Matrix derivative of $Tr(A\log(X))$

Question

I'm trying to work out the derivative of $Tr(A\log(X))$ with respect to $X$. Assume $X$ is positive so the $\log$ is well defined. I know that

$$Tr(A\log(X)) = A^\dagger: \log(X)$$

but what I should be doing is to express it in the form $F : X$ similar to this answer so that I can take $d Tr(A(\log(X)) = F: dX$. How can I proceed?

Answer 1

$ \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\Diag#1{\op{Diag}\LR{#1}} \def\diag#1{\op{diag}\LR{#1}} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} $For typing convenience, define $$F = \log(X)$$ Then write the function and calculate its differential $$\eqalign{ \phi &= A:F \qiq d\phi &= A:\c{dF} \\ }$$ A positive definite matrix can be diagonalized with orthogonal factors $$\eqalign{ X &= QBQ^T,\quad B=\Diag b,\quad Q^TQ=I \\ }$$ The Daleckii-Krein theorem says that $$\eqalign{ Q^T dF\:Q &= R\odot\LR{Q^T dX\:Q} \\ dF &=Q\LR{R\odot\LR{Q^T dX\:Q}}Q^T \\ }$$ where $(\odot)$ denotes the Hadamard product.

Substituting into the previous differential yields $$\eqalign{ d\phi &= A:\c{Q\LR{R\odot\LR{Q^T dX\:Q}}Q^T} \\ &= Q\LR{R\odot\LR{Q^TAQ}}Q^T:dX \\ \grad{\phi}{X} &= Q\LR{R\odot\LR{Q^TAQ}}Q^T \\ }$$ All that remains is to calculate the symmetric $R$ matrix which lies at the heart of the theorem $$\eqalign{ R_{ij} = \begin{cases} {\Large\frac{f(b_i)-f(b_j)}{b_i-b_j}} \qquad{\rm if}\;b_i\ne b_j \\ \\ \quad f'(b_j) \qquad\qquad {\rm otherwise} \\ \end{cases} }$$ NB:$\:$ For the current problem $$f(b_j) = \log(b_j),\qquad f'(b_j) = \frac{\tt1}{b_j}$$

Answer 2

$$\DeclareMathOperator{\Tr}{Tr}$$ Let $f$ be a linear map from matrices to matrices. write $f^{\mathsf{T}}$ for the unique linear map from matrices to matrices such that $$\Tr f^{\mathsf{T}}(A)B=\Tr A f(B)$$ For a differentiable function $\phi$, write $\mathrm{d}X\mapsto \phi'_X(\mathrm{d}X)$ for its Fréchet derivative at $X$ as a matrix function, so that $$\mathrm{d}\phi(X)=\phi'_X(\mathrm{d}X)\text{.}$$ Then $$\begin{split}\mathrm{d}\Tr A\,\phi(X)&=\Tr A\,\mathrm{d}\phi(X)\\ &=\Tr A\,\phi_X'(\mathrm{d}X)\\ &=\Tr (\phi'_X)^{\mathsf{T}}(A)\mathrm{d}X\text{.} \end{split}$$ If you'd like a bit more explicitness, combine $$\log X=\int_0^{\infty}\frac{\mathrm{d}t}{t}\left(\frac{1}{1+t}-\frac{1}{1+tX}\right)$$ with $$\mathrm{d}\left(\frac{1}{1+tX}\right)=-\frac{1}{1+tX}t\mathrm{d}X\frac{1}{1+tX}$$ to get $$\mathrm{d}\Tr A \log X =\Tr \left(\int_0^{\infty}\frac{1}{1+tX}A\mathrm{d}t\frac{1}{1+tX}\right)\mathrm{d}X\text{.}$$ Or, if you like, $$\int_0^{\infty}\frac{1}{1+tX}A\mathrm{d}t\frac{1}{1+tX}=\int_0^{\infty}\mathrm{d}t\int_0^1\mathrm{d}s \mathrm{e}^{-stX}A\mathrm{e}^{stX}\mathrm{e}^{-tX}\text{.} $$