Solution for $\mathrm{e}^{aX}=b I$ besides $X \propto I$?

Question

Is there any solution of the equation $\mathrm{e}^{aX}=b I$, where $a$ and $b$ are numbers, $X$ is a matrix and $I$ is the identity matrix. One solution is of course $X=cI$, with some number $c$, because

$$ \mathrm{e}^{acI}=a \sum_k \frac{(cI)^k}{k!}=aI \sum_k \frac{(c)^k}{k!}=aI \mathrm{e}^c = b I $$ for $a \mathrm{e}^c=b$.

Are there more or is this the only solution?

Answer 1

Let $$ D=\mathrm{diag}(2k_1\pi \mathrm{i},\ldots,2k_n\pi \mathrm{i}) \in\mathbb C^{n\times n}, $$ where $k_1,\ldots, k_n\in\mathbb Z$, and $U\in\mathbb C^{n\times n}$ invertible. Then, if $X=U^{-1}DU$, $$ \mathrm{e}^{X}=I. $$ If you are looking for a matrix with real elements, try $$ X=\left(\begin{array}{cc} 0 & 2\pi \\ -2\pi & 0\end{array}\right). $$ Then, also $\mathrm{e}^X=I$.