Given matrix $e^{tA}$, find $A$

Question

I tried answering the following question in an old exam:

Given matrix:

$$ e^{tA} = \begin{bmatrix}\frac{1}{2}(e^t+e^{-t}) & 0 & \frac{1}{2}(e^t-e^{-t})\\0 & e^t & 0\\\frac{1}{2}(e^t-e^{-t}) & 0 & \frac{1}{2}(e^t+e^{-t})\end{bmatrix}$$

Calculate A.

Where $$e^{tA} = \sum_{k=0}^{\infty} \frac{t^k}{k!}A^k$$

Hope I can get some help with this, as I don't think I can find the right way to approach it myself.

Answer 1

$$A=\left.\frac d{dt}e^{tA}\right|_{t=0}=\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}.$$

Answer 2

The result given by José is fine if you know that the matrix is for sure obtainable as $\exp(t A)$.

If you are doubtful (maybe it is a trick questions), you can use the alternative $$A = e^{-t A}\left( \frac{d}{dt} e^{t A}\right) = \begin{pmatrix} 0& 0 & 1\\ 0 & 1 &0\\ 1 &0 &0 \end{pmatrix} $$ where you can check explicitly that $t$ drops out.

Answer 3

You may find $A$ from $e^{tA}$ in two steps.

Take derivative of the matrix, $$ \frac {d}{dt} e^{tA} = Ae^{tA} $$

and let $t=0$ you will get $$ Ae^{0A}=AI=A $$