Eigenvalues of a matrix $A$ and $e^{A}$

If $A$ is upper triangular, then it's easy: then $e^A$ is upper triangular too, and the diagonal elements of $e^A$ are the exponentials of the diagonal elements of $A$.

But this is always the case, because we can choose a basis in which $A$ is in Jordan canonical form (which is in particular upper triangular).

So the eigenvalues of $A$ are logarithms of the eigenvalues of $e^A$.

But beware that the eigenvalues of $A$ can be complex even if $A$ and $e^A$ are both real, and they are not necessarily the principal logarithms of $e^A$'s eigenvalues. For example, if $A=\pmatrix{0&2\pi\\-2\pi&0}$ then $e^A=I$, but the eigenvalues of $A$ are $\pm 2\pi i$.

The exponential of a Jordan block $$ \begin{bmatrix} \lambda&1&0&0&\dots\\ 0&\lambda&1&0&\dots\\ 0&0&\lambda&1&\dots\\ 0&0&0&\lambda&\dots\\ &&\vdots&&\ddots \end{bmatrix}\tag{1} $$ is $$ e^\lambda\begin{bmatrix} 1&\frac{1}{1!}&\frac{1}{2!}&\frac{1}{3!}&\dots\\ 0&1&\frac{1}{1!}&\frac{1}{2!}&\dots\\ 0&0&1&\frac{1}{1!}&\dots\\ 0&0&0&1&\dots\\ \vdots&\vdots&\vdots&\vdots&\ddots \end{bmatrix}\tag{2} $$ Since the Jordan Normal Form of a matrix is composed of Jordan blocks $(1)$ along the diagonal, and similarly placed square blocks along the diagonal do not interact when added or multiplied, the exponential of the Jordan Normal Form consists of exponential blocks $(2)$ along the diagonal.

If $A=PJP^{-1}$, then $e^A=Pe^JP^{-1}$. Thus, if the eigenvalues of $A$ are $\{\lambda_j\}$, the eigenvalues of $e^A$ are $\{e^{\lambda_j}\}$. However, just as with logarithms, if the eigenvalues of $e^A$ are known, the eigenvalues of $A$ are known up to an integral multiple of $2\pi i$.