matrix exponential property Take $\alpha\in\mathrm{C}$ and $A$ a complex matrix.
Is is true that :
$$\exp(\alpha A) = (\exp{A})^\alpha$$
My intuition tells me that this is true. But I can't prove it and I can't find
this property anywhere. If this property isn't true, could you give me a
counter-example.
 A: Assume that $\exp(A)$ has no eigenvalues in $(-\infty,0]$ and let $\log(.)$ be the principal logarithm. Then we can define $(\exp(A))^{\alpha}=\exp(\alpha \log(e^A))$. For the sake of simplicity, assume that $A=diag(\lambda_j)$; then $\log(e^A)=diag(\log(e^{\lambda_j}))=diag(\lambda_j+2k_ji\pi)$ where $k_j\in\mathbb{Z}$. Finally $\exp(\alpha\log(e^A))=diag(\exp(\alpha(\lambda_j+2k_ji\pi))=diag(\exp(\alpha\lambda_j)\exp(2k_j\alpha i\pi))=$
$\exp(\alpha A)diag(\exp(2k_j\alpha i\pi))$, that is not $\exp(\alpha A)$ when some $k_j$ is not $0$.
EDIT. Answer to PinkFloyd. Let $\theta\in \mathbb{R}\setminus \pi(1+2\mathbb{Z})$. If $A=\begin{pmatrix}0&-\theta\\\theta&0\end{pmatrix}$ then  $e^A=\begin{pmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{pmatrix}$ and $\log(e^A)=\begin{pmatrix}0&-\theta-2k\pi \\ \theta+2k\pi &0\end{pmatrix}$ where $k\in \mathbb{Z}$ is s.t. $\theta+2k\pi\in(-\pi,\pi)$.
A: Hint: write the matrix exponential in the form
$$
e^A=\sum_{k=0}^{\infty}\frac{1}{k!}A^k
$$
