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?



Matrix logarithms are not unique
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.


