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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) = 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?

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2 Answers

up vote 7 down vote accepted

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.

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Jordan normal form to the rescue... –  john mangual Jan 9 '10 at 2:59
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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 $2\pi i$.

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Moreover, Jordan blocks of $A$ and $B$ can be put in bijection such that corresponding blocks have equal size and eigenvalues differ by a multiple of $2\pi i$. This is a criterion for $e^A$ to be similar to $e^B$. –  t3suji Jan 9 '10 at 13:05
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