10.4: The Matrix Exponential via the Laplace Transform
You may recall from the Laplace Transform module that may achieve \(e^{at}\) via
\[e^{at} = \mathscr{L}^{-1}(\frac{1}{s-a}) \nonumber\]
The natural matrix definition is therefore
\[e^{At} = \mathscr{L}^{-1} ((sI-A)^{-1}) \nonumber\]
where \(I\) is the n-by-n identity matrix.
The easiest case is the diagonal case, e.g.,
\[A = \begin{pmatrix} {1}&{0}\\ {0}&{2} \end{pmatrix} \nonumber\]
for then
\[(sI-A)^{-1} = \begin{pmatrix} {\frac{1}{s-1}}&{0}\\ {0}&{\frac{1}{s-2}} \end{pmatrix} \nonumber\]
and so
\[e^{At} = \begin{pmatrix} {\mathscr{L}^{-1} (\frac{1}{s-1})}&{0}\\ {0}&{\mathscr{L}^{-1} (\frac{1}{s-2})} \end{pmatrix} \nonumber\]
As a second example let us suppose
\[A = \begin{pmatrix} {0}&{1}\\ {-1}&{0} \end{pmatrix} \nonumber\]
and compute, in matlab,
>> inv(s*eye(2)-A) ans = [ s/(s^2+1), 1/(s^2+1)] [-1/(s^2+1), s/(s^2+1)] >> ilaplace(ans) ans = [ cos(t), sin(t)] [-sin(t), cos(t)]
If
\[A = \begin{pmatrix} {0}&{1}\\ {0}&{0} \end{pmatrix} \nonumber\]
then
>> inv(s*eye(2)-A) ans = [ 1/s, 1/s^2] [ 0, 1/s] >> ilaplace(ans) ans = [ 1, t] [ 0, 1]