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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.

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  • $\begingroup$ What do you mean by raising a matrix to a noninteger complex power? In general this sort of operation requires a choice a branch cut. $\endgroup$ – Travis Willse Nov 4 '15 at 10:56
  • $\begingroup$ @Travis ok, I should have been more precise, I'm a physicist who deals with quantum operators. One way to represent these operators is by using a matrix. it is quite common to see something like $exp{ \frac{i}{\hbar} p }$ where $p$ is the impulstion operator. $\endgroup$ – PinkFloyd Nov 4 '15 at 10:59
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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. $\theta+2k\pi\in(-\pi,\pi)$.

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  • $\begingroup$ so you proved that for this particular matrices, the formula I wrote in my question is wrong. so for complex matrices and complex $\alpha$ the fomula doen't apply. what about real matrices and $\alpha$ ? $\endgroup$ – PinkFloyd Nov 5 '15 at 12:04
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Hint: write the matrix exponential in the form $$ e^A=\sum_{k=0}^{\infty}\frac{1}{k!}A^k $$

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  • $\begingroup$ sure, I did that, but this doesn't help at all for $\alpha\notin\mathrm{N}$ $\endgroup$ – PinkFloyd Nov 5 '15 at 12:19

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