TITLE: Is there any relation between the eigenvalues of possibly non-Hermitian matrix A and those of exp(A)?
QUESTION [5 upvotes]: Is there any relation between the eigenvalues of possibly non-Hermitian matrix A and those of exp(A)?
For hermitian matrices, they are just exponentials of the corresponding values. But in general, any relation?
Thanks.

REPLY [4 votes]: If you are willing to forget about problems of convergence for a moment, think first with polynomials: let
$$
f(t) = a_0 + a_1 t + a_2 t^2 + \cdots + a_n t^n  
$$
and $A$ any square matrix. Then
$$
f(A) = a_0 I + a_1 A + a_2 A^2 + \cdots + a_n A^n \ .
$$
Assume $v$ is an eigenvector of $A$ with eigenvalue $\lambda$: $Av = \lambda v$. Then
$$
\begin{align}
f(A)v &= a_0 Iv + a_1 Av + a_2 A^2v + \cdots + a_n A^nv \\
      &= a_0 v + a_1 \lambda v + a_2 \lambda^2 v + \cdots + a_n \lambda^n v \\
      &= \left( a_0 + a_1 \lambda + a_2 \lambda^2 + \cdots + a_n \lambda^n \right) v \\
      &= f(\lambda ) v \ .
\end{align}
$$
That is, $v$ is also an eigenvector of $f(A)$ with eigenvalue $f(\lambda )$.
The same result holds for analytic functions, such $\exp$: if
$$
f(t) = \sum_{n=0}^\infty a_n t^n
$$
then also $v$ is an eigenvector of $f(A)$ with eigenvalue $f(\lambda )$.
So, for no matter what kind of square matrix $A$, if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, $v$ is an eigenvector of $\exp (A)$ with eigenvalue $\exp (\lambda )$ too.