Has a matrix block diagonal structure if and only if its exponential has it as well?

Obviously if $\mathbf{A}=\begin{bmatrix}\mathbf{C} & \mathbf{0} \\ \mathbf{0} & \mathbf{D}\end{bmatrix}$ then $e^{A}=\begin{bmatrix}\mathbf{e^C} & \mathbf{0} \\ \mathbf{0} & \mathbf{e^D}\end{bmatrix}$.

Given any invertable matrix X we can express it as $X=e^A$ (but complex matrix A is not unique until X is not from one-parameter subgroup of GL(n,$\mathbb{C}$)). Anyway: if given X has a block diagonal structure then A has to have block diagonal structure as well?

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Perhaps you could try showing something like this : can any invertible matrix can be expressed as the logarithm of some other matrix? I don't think you'll have uniqueness either but maybe it's one way to go. –  Patrick Da Silva Feb 15 '12 at 15:35
No. A counterexample is $\mathbf A=\pmatrix{0&2\pi\mathrm i\\2\pi\mathrm i&0}$, with $\mathrm e^{\mathbf A}=I$.