Why does the Laplace transform of a matrix exponential $f(t) = e^{At}$ satisfy $sF(s) = AF(s) + I$

Question

Where $A$ is some $n \times n$ matrix

Suppose I am given $f(t) = e^{At}$, then $\dot f(t) = Ae^{At}$, so $L(\dot f(t)) = sF(s) = AF(s)$

Why does $sF(s) = AF(s) + I$ hold true. (more specifically, where does the $I$ come from?)

Answer 1

We can write $f(t)=e^{At}$ as

$$e^{At}=\sum_{n=0}^{\infty}\frac{t^nA^n}{n!}$$

Then, assuming that $||A||<s, $the Laplace Transform of $e^{At}$ becomes

$$\begin{align} F(s)&=\mathscr{L}\left(e^{At}\right)(s)\\\\ &=\int_0^{\infty} e^{At}e^{-st}dt\\\\ &=\sum_{n=0}^{\infty}\frac{A^n}{n!}\int_0^{\infty}t^ne^{-st}dt\\\\ &=\sum_{n=0}^{\infty}\frac{A^n}{n!}\frac{\Gamma(n+1))}{s^{n+1}}\\\\ &=\sum_{n=0}^{\infty}\frac{A^n}{s^{n+1}}\\\\ &=\frac1s\left(I-\frac1sA\right)^{-1} \tag 1 \end{align}$$

Now, multiplying both sides of $(1)$ by $\left(I-\frac1sA\right)$ reveals

$$\left(I-\frac1sA\right)F(s)=I$$

whereupon rearranging terms and recognizing that $IF(s)=F(s)$, we obtain

$$\bbox[5px,border:2px solid #C0A000]{sF(s)=AF(s)+I}$$

as expected!

Answer 2

Let's take the scalar case for simplicity. $f(t)=e^{at}$ and $a<0$, so $$ F(s)=\int_0^\infty e^{at}e^{-st}\,dt=\left[\frac{e^{(a-s)t}}{a-s}\right]_0^\infty=\frac{1}{s-a}, \qquad \text{Re}\,s\ge 0. $$ Now $$ sF(s)=\frac{s}{s-a}=\frac{s-a+a}{s-a}=1+\frac{a}{s-a}=1+aF(s). $$ What this $1$ came from? It comes from $f(0)$: $L(\dot f)=sF(s)-f(0)$. For matrices is just the same.