Laplace transform of a matrix exponential

[A_B]

Homework Statement

show that the Laplace transform of e^(At) = (sI - A)^(-1)

<br /> \mathcal{L}\left\{ e^{At} \right\}(s) = \left(sI - A \right)^{-1}<br />

The Attempt at a Solution

I find
<br /> \left( e^{At} \right)_{ij} = \sum_{k=0}^{\infty} \frac{(A^k)_{ij}t^k}{k!}<br />

and since
<br /> \mathcal{L}\left\{ (A^k)_{ij}t^k \right\}(s) = \frac{k!}{s^{k+1}} (A^k)_{ij}<br />

we have
<br /> \mathcal{L}\left\{\left( e^{At} \right)_{ij}\right\}(s) = \sum_{k=0}^{\infty} \frac{(A^k)_{ij}}{s^{k+1}}<br />

and there I'm stuck.

Thanks
A_B

 

[Dick]

First show it's true for a diagonal matrix D. Then show it's still true if A is diagonalizable, i.e. PAP^(-1)=D for some invertible matrix P.

 

[Ray Vickson]

Homework Statement

show that the Laplace transform of e^(At) = (sI - A)^(-1)

<br /> \mathcal{L}\left\{ e^{At} \right\}(s) = \left(sI - A)^{-1} \right)<br />

The Attempt at a Solution

I find
<br /> \left( e^{At} \right)_{ij} = \sum_{k=0}^{\infty} \frac{(A^k)_{ij}t^k}{k!}<br />

and since
<br /> \mathcal{L}\left\{ (A^k)_{ij}t^k \right\}(s) = \frac{k!}{s^{k+1}} (A^k)_{ij}<br />

we have
<br /> \mathcal{L}\left\{\left( e^{At} \right)_{ij}\right\}(s) = \sum_{k=0}^{\infty} \frac{(A^k)_{ij}}{s^{k+1}}<br />

and there I'm stuck.

Thanks
A_B

(sI-A)^{-1}=<br /> \frac{1}{s}(I-\frac{1}{s}A)^{-1}.

RGV