) to exist. When \(A\) is not diagonalizable, try to find a general form of \(A^k\text{.}\)

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*}

A very intriguing property of \(e^{tA}\) is that

\begin{align*} \frac d{dt} e^{tA} \amp = Ae^{tA}. \end{align*}

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*}

So, \(\mathbf x(t) = e^{tA} \mathbf x_0\) is the general solution of \(\mathbf x' = A \mathbf x\) with \(\mathbf x(0) = \mathbf x_0\text{.}\)