Is there a diagonalizable matrix $A \neq B$ such that $e^A = e^B$?

Question

Is there a diagonalizable matrix $A \neq B$ such that $e^A = e^B$?

My initial thought was that if this is possible, then $$ e^A e^{-A} = e^B e^{-A} $$ so $$ I = e^B e^{-A}. $$ This means the statement is true if there exists a nonzero matrix $C$ such that $e^C = I$. But is this possible? My intuition tells me no, but how can it be proved?

Answer 1

If by diagonalizability, you mean diagonalizability over $\Bbb C$ of real (or complex) matrices, we have $$\exp \pmatrix{0 & -2 \pi k \\ 2 \pi k & 0} = I_2$$ for all integers $k$.

On the other hand, if one insists on diagonalizability over $\Bbb R$ of real matrices, the answer is no, as for any real diagonal matrix $\Delta$ with diagonal entries $\lambda_a$, $\exp \Delta$ is diagonal, with diagonal entries $\exp \lambda_a$, so $\exp \Delta = I$ implies that $\Delta = 0$.

It follows from the transcendence of $\pi$ and the fact that $\ker(\exp: \Bbb C \to \Bbb C) = \{2 \pi i k : k \in \Bbb Z\}$ that there are no examples of rational matrices diagonalizable over $\Bbb C$ with the claimed property.

Answer 2

You can have the $1\times 1$ matrices $[0],[2\pi i]$.

Answer 3

You can have the $2 \times 2$ matrices $I=\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$ and $J=\left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right]$, compare $aI+bJ$ and $aI+(b+2\pi k)J$ for any $k \in \mathbb{Z}$.