Is $A=0$ the only solution to $\exp(A)=I$?

Question

So obviously one solution to $\exp(A)=I$ is $A=0$, however is it the only solution? And also If $\exp(A)$ is diagonalizable does this mean $A$ is diagonalizable?

Answer 1

No, they are infinitely many solutions of $\exp(A)=I$. For all integer $k$ consider the following matrix: $$A_k:=\begin{pmatrix}0&-2k\pi\\2k\pi&0\end{pmatrix}.$$ In fact, $\exp(A)=I$ if and only if $A$ is diagonalizable over $\mathbb{C}$ and $\textrm{Sp}(A)\subseteq 2i\pi\mathbb{Z}$. The converse is easy to prove and the direct implication follows from Jordan-Chevalley decomposition.

Regarding your other question, the answer is yes. This also follows from Jordan-Chevalley decomposition.

For both of your questions, the key observation is that if $A=D+N$ is the Jordan-Chevalley decomposition of $A$, then $\exp(A)=\exp(D)+\exp(D)(\exp(N)-I)$ is the Jordan-Chevalley decomposition of $\exp(A)$.

Answer 2

For the first question: not true in general. As a counter example consider the matrix with complex entries: $$ A=2k\pi i\begin{bmatrix} m&0\\0&n \end{bmatrix} $$ with $k,m,n$ integers.

or the matrix with real entries : $$ A=2k\pi \begin{bmatrix} 0&-1\\1&0 \end{bmatrix} $$