TITLE: Does the exponential of a matrix commute with the matrix?
QUESTION [8 upvotes]: Can someone give me an idea for the proof that for every $t\in \mathbb{C}$ we have $e^{tA}\cdot A = A \cdot e^{tA} =$ ?
I couldn't find a counterexample, so my gues is, that it would be true, but I'm not sure even how to begin the proof.

REPLY [15 votes]: $$e^{tA}\cdot A = \left(\sum_{k=0}^\infty \frac{t^kA^k}{k!}\right)\cdot A$$
$$= \sum_{k=0}^\infty \frac{t^kA^{k+1}}{k!}$$
$$= A \cdot \left(\sum_{k=0}^\infty \frac{t^kA^k}{k!}\right)$$
$$= A \cdot e^{tA}$$