# On the uniqueness of the real logarithm of a real matrix

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?

Nicolas

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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 $\log J=\operatorname{LOG} J$.