$\newcommand{\C}{\mathbb{C}}$
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\ga}{\gamma}$
$\newcommand{\al}{\alpha}$
Let $A$ be an $n \times n$ real matrix. Assume $e^A$ is a diagonal matrix. Does this imply $A$ is diagonal?
If not, then for which matrices, their exponential is diagonal? Can we obtain some nice characterization?
(The complex case might also be interesting)
Update:
Define $G=\C^* = \C \setminus \{0\}$ to be the group of nonzero comlplex numbers, with multiplication.
Let $H=\{\begin{pmatrix}a&b\\-b&a\end{pmatrix}|a,b \in \R \, , \, a^2+b^2 \neq 0 \}$ be a group of $2 \times 2$ real matrices (with the operation of matrix multiplication).
Look at the following group isomorphism:
$\phi:G \to H, \phi(a+ib)=aI+bJ$, where $I=\begin{pmatrix}1&0\\0&1\end{pmatrix} \, , \, J = \begin{pmatrix}0&1\\-1&0\end{pmatrix}$
By the theory of Lie groups, we know:
$$(1): \, \, \phi(\exp^G(v))=\exp^H(\phi_*(v))$$ Since $H$ is a subgroup of $GL_2(\R)$ $\exp^H$ is just the usual matrix exponential.
Note that $T_eG \cong \C$, We claim, $\exp^G(z)=e^z$ (where $e^z$ is the standard complex exponential).
Proof:
Let $v \in T_eG = \C$. Define $\ga:I \to \C^*$, $\ga(t)=e^{tv}$. Then $\ga$ satisfies $\ga(0)=1,\dot \ga (0) = v, \ga(t+s)=\ga(t)\cdot\ga(s)$, so $\ga$ is a one-parameter subgroup in $G=\C^*$ with initial velocity $v$.
By definition, $\exp^G(v)=\ga(1)=e^v$, as required.
Hence, equation $(1)$ becomes $(1'):$
$$(1'): \, \, \phi(e^v)=\exp(\phi_*(v))$$
Let us calculate $\phi^*=(d\phi)_e:T_eG \to T_IH \subset M_2$.
Let $v=x+iy \in T_eG=\C$. Define $\al(t)=1+tv=(1+tx)+i(ty),\dot \al(0)=v$, and
$$\phi^*(v)=(d\phi)_e(v)=\frac{d}{dt}\big(\phi(\al(t))\big)|_{t=0}= \frac{d}{dt}\big( \begin{pmatrix}1+tx&ty\\-ty&1+tx\end{pmatrix}\big)|_{t=0}=\begin{pmatrix}x&y\\-y&x\end{pmatrix}$$
So, Hence, equation $(1')$ becomes $(1''):$
$$(1''): \, \, \phi(e^{(x+iy)})=\exp(\begin{pmatrix}x&y\\-y&x\end{pmatrix})$$,
Finally, since $\phi(e^{(x+iy)})=\phi(e^x\cos y+e^x\sin y)=\begin{pmatrix}e^x\cos y&e^x\sin y\\-e^x\sin y&e^x\cos y\end{pmatrix} $
we get the following formula:
$$\exp(\begin{pmatrix}x&y\\-y&x\end{pmatrix}) = \begin{pmatrix}e^x\cos y&e^x\sin y\\-e^x\sin y&e^x\cos y\end{pmatrix} $$
In particualr, taking $x=0,y=t$ we get:
$$\exp(\begin{pmatrix}0&t\\-t&0\end{pmatrix}) = \begin{pmatrix}\cos t&\sin t\\-\sin t&\cos t\end{pmatrix} $$
Best Answer
If $e^A$ is diagonal, then $A$ is not necessarily diagonal.
Take $A=\begin{pmatrix}0&\pi\\-\pi&0\end{pmatrix}$. Then $e^A=-I_2$.
EDIT. Anwer to Asaf. Assume that $e^A=diag(\lambda_i)$ where the $(\lambda_i)$ are non-zero distinct; since $A$ and $e^A$ commute, $A$ is diagonal; then necessarily $e^A$ admits at least one double eigenvalue. Moreover, we can prove that if $e^A$ is diagonalizable, then $A$ is also diagonalizable.
Thus, when $n=2$, $e^A=\lambda I_2$ with $\lambda\not= 0$; thus $spectrum(A)=\{u,v\}$ where $u\not= v$ and $e^u=e^v=\lambda$, that implies $v=u+2ki\pi$ with $k\in \mathbb{Z}^*$.