Eigenvalues of a matrix $A$ and $e^{A}$ If I know the eigenvalues of $e^{A}$, what can I say about the eigenvalues of $A$ itself?
 A: 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 $2\pi i$.
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 $\pm 2\pi i$.
A: This process May be right ,,,,
$$AX=\lambda  X$$
$$A^nX=\lambda^n X$$
$$e^A=I +\frac{A}{1!}+\frac{A^2}{2!}+\ldots \tag{1}$$
Now multiplying $X$ vector  on both sides of equation (1), we get
$$e^A \cdot X=X+\frac{AX}{1!}+\frac{A^2X}{2!}+ \ldots$$
or,$$e^A\cdot X=X+\lambda\frac{X}{1!} + \lambda^2\frac{X}{2!}+\ldots$$
Or,$$e^A\cdot X=(1+\frac{\lambda}{1!}+\frac{\lambda^2}{2!}+ \ldots )X$$
or,
$$e^A\cdot X=e^{\lambda}X$$
Thus eigenvalues of $e^A$ are $e^{\lambda}$
