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Here's the textbook definition of the matrix exponential. Let $A$ be a $n\times n$ matrix, then

$$e^A=\sum_{n=0}^{\infty} \dfrac{1}{n!}A^{n}.$$

It's not quite clear to me how to work this series out. Some properties are quite intuitive and resemble the well-known $e^x$, where $x$ is a real number. I'd like to know if, for some constant positive integer $t$,

$$\left(e^A\right)^t=e^{tA}.$$

I've crunched some numbers and it seems to hold, but I have no clue on how to prove it. Is this true?

Rodrigo de Azevedo
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vshp11
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    What does $(e^A)^t$ mean when $t$ is not an integer? – Ted Shifrin Dec 10 '23 at 21:46
  • Beats me. I just thought it would make the prompt extra clear. – vshp11 Dec 10 '23 at 21:48
  • @vshp11 Ted's point (or an immediate corollary of it) is that $e^{t A}$ is the only sensible way to interpret $(e^A)^t$ for noninteger $t$ in the first place. It follows from the definition that the two expressions already coincide for integer $t$. – Travis Willse Dec 10 '23 at 21:52
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    The proof follows from the more general fact that for $A$ and $B$ commuting matrices, $e^{A+B} = e^A e^B$. https://math.stackexchange.com/questions/370817/commuting-in-matrix-exponential – Alex Dec 10 '23 at 21:55
  • @Alex That's great! I see it now. Should I answer my own question or just close it? I'm quite new here. – vshp11 Dec 10 '23 at 21:59
  • You could ask the same question without matrices. Recall from calculus (perhaps) that for general $a>0$, $a^x$ is defined to be $e^{x\ln a}$. Then you can check that $(e^x)^y = e^{xy}$ holds. – Ted Shifrin Dec 10 '23 at 21:59
  • @vshp11 I'm new too and don't know either ;) – Alex Dec 10 '23 at 22:02
  • @vshp11 If you think you can write a good, clear answer to the question then it would be nice if you do it. That way others will be able to learn from it. – Mark Dec 10 '23 at 22:09
  • Since $(e^a)^b$ doesn't mean anything when $a,b$ are matrices, you'll have a hard time getting it equal to anything. $M^B$ is not defined for matrices $M,B.$ – Thomas Andrews Dec 10 '23 at 22:14

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As mentioned in the comments, the proof follows from the more general fact: if $A$ and $B$ are commuting matrices, then $e^{A+B}=e^{A}e^{B}$.

Let $t$ be a positive integer and $A$ a $n\times n$ matrix. Obviously, $A$ commutes with itself, thus, by the fact pointed out above,

$$(e^{A})^t=\underbrace{e^A\dots e^A}_\text{t times}=e^{A+\dots +A}=e^{tA}.$$

Therefore, the equality holds.

vshp11
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