Let $\mathcal {L}^{-1}[\cdot]$ be an inverse Laplace transform. Let $A$ be a square matrix, and $I$ an identity matrix.
Based on the fact that $\mathcal {L} ^{-1} [{(sI-A)}^{-1}] = e ^{tA}$, how can we compute the following inverse laplace transform?
$$\mathcal{L}^{-1} [{(sI-A)}^{-1} \cdot x \cdot y^T \cdot {(sI-A)}^{-1}],$$
where $x$ and $y$ are column vectors.
Best Answer
Denote the convolution $f*g(u)=\int_0^u f(\tau)g(u-\tau)d\tau$. For matrix functions $A$ and $B$, we have
$$A*B(u)=\int_0^u \sum_j A_{ij}(\tau)B_{jk}(u-\tau)d\tau=\sum_j A_{ij}*B_{jk}(u). \tag{$\circ$}$$
Also, we have the rule
$$\mathcal{L}\{f*g\}(s)=\mathcal{L}\{f\}(s)\cdot\mathcal{L}\{g\}(s)=F(s)G(s). \tag{a}$$
Inverting this, we deduce
$$\mathcal{L}^{-1}\{F(s)G(s)\}(x)=f*g(u). \tag{b}$$
Use summation notation and linearity of the operator (and remember $(\circ)$) to prove that $\rm(a)$ and $\rm(b)$ also apply to matrix functions as well. This will help in your problem if you set
$$F(s)=(sI-A)^{-1}x, \\ G(s)=y^T(sI-A)^{-1}.$$
Check with summation notation that, again by linearity,
$$\mathcal{L}\{A(u)C\}=\mathcal{L}\{A(u)\}C, \\ \mathcal{L}\{CA(u)\}=C\mathcal{L}\{A(u)\} \tag{c}$$
for matrix functions $A(u)$ and constant matrices $C$. Notice how $(c)$ can be reworked for the inverse transform just as well. This will help with the constant matrices $x$ and $y^T$.
(Throughout this answer, I have taken matrices to be of arbitrary dimensions - outside of the multiplications being well-defined in each instance.)
This doesn't actually do the work for you, but it tells you what you need to do. The final thing you'll need to do is a convolution involving (but not quite "of," due to $x$ and $y$ in the mix) matrix exponentials. Remember that if $S,T$ are commuting matrices, $e^{S+T}=e^Se^T$ is valid.