Scalar powers of a matrix exponential

[Hiero]

Starting from the definition of a matrix exponential as a power series, how would we show that $(e^A)^n=e^{nA}$?

I know how to show that if A and B commute then $e^Ae^B = e^{A+B}$ and from this we can show that the first identity is true for integer values of n, but how can we show it’s true for any real value of n?

 

[fresh_42]

Define $\left(e^A\right)^n$ in case $n\in \mathbb{R}.$

 

[Hiero]

Define $\left(e^A\right)^n$ in case $n\in \mathbb{R}.$

I was naively imagining the nth power of the infinite sum, but I suppose to define the nth power of a matrix in the first place we would have to use the series definition? So it’s just by definition?

 

[fresh_42]

I don't know what the the left hand side should be. It makes no sense. What is $\begin{bmatrix}-1&3\\1&-5\end{bmatrix}^\pi\;?$

 

[Hiero]

I don't know what the the left hand side should be. It makes no sense. What is $\begin{bmatrix}-1&3\\1&-5\end{bmatrix}^\pi\;?$

That’s what I mean, we would have to use the exponential to define it like $e^{\pi log(\begin{bmatrix}-1&3\\1&-5\end{bmatrix}\;)}$ making the “identity” a definition.

I was overlooking that point, thanks.

 

[Hiero]

I’m still curious about the scalar analog of the question. I know it’s pointless but I’m just wondering if someone has a solution to the following because I haven’t a clue how to approach it:

Suppose we defined $e^x$ by its power series $e^x=1+x+0.5x^2+...+\frac{x^n}{n!}+...$ (where x is a scalar now not a matrix)

Starting from that definition, how would we prove that $(e^x)^n=e^{nx}$ for any real n?

I’m just curious; there ought to be a way. So if you smart people are bored, then that is my problem to you.

 

[fresh_42]

I would prove that $e^x$ is a solution of $y'=y,y'(0)=1$ which is obvious, and then show that $e^{nx}$ and $(e^x)^n$ both solve $y'=ny, y'(0)=n$, and see if Picard Lindelöff applies for uniqueness.