Suppose $e^A = A$, prove that $A$ is diagonalizable Suppose $e^A = A$, prove  that $A$ is diagonalizable, where A is a matrix.
What I have tried to do is write $A= D + N$,  where $D$ is diagonalizable, $N$ is nilpotent and $DN = ND$.
Since $N$ is nilpotent, there exist a minimal $n$ such that $N^n=0$.
Then $e^A=e^{D+N}=e^De^N=e^D(I+N+\frac{N^2}{2}+...+\frac{N^{n-1}}{(n-1)!})=A=D+N$.
If we times $N^{n-1}$ on both side, then what remain is $e^DN^{n-1}=DN^{n-1}$.
And then I don't know how to carry on.
Please help! Do I need a new method to do this question? Thanks a lot!
 A: You write
$$D+N=e^{D}e^{N}=e^{D}+e^{D}(e^{N}-I)$$
As the matrix $e^{D}(e^{N}-I)$ is nilpotent (because it is of the form $e^{D}NQ(N)$ and everything commutes), $e^D$ is diagonalizable (because $D$ is), and these two matrices commutes, from uniqueness in Dunford decomposition, you get:
$$D=e^D, \;\;\;\;N=e^{D}(e^{N}-I)$$
Therefore you get, multiplying by $N^{k-1}$ where $k$ is the smallest integer such that $N^{k+1}=0$ (and assuming by contradiction $k\geq 1$):$$N^k=DN^k,$$ in other words
$$(I-D)N^k=0.$$
Now, is $(I-D)$ inversible? Yes, as $D=e^D$, 1 cannot be an eigenvalue for $D$, and you conclude $N^k=0$, which by definition of $k$, constitute a contradiction, and finish the proof as $k=0$ and $N=0$.
A: It is sufficient to look at a Jordan  block $J$ of $A$. Such a block, of size $r\geq1$, has an eigenvalue $\lambda$ of $A$ along its main diagonal,  ones in the upper secondary diagonal, and otherwise zeros:
$$J=\lambda I+N\ .$$
As $\lambda I$ commutes with $N$ the condition $J=e^J$ leads to
$$\lambda I+N=e^{\lambda I}\cdot e^N=e^\lambda\left(I+N+{N^2\over 2!}+{N^3\over 3!}+\ldots\right)\ .\tag{1}$$
Looking at the diagonal elements here we see that $\lambda\in{\mathbb C}$ has to satisfy $$\lambda=e^\lambda \ .\tag{2}$$ 
Now we look at the secondary diagonal in $(1)$ and find that $\lambda$ has to satisfy $1=e^\lambda \cdot 1$ as well, unless $r=1$.  The equation $(2)$ has an infinity of solutions, but none of them is $=2k\pi i$ with integer $k$. It follows that in fact $r=1$, and this implies that all Jordan blocks of $A$ have size $1$, hence $A$ is diagonalizable.
A: By elementary divisor theorem (and the characterization of diagonalizability with the minimal polynomial) we only have to show that the equation 
$$ e^X = X $$
in 
$$ \Bbb{C}[X] / (X -\lambda)^n $$
implies that $n=1$.
Using that $e^X = X$ we know that $ e^{(X-\lambda)^n} = {e^X}^n {e^{\lambda}}^n = X^n {e^{\lambda}}^n$ in $\Bbb{C}[X]$, but $e^{(X-\lambda)^n} = 1$ in $ \Bbb{C}[X] / (X -\lambda)^n$ which is a contradiction since $X-e^{-{{\lambda}^{n}}} \ne (X -\lambda)^n$.
