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10.4: The Matrix Exponential via the Laplace Transform

  • Page ID
    21864
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    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]	       
    

    This page titled 10.4: The Matrix Exponential via the Laplace Transform is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by Steve Cox via source content that was edited to the style and standards of the LibreTexts platform.