Diagonalization of Matrix Exponential Given $A \in M_{n \times n}(\mathbb{C})$, how do I show that $e^{A} = I_{n} \iff$ $A$ is diagonalizable with eigenvalues in $2\pi i \mathbb{Z}$. I know that if $A = PDP^{-1}$ then $e^{A} = Pe^{D}P^{-1}$, but I am not sure how this helps. Moreover, the canonical definition is $e^{A} = \sum_{j = 0}^{\infty} \frac{1}{j!}A^{j}$. I suppose given what I have said that $A$ is diagonalizable with eigenvalues in $2 \pi i \mathbb{Z}$ implies that $e^{A} = I_{n}$, but what about the forward implication? 
 A: Partial answer :
$\DeclareMathOperator{\tr}{\textrm{trace}}$
If $\exp(A) = I_n$.
Let be $\lambda$ an eigenvalue of $A$.
There is an eigenvector $x$ such that $Ax = \lambda x$.
So that: $\forall j \in \mathbb{N}, \dfrac{A^j}{j!}x = \dfrac{\lambda^j}{j!} x$.
Now, we have: $\exp(A)x = \exp(\lambda) x$ by sum of the previous relation.
But, $\exp(A) = I_n$, so that: $I_n x = x = \exp(\lambda) x$.
Thus: $\exp(\lambda) = 1$.
Thus: $\lambda = 2\pi i k, k \in \mathbb{Z}$.
Thus: all eigenvalues of $A$ are contained in $2\pi i\mathbb{Z}$.
A: Every matrix can be put in Jordan canonical form, i.e. there exist an (invertible) $S$ such that
$$
S^{-1} A S = D+ N
$$
where $D$ is diagonal, $N$ is nilpotent (with a certain structure) and they commute with each other. 
Hence you have
\begin{align}
e^{A} &= I \\
\Leftrightarrow S^{-1} e^{A} S &= I \\
\Leftrightarrow e^{D+N} &= I
\end{align}
Now it turns out that 
$$
e^{D+N} = \sum_{i=1}^s \left [ e^{\lambda_i} |i\rangle \langle i| + \sum_{n=1}^{m_i-1} \frac{e^{\lambda_i}}{n!} D_i^n \right ]
$$
where $s$ is the dimension of the matrices, $|i\rangle\langle i|$ is the matrix with elements $a,b$ equal to $\delta_{a,b} \delta_{a,i}$, $D_i$ is a Jordan block and $m_i$ is the dimension of the Jordan block. 
The point now is that all the matrices $D_i$ are upper triangular and so are their powers. So in order to have 
$$
e^{D+N} = I
$$
you must have $\lambda_i = 2\pi n_i $ with $n_i \in \mathbb{Z}$ and $m_i=1$ (i.e. there are no Jordan blocks and the matrix is diagonalizable). 
You can read more on Jordan canonical form (and exponential) in this other MSE question:
Matrix exponential for Jordan canonical form
A: For a somewhat lighter weight solution that just uses basic results about
0.) the preimage of 1 under the exponential map is $0 + 2\pi i n$ which implies these are the eigenvalues of $A$  (integer n)
1.) block triangular matrix multiplication
2.) linear independence of powers of nilpotent matrices
3.) with scalars in $\mathbb C$, any matrix is similar to an upper triangular one, say using Schur's   Triangularization Theorem (though Jordan form works here of course)
4.) for commuting matrices $A$ and $B$, $e^{A+B} = e^A e^B$.  This is typically developed in any text introducing the matrix exponential (and of course is implied e.g. by the Lie Product Formula).  
for (1)
The needed fact is that
$\begin{bmatrix}
 R & *\\ 
\mathbf 0 & Y
\end{bmatrix}^k = \begin{bmatrix}
R^k & *\\ 
\mathbf 0 & Y^k
\end{bmatrix}$
by direct multiplication, where $*$ denotes entries we are not concerned with    
for (2)
for any non-zero n x n nilpotent matrix we have $N^j = \mathbf 0$ for some $1 \lt j \leq n$, where $N^{j-1} \neq 0$
then the powers $N^k$ for $1\leq k \lt j$ are linearly independent 
i.e.
$\sum_{k=1}^{j-1} \alpha_k N^k = \mathbf 0 \longrightarrow$ each $\alpha_k = 0$
to prove this consider that if this weren't true, then there is some non-trivial linear combination equal to zero and we can isolate the lowest power with non-zero coefficient (call the power $m$) and write it as a linear combination of powers $\geq m+1$.  Multiply each side by $N^{j-1-m}$ and we have $\alpha_m N^{j-1} = \mathbf 0$, a contradiction.  
It follows that 
$e^{N}-I = \sum_{k=1}^\infty \frac{N^k}{k!} = \sum_{k=1}^j \frac{1}{k!}N^k \neq \mathbf 0$
and per what follows in 3, we find $N$ is similar to a strictly upper triangular matrix and $e^N$ is similar to a non-diagonal upper triangular matrix with ones on the diagonal.  
for (3) 
by Schur's Triangularization theorem, any matrix in $\mathbb C$ is unitarily similar to an upper triangular one.  The decomposition need not be unique -- it is sufficient for our purposes.  
main argument:
if $e^A = I$ and $A$ is diagonalizable, then the argument works as outlined in the original post.  
now suppose for a contradiction that $A$ is defective but it is still true that $e^A = I$.
We may select any eigenvalue with insufficient geometric multiplicity and name it $\lambda_1$ with algebraic multiplicity of $r$ but
$1 \leq \text{geometric multiplicity of } \lambda_1 =g\lt r$ 
Applying Schur Triangularization we may select the block in the top left corner to be an $r x r$ upper triangular matrix with $\lambda_1$ on the diagonal.  So
$U^{-1} A U = U^* A U = \begin{bmatrix}
 R_r & *\\ 
\mathbf 0 & Y_{n-r}
\end{bmatrix} = \begin{bmatrix}
 R_r & *\\ 
\mathbf 0 & *
\end{bmatrix}$
where we need only focus on the top left block matrix  
Then
$\big(U^* A  U\big)^k = U^* A^k  U = \begin{bmatrix}
 R_r^k & *\\ 
\mathbf 0 & *
\end{bmatrix}$  and  
$I = U^* I U = U^* e^{A}  U = e^{U^*A  U} = \begin{bmatrix}
 e^{R_r} & *\\ 
\mathbf 0 & *
\end{bmatrix} = \begin{bmatrix}
 I_r & *\\ 
\mathbf 0 & *
\end{bmatrix}$ 
But $R_r = \lambda_1 I + N$ i.e. a scaled form of the identity matrix and a non-zero nilpotent (strictly upper triangular) matrix. These commute. 
$N$ is strictly upper triangular by construction, and non-zero because the the geometric multiplicity of $\lambda_1$ is strictly less than the algebraic multiplicity.  (i.e. we have $g \lt r$ linearly independent eigenvectors with eigenvalue $\lambda_1$ and we may, via gramm schmidt, select them to be mutually orthonormal but for the $r-g$ remaining eigenvalues there must be non-zero components above the diagonal -- if there wasn't, then $\text{dim null}\big(\lambda_1 I- A\big) \gt r$.)  
Thus
$e^{R_r} = e^{\lambda_1 I + N} = e^{\lambda_1 I}e^{N} = e^{\lambda_1} I e^{N}= e^{\lambda_1}  e^{N} = 1 \cdot e^{N} = e^N \neq I$
which is a contradiction that follows from (2)  
Jordan Canonical Form would streamline the above, though it involves a lot more machinery.  
