# Closed form solution for ${d \operatorname{Tr}(A\log(X))\over dX}$

I need a closed form solution for ${d \operatorname{Tr}(A\log(X))\over dX}$. Here $A, X$ are $n\times n$ matrices, $\log$ is the matrix logarithm. ${d \operatorname{Tr}(\log(X))\over dX}=X^{-1}$, but ${d \operatorname{Tr}(A\log(X))\over dX}$ seems to me much more complicated. This is not equal to $A X^{-1}$ as it may seem. The best I derived so far is the following: we have $$\log(X)=-\sum_{i=1}^\infty {1\over i}(I-X)^i$$ Let's define $Z=I-X$ and then $$d \log(X)=\log(X+dX)-\log(X)=\sum_{i=1}^\infty {1\over i} \sum_{k=1}^i Z^{i-k}dX Z^{k-1}=\sum_{k=1}^\infty \big(\sum_{i=k}^\infty {1\over i}Z^{i-k} dX \big)Z^{k-1}$$
So $$d \operatorname{Tr}(A\log(X)) = \operatorname{Tr}(Ad\log(X))=\sum_{k=1}^\infty \operatorname{Tr}\big(Z^{k-1} A \sum_{i=k}^\infty {1\over i}Z^{i-k}dX\big)$$ Thus $${d \operatorname{Tr}(A\log(X))\over dX}=\sum_{k=1}^\infty \big(Z^{k-1} A\sum_{i=k}^\infty{1\over i}Z^{i-k}\big)=\sum_{k=1}^\infty \big((I-X)^{k-1} A\sum_{i=k}^\infty{1\over i}(I-X)^{i-k}\big)$$ It is reduced to $AX^{-1}$ only in the case when $AX=XA$

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Hm, I have doubts that your formula has a closed form. Might be useful to check at mathoverflow, your problem looks more like a research-level problem. – TZakrevskiy Sep 6 '13 at 17:07