If $S$ is invertible and $A$ a $n\times n$ matrix, how to prove that
$$Se^AS^{-1}=e^{SAS^{-1}}?$$
I used
$$e^{SAS^{-1}}=I+SAS^{-1}+\frac{(SAS^{-1})^2}{2!}+\cdots= S\left(I+A+\frac{A^2}{2!}+\cdots\right)S^{-1}=Se^AS^{-1}$$
Does it work or is there another way to prove this equality?
$$ \text{Is } S\left( \cdots + \frac{A^n}{n!} + \cdots \right) S^{-1} \text{ equal to } \cdots+\frac{(SAS^{-1})^n}{n!} +\cdots \text{ ?} $$ $$ \text{Is } SA^n S^{-1} = (SAS^{-1})^n \text{ ?} $$ \begin{align} (SAS^{-1})^n & = (SAS^{-1}) (SAS^{-1}) \cdots\cdots (SAS^{-1}) (SAS^{-1}) \\[10pt] & = SA(S^{-1}S) A(S^{-1}S) \cdots\cdots (S^{-1}S)A(S^{-1}S) AS^{-1} & & \text{ by associativity} \\[10pt] &= S(AAA\cdots A)S^{-1}. \end{align}
