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.
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4It suffices to observe that the subspace of all matrices commuting with a fixed matrix is closed. – Qiaochu Yuan Nov 12 '11 at 18:11
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$$e^{tA}\cdot A = \left(\sum_{k=0}^\infty \frac{t^kA^k}{k!}\right)\cdot A$$ $$= \sum_{k=0}^\infty \frac{t^kA^{k+1}}{k!}$$ $$= A \cdot \left(\sum_{k=0}^\infty \frac{t^kA^k}{k!}\right)$$ $$= A \cdot e^{tA}$$
TonyK
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