How to prove that $(\exp (\textbf{A}))^{-x}$ is equal to $\exp (-x \textbf{A})$

Question

If $\textbf{A}$ is a square matrix, how can I prove that, by using the power series of matrices that the above equality holds? Note that the $x \in \mathbb{N}$ and $\textbf{A}$ is a square matrix.

Answer 1

If $AB=BA$ it follows by induction on $n$ that the binomial theorem holds for $(A+B)^n$ ($n\in\Bbb N$). Now if you simply mulitply the two power series and collect terms this shows that$$e^{A+B}=e^Ae^B\quad(AB=BA).$$

By induction on $n$ this shows that $$\left(e^A\right)^n=e^{nA}\quad(n\in\Bbb N).$$

It also follows that $e^Ae^{-A}=e^0=I,$ so $$\left(e^A\right)^{-1}=e^{-A}.$$

Hence for $n\in\Bbb N$ you have $$\left(e^A\right)^{-n}=\left(\left(e^A\right)^n\right)^{-1}=\left(e^{nA}\right)^{-1}=e^{-nA}.$$

Answer 2

Hint: $exp(A+B)=exp(A)exp(B)$ if $A$ commutes with $B$