Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 0 down vote accepted

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.

share|cite|improve this answer
Thank you so much! That's very helpful! – John Smith May 11 '12 at 15:05
hi, could you kindly give some hints on this laplace transform ( Thanks! – John Smith May 12 '12 at 3:30
@JohnSmith: Sorry, I'm not seeing a general way to simplify that (without perhaps narrower contextual information). – anon May 12 '12 at 11:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.