Computing a matrix from its exponential

Question

Given the exponential of a matrix,

$e^{At}$, what is the best way to compute A? Thinking about complex-valued 3x3 matrices here.

Answer 1

If the matrix $A$ is solution of the equation $$e^{tA}=A_2\,e^{2t}+A_1\,e^t$$ and since the left and right part both admit a derivative relatively to $t$ we must have : $$A\,e^{tA}=2\,A_2\,e^{2t}+A_1\,e^t$$ and at the limit $\,t\to 0\;$ your solution can only be $\,A=2\,A_2+A_1$.

Since it may appear rather surprising that $\;e^{\large{t\,(2\,A_2+A_1)}}=A_2\,e^{2t}+A_1\,e^{t}\;$ (as it was for me!) I had to find why this was true.

Notice that your matrices $\,A_2:=\left( \begin{array}{ccc} 2 & 1 & -1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array}\right) $ and $A_1:=\left( \begin{array}{ccc} -1 & -1 & 1 \\ -1 & -1 & 1 \\ -3 & -3 & 3 \end{array}\right) \;$ are rather special since verifying $\,A_1+A_2=1,\ \,A_1 A_2=A_2 A_1=0\;$ and $\,A_1^n=A_1$ and $\,A_2^n=A_2\,$ for any positive integer $n$.

From this we deduce that $\;e^{t\,A_i}=1+A_i (e^{t}-1)\;$ and for any real values $x$ and $y$ : $$e^{\large{\,x\,A_2+y\,A_1}}=1+A_2 (e^{x}-1)+A_1 (e^{y}-1)=A_2e^x+A_1 e^y$$