Let $X\in \mathbb{C}^{n\times n}$ and $I$ is identity matrix , than if: $$ \forall t\in \mathbb{R}\quad e^{Xt} = I $$ than $$ X = 0. $$
I'm looking for short and slick proof of this statement.
The only way I can think of to prove this is take Jordan normal form of $X = SJS^{-1}$. $J$ can be writen as $J = D+N$ where $D$ is diagonal and $N$ nillpotent(those ones on superdiagonal) and $DN = ND$ so $e^{D+N} = e^D e^N$.
So we want to show that if $e^{(D+N)t} = I$ than $D=N=0$. Because $J$ is block diagonal we can work on those blocks separately. So we can assume that $D = \lambda I$ and $N$ is zero or has ones everywhere on superdiagonal.
If I'm right than for nonzero $N$ we have $$ (e^{Nt})_{ij} = \frac{t^{j-i}}{(j-i)!} \qquad i\leq j $$ zero everywhere else.
If $N$ is nonzero than: $$ e^{(D+N)t}_{ij} = e^{\lambda t} \frac{t^{j-i}}{(j-i)!} \qquad i\leq j $$
So this would have to hold $$ e^{\lambda t} \frac{t^{j-i}}{(j-i)!} = \delta_{ij} \qquad i\leq j $$ That is not possible thus $N$ has to be zero.
If $N$ is zero than we get $e^{\lambda t} = 1$. So $\lambda = 0$.
I don't like this proof, there has to be simpler one.
We have
$$\frac{d}{dt} e^{tX} = X e^{tX},$$
so if $e^{tX} \equiv I$, evaluating the derivative at $t = 0$ yields
$$\frac{d}{dt} e^{tX}\bigl\lvert_{t=0} = X = 0.$$