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?
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 $A=\pmatrix{0&amp;2\pi\\-2\pi&amp;0}$ then $e^A=I$, but the eigenvalues of $A$ are $\pm 2\pi i$.
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 $e^{\lambda}$