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.

Example $\PageIndex{1}$

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

$$e^{At} = \begin{pmatrix} {\mathscr{L}^{-1} (\frac{1}{s-1})}&{0}\\ {0}&{\mathscr{L}^{-1} (\frac{1}{s-2})} \end{pmatrix} \nonumber$$

Example $\PageIndex{2}$

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)]
	       

Example $\PageIndex{3}$

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]