Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was wondering about the uniqueness claim in the paper, on the exitence and uniqueness of the real logarithm of a matrix, to answer the questions but I have not been able to understand the sufficiency part of the proof of Theorem 2.

Can someone provide guidance?


share|cite|improve this question
Could you perhaps summarize the claim you're asking about in your question, such that readers here won't have to download an off-site article in order to figure out whether your question is even one they have a chance of answering? – Henning Makholm Feb 3 '13 at 1:56

We have $\log C = S(\log J)S^{-1}$ if $C=SJS^{-1}$ is an eigen-decomposition. Although $J$ is a Jordan form, $\log J$ is merely a direct sum of matrix blocks, but not necessarily a Jordan form. What the paper hasn't mentioned is that every block inside $\log J$ is similar to a Jordan block of $\log C$. It follows that if $\log C$ is taken to be real, all nonreal blocks of $\log J$ must occur in conjugate pairs.

Now, the multivalued function $\log J$ is defined blockwise as $\log J_k = \operatorname{LOG} J_k + i2\pi q_kI$, where $\operatorname{LOG} J_k$ is uniquely defined but $q_k$ can take any integer value. If the conditions in theorem 2 are met, so that all eigenvalues of $C$ are positive, then every $\operatorname{LOG} J_k$ is real. So, if some $\log J_k$ is taken to be $\operatorname{LOG} J_k + i2\pi q_kI$ for $q_k\not=0$, then $\log J_k$ is a nonreal block that has a conjugate counterpart $\log J_h=\overline{\log J_k}=\operatorname{LOG} J_k - i2\pi q_kI$. But then by taking matrix exponentials, we would get $J_k = J_h$. So, the Jordan block $J_k$ appears at least twice in $J$, which is a contradiction to the assumptions of theorem 2. Therefore each $\log J_k$ must be taken as $\operatorname{LOG} J_k$ and in turn the choice of $\log J$ is unique (namely, $\operatorname{LOG} J$).

Having fixed $\log J$, the logarithm $\log C$ may still be multivalued. This is because the decomposition $SJS^{-1}$ is not unique. Suppose $C=SJS^{-1}=\widetilde{S}J\widetilde{S}^{-1}$ for some $\widetilde{S}\not=S$. In general, $S(\log J)S^{-1}$ and $\widetilde{S}(\log J)\widetilde{S}^{-1}$ may be different. However, in order that $SJS^{-1}=\widetilde{S}J\widetilde{S}^{-1}$, we must have $\widetilde{S}=SK$ for some $K$ that commutes with $J$. Therefore, given the choice of $\log J$ is fixed, $\log C$ is unique if and only if $S(\log J)S^{-1}=SK(\log J)K^{-1}S^{-1}$, or equivalently, $(\log J)K=K(\log J)$ for every $K$ that commutes with $J$. Now, the proof of theorem 2 has cited a result ("[2, p.220]"), which says that this is true when $\log J=\operatorname{LOG} J$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.