# Scalar powers of a matrix exponential

• I
• Hiero

#### 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?

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?

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\;?##

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.

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.

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.