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]