Exponential Functions

The most natural way to define eM is to imitate the power series for ex :

In general,

etM = $\displaystyle \sum_{{k=0}}^{{\infty }}$$\displaystyle {\frac{{\left( tM\right) ^{k}}}{{k!}}}$

To evaluate the expression eM for a matrix M, leave the insertion point in the expression eM and choose Evaluate, as shown in the following examples. First define

A = $\displaystyle \left[\vphantom{
\begin{array}{cc}
1 & 2 \\
0 & 3
\end{array}
}\right.$$\displaystyle \begin{array}{cc}
1 & 2 \\
0 & 3
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{cc}
1 & 2 \\
0 & 3
\end{array}
}\right]$, B = $\displaystyle \left[\vphantom{
\begin{array}{cc}
1 & 2 \\
0 & 1
\end{array}
}\right.$$\displaystyle \begin{array}{cc}
1 & 2 \\
0 & 1
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{cc}
1 & 2 \\
0 & 1
\end{array}
}\right]$, C = $\displaystyle \left[\vphantom{
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
}\right.$$\displaystyle \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
}\right]$, D = $\displaystyle \left[\vphantom{
\begin{array}{ccc}
1 & 3 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}
}\right.$$\displaystyle \begin{array}{ccc}
1 & 3 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{ccc}
1 & 3 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}
}\right]$

$\blacktriangleright$ Evaluate

eA =  $\left[\vphantom{
\begin{array}{cc}
\frac{2}{3}e^{-1}+\frac{1}{3}e^{5} & \fra...
...}-\frac{2}{3}e^{-1} & \frac{1}{3}e^{-1}+\frac{2}{3}e^{5}
\end{array}
}\right.$$\begin{array}{cc}
\frac{2}{3}e^{-1}+\frac{1}{3}e^{5} & \frac{1}{3}e^{5}-\frac{...
...{2}{3}e^{5}-\frac{2}{3}e^{-1} & \frac{1}{3}e^{-1}+\frac{2}{3}e^{5}
\end{array}$$\left.\vphantom{
\begin{array}{cc}
\frac{2}{3}e^{-1}+\frac{1}{3}e^{5} & \fra...
...}-\frac{2}{3}e^{-1} & \frac{1}{3}e^{-1}+\frac{2}{3}e^{5}
\end{array}
}\right]$ 6pt

etA =  $\left[\vphantom{
\begin{array}{cc}
\frac{2}{3}e^{-t}+\frac{1}{3}e^{5t} & \vs...
...-\frac{2}{3}e^{-t} & \frac{1}{3}e^{-t}+\frac{2}{3}e^{5t}
\end{array}
}\right.$$\begin{array}{cc}
\frac{2}{3}e^{-t}+\frac{1}{3}e^{5t} & \vspace{6pt}\frac{1}{3...
...}{3}e^{5t}-\frac{2}{3}e^{-t} & \frac{1}{3}e^{-t}+\frac{2}{3}e^{5t}
\end{array}$$\left.\vphantom{
\begin{array}{cc}
\frac{2}{3}e^{-t}+\frac{1}{3}e^{5t} & \vs...
...-\frac{2}{3}e^{-t} & \frac{1}{3}e^{-t}+\frac{2}{3}e^{5t}
\end{array}
}\right]$ 6pt

eA+B =  $\left[\vphantom{
\begin{array}{cc}
e^{2} & 2e^{4}-2e^{2} \\
0 & e^{4}
\end{array}
}\right.$$\begin{array}{cc}
e^{2} & 2e^{4}-2e^{2} \\
0 & e^{4}
\end{array}$$\left.\vphantom{
\begin{array}{cc}
e^{2} & 2e^{4}-2e^{2} \\
0 & e^{4}
\end{array}
}\right]$ 6pt

eAeB =  $\left[\vphantom{
\begin{array}{cc}
e^{2} & e^{2}+e^{4} \\
0 & e^{4}
\end{array}
}\right.$$\begin{array}{cc}
e^{2} & e^{2}+e^{4} \\
0 & e^{4}
\end{array}$$\left.\vphantom{
\begin{array}{cc}
e^{2} & e^{2}+e^{4} \\
0 & e^{4}
\end{array}
}\right]$ 6pt

DetCD-1 =   $\left[\vphantom{
\begin{array}{ccc}
1 & t & \frac{1}{2}t^{2}+3t \\
0 & 1 & t \\
0 & 0 & 1
\end{array}
}\right.$$\begin{array}{ccc}
1 & t & \frac{1}{2}t^{2}+3t \\
0 & 1 & t \\
0 & 0 & 1
\end{array}$$\left.\vphantom{
\begin{array}{ccc}
1 & t & \frac{1}{2}t^{2}+3t \\
0 & 1 & t \\
0 & 0 & 1
\end{array}
}\right]$ 6pt

eDtCD-1 =  $\left[\vphantom{
\begin{array}{ccc}
1 & t & \frac{1}{2}t^{2}+3t \\
0 & 1 & t \\
0 & 0 & 1
\end{array}
}\right.$$\begin{array}{ccc}
1 & t & \frac{1}{2}t^{2}+3t \\
0 & 1 & t \\
0 & 0 & 1
\end{array}$$\left.\vphantom{
\begin{array}{ccc}
1 & t & \frac{1}{2}t^{2}+3t \\
0 & 1 & t \\
0 & 0 & 1
\end{array}
}\right]$ 6pt

Note that one of the properties of exponents that holds for real numbers fails for matrices. The equality eA+B = eAeB requires that AB = BA, and this property fails to hold for the matrices in the example. However, exponentiation preserves the property of similarity, as demonstrated by DetCD-1 = eDtCD-1.