TITLE: Eigenpairs of $A$ and Eigenpairs of $\exp(A)$
QUESTION [2 upvotes]: I would like to understand under which assumptions the following statement holds
\begin{equation}
(\lambda,v) \text{ is an eigenpair of the matrix } A \Leftrightarrow (\exp(\lambda\Delta),v) \text{ is an eigenpair of the matrix } \exp(A\Delta)
\end{equation}
where $\Delta$ is a positive real constant and $A$ is a $\mathbb{R}^{n\times n}$-valued matrix. 
$\Rightarrow$ can be proved by induction, see for example Eigenvector and eigenvalue for exponential matrix.
I am wondering under which assumption $\Leftarrow$ holds. Indeed, $\log(\exp(A\Delta))$ is not uniquely defined as the matrix $A$.
Example where I got stuck in the proof.  To prove $\Leftarrow$, I assume $(\exp(\Delta\lambda),v)$ eigenpair of $\exp(A\Delta)$, therefore $\exp(A\Delta)v=\exp(\Delta\lambda)v$. Considering the definition of the exponential matrix and the taylor expansion of $\exp(\lambda\Delta)$ we get 
\begin{equation}
\sum_{k=0}^\infty \frac{\Delta^kA^k}{k!}v=\sum_{k=0}^\infty \frac{\Delta^k\lambda^k}{k!}v
\end{equation}
which is equivalent to 
\begin{equation}
\sum_{k=0}^\infty \frac{\Delta^k\left(A^k-\lambda^kI_n\right)v}{k!}=0.
\end{equation}
If I assume by contradiction that $\lambda$ is not an eigenvalue of $A$, how can I prove that the previous series is different from $0$?

REPLY [0 votes]: Suppose wlog that $A$ is triangular with complex values, up to a conjugation. Then if its diagonal (which is the list of complex eigenvalues with multiplicity) is $\lambda_1, \ldots, \lambda_n$, the diagonal of $\text{exp}(A)$ will be $\text{exp}(\lambda_1), \ldots, \text{exp}(\lambda_1)$ as a simple calculation yields. The eigenvectors are the standard vectors of the basis in both cases.
On balance, if you are interested to complex eigenvalues your question is true: there is a correspondence between the eigenvalues of $A$ and $\text{exp}(A)$ given by $\lambda \mapsto e^{\lambda}$. On the other hand, if you only look at real eigenvalues, something could disappear. Consider the matrix $(0 \ \ \pi , \pi \ \ 0)$: the eigenvalues are $\pm i\pi$, so nothing real. But the exponential diagonalize as $(-1 \ 0 , 0 \ -1)$
Maybe I misunderstood what you were talking about?
{"set_name": "stack_exchange", "score": 2, "question_id": 3394584}