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For positive definite matrices $A$ and $B$, we have the following identity $$\text{tr}\log(A)+\text{tr}\log(B)=\text{tr}\log(AB).$$

What can be said about the case for complex matrices $A$ and $B$? Assume they are still invertible.

fewfew4
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    Does $\log(zw)=\log z+\log w$ hold for nonzero complex numbers $z,w$? In particular, what is $\log z$? – Ted Shifrin Oct 09 '21 at 17:53
  • I see, does the problem occur when taking the product of $z$ and $w$ sends you to a number that crosses the negative real axis? – fewfew4 Oct 09 '21 at 17:57
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    What do you mean by "crosses"? Certainly if $zw$ lands on the negative real axis, you have a problem. But even if it does not, you can still have problems. – Ted Shifrin Oct 09 '21 at 18:01
  • What I meant was if $w$ and $z$ are on the same branch, $zw$ might be on a different branch? – fewfew4 Oct 09 '21 at 18:08
  • Might there be a simplification if $B=A^{\dagger}$? – fewfew4 Oct 09 '21 at 18:09
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    Yes, that's precisely the problem with the case of numbers. How are you defining $\log A$ for an invertible matrix $A$? Even if it is diagonalizable, you're going to run into the branch problem we've been discussing, aren't you? – Ted Shifrin Oct 09 '21 at 18:13
  • Would you have to assume the eigenvalues all lie on the same branch? Or at least make some choice. In the case of $B=A^{\dagger}$, as long as $A$ does not have real non-positive eigenvalues there doesn't seem to be any branch issues. – fewfew4 Oct 09 '21 at 18:35
  • But you still have the crossing issues, don't you, if you allow general eigenvalues? – Ted Shifrin Oct 09 '21 at 18:54
  • You're right, I think I would need to further assume $A$ is normal. Then the product of the eigenvalues $\lambda$ and $\lambda ^*$ will always avoid "crossing". – fewfew4 Oct 09 '21 at 18:56
  • No, that's not right. The eigenvalues of a unitary matrix (which is certainly normal) are complex numbers of modulus $1$. – Ted Shifrin Oct 09 '21 at 19:05
  • To be specific, what do you do if $\lambda = -1$? Then you won't have a log. – Ted Shifrin Oct 09 '21 at 19:14
  • Right, so $A$ would have to be normal, with no real non-positive eigenvalues. – fewfew4 Oct 09 '21 at 19:22

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If matrix logarithm is defined as a primary matrix function and the principal branch of logarithm is selected, the trace identity is still valid when all eigenvalues of $A,B$ and $AB$ are positive (we don't need $A$ or $B$ to be positive definite). This is because $\operatorname{tr}\log(M)=\log(\det(M))$ when $M$ has a positive spectrum.

If matrix logarithm is defined as a primary matrix function but at least one of $A,B$ or $AB$ has a non-positive eigenvalue, it can happen that the trace identity always fails regardless of the branch of logarithm selected. For example, let $\omega=\exp(2\pi i/3),\,A=\operatorname{diag}(1,\omega,\omega^2)$ and $B=\omega I_3$. Then $AB=\operatorname{diag}(\omega,\omega^2,1)$ and \begin{aligned} \log(A)&=\operatorname{diag}\left(\log(1),\log(\omega),\log(\omega^2)\right),\\ \log(B)&=\operatorname{diag}\left(\log(\omega),\log(\omega),\log(\omega)\right),\\ \log(AB)&=\operatorname{diag}\left(\log(\omega),\log(\omega^2),\log(1)\right). \end{aligned} Hence \begin{aligned} \operatorname{tr}\log(AB) &=\log(1)+\log(\omega)+\log(\omega^2)\\ &\ne\log(1)+4\log(\omega)+\log(\omega^2)\\ &=\operatorname{tr}\log(A)+\operatorname{tr}\log(B), \end{aligned} because $\log(\omega)$ is nonzero for all branches of logarithm.

user1551
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