Matrix logarithms are not unique In my ODE class, we proved that if $\exp(L) = \exp(L')$ then the eigenvalues are congruent mod $2 \pi i$.  Here, $L$ and $L'$ are two $n \times n$ matrices.  I wanted to know if something more precise was true.
In a way, we should expect that matrix logs are multiple valued since this is the case in $\mathbb{C}$.
$$\log(r e^{i \theta}) = \log r + i\theta + 2 \pi i k$$
with $k \in \mathbb{Z}$.  In this way we can construct an branched infinite cover of the complex plane.
We'll define multiplicity mod $2 \pi i$ of an eigenvalue $\lambda$ to be the number of eigenvalues congruent to $\lambda$ mod $2 \pi i$ up to multiplicity.  If $\exp(L) = \exp(L')$ are the spectra of $L$ and $L'$ the same including multiplicity mod $2 \pi i$?
To put this another way, I could imagine two $5 \times 5$ matrices $\exp(L) = \exp(L')$ where

*

*the spectrum of $L$ is $(\lambda_1, \lambda_1, \lambda_2, \lambda_2 + 2 \pi i, \lambda_2 + 4 \pi i)$
while

*

*the spectrum of $L'$ is $(\lambda_1, \lambda_1, \lambda_1, \lambda_2 + 2 \pi i, \lambda_2 + 4 \pi i)$
Here the multiplicities mod $2 \pi i$ are different.  One would be $(2,3)$ while the other would be $(3,2)$.  Could $\exp(L) = \exp(L')$ in this case?
 A: No, they cannot.  Note that $\exp(PAP^{-1})=P\exp(A)P^{-1}$, so wlog, both are in Jordan form.  Then, we can compute by exponentiating Jordan blocks, and the first will have a two by two block (or two one by one) depending on whether it is diagonal or not $\delta=0,1$, and three $\exp(\lambda_2)$ eigenvalues.  The second has a 3 by 3 block and two $\exp(\lambda_2)$ eigenvalues.  These will exponentiate to new Jordan blocks for the eigenvalus $\exp(\lambda_1)$, and so the two matrices have different spectra, and so cannot be the same.
A: Charles deals with your final question regarding your example. The same idea of considering Jordan forms applies to your general initial question. 
If $A$ is a matrix with eigenvalues $\lambda_1,\dots,\lambda_n$ (repeated according to their multiplicities), then the eigenvalues of $e^A$ are the numbers $e^{\lambda_1},\dots,e^{\lambda_n}$. It follows from this that if $B$ is another matrix such that $e^A=e^B$, then the eigenvalues of $A$ and of $B$ are the same, counting multiplicities modulo $2\pi i$.
