Question

In my ODE class, we proved that if exp(L) = exp(L') then the eigenvalues are congruent mod 2πi. Here L, L' are two nxn 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(re^iθ) = log r + iθ + 2πik with $k \in \mathbb{Z}$. In this way we can construct an branched infinite cover of the complex plane.

We'll define multiplcity mod 2πi of an eigenvalue λ to be the number of eigenvalues congruent to λ mod 2πi up to multiplicity. If exp(L) = exp(L') are the spectra of L and L' the same including multiplicity mod 2πi?

To put this another way, I could imagine two 5x5 matrices exp(L) = exp(L') where

• the spectrum of L is (λ₁, λ₁, λ₂, λ₂ + 2πi, λ₂ + 4πi)

while

• the spectrum of L' is (λ₁, λ₁, λ₁, λ₂ + 2πi, λ₂ + 4πi)

Here the multiplicities mod 2πi are different. One would be (2,3) while the other would be (3,2). Could exp(L) = exp(L') in this case?

Answer 1

No, they cannot. Note that $exp(PAP^{-1})=Pexp(A)P^{-1}$, 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 $exp(\lambda_2)$ eigenvalus. 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.

Answer 2

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 $2\pi i$.