The finite-temperature matrix entropy has its interior trace derivative

For $x>0$, [F1] gives ${\int }_0^R((1+s{)}^{-1}-(x+s{)}^{-1})ds=\log (1+R)-\log (x+R)+\log x$. The difference of the first two terms tends to zero: it is ${\int }_{x+R}^{1+R}dt\mathrm{/}t$, whose absolute value is at most $\mathrm{\mid }1-x\mathrm{\mid }\mathrm{/}(R+\min \{1,x\})$. Thus $\log x={\int }_0^{\mathrm{\infty }}((1+s{)}^{-1}-(x+s{)}^{-1})ds$. Diagonalizing a positive-definite Hermitian $X$ by [F4] gives the same identity for $\log X$. If $X⪰mI$ for some $m>0$, the resolvent $R_s=(X+sI{)}^{-1}$ has norm at most $(m+s{)}^{-1}$.

Let $H$ be small enough that $\mathrm{\parallel }H{\mathrm{\parallel }}_2<m\mathrm{/}2$, so $X+H⪰(m\mathrm{/}2)I$. For each $s\ge 0$, the resolvent identity [F2] gives $R_s(X+H)-R_s(X)=-R_s(X+H)H{R}_s(X)$. The induced operator norm is submultiplicative: for any matrices $A,B$ and unit vector $x$, $\mathrm{\parallel }ABx{\mathrm{\parallel }}_2\le \mathrm{\parallel }A{\mathrm{\parallel }}_2\mathrm{\parallel }Bx{\mathrm{\parallel }}_2\le \mathrm{\parallel }A{\mathrm{\parallel }}_2\mathrm{\parallel }B{\mathrm{\parallel }}_2$, hence $\mathrm{\parallel }AB{\mathrm{\parallel }}_2\le \mathrm{\parallel }A{\mathrm{\parallel }}_2\mathrm{\parallel }B{\mathrm{\parallel }}_2$ by [F6]. Using the integral representation from step 1.1, the remainder after the candidate linear term is $\log (X+H)-\log X-{\int }_0^{\mathrm{\infty }}{R}_s(X)H{R}_s(X)ds=-{\int }_0^{\mathrm{\infty }}{R}_s(X+H)H{R}_s(X)H{R}_s(X)ds\mathrm{.}$ By step 1.1, $\mathrm{\parallel }{R}_s(X+H){\mathrm{\parallel }}_2\le (m\mathrm{/}2+s{)}^{-1}$ and $\mathrm{\parallel }{R}_s(X){\mathrm{\parallel }}_2\le (m+s{)}^{-1}$. Therefore the remainder has norm at most $\mathrm{\parallel }H{\mathrm{\parallel }}_2^2{\int }_0^{\mathrm{\infty }}\frac{ds}{(m\mathrm{/}2+s)(m+s{)}^2}\le \frac{2\mathrm{\parallel }H{\mathrm{\parallel }}_2^2}{m^2}=o(\mathrm{\parallel }H{\mathrm{\parallel }}_2)\mathrm{.}$ This proves the real Fr'echet derivative formula $D\log (X)[H]={\int }_0^{\mathrm{\infty }}{R}_{s}H{R}_{s}ds$.

By the product rule [F5] and trace differential [F3], $D\mathrm{tr}(X\log X)[H]=\mathrm{tr}(H\log X)+{\int }_0^{\mathrm{\infty }}\mathrm{tr}(X{R}_{s}H{R}_s)ds$. The trace identity $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ follows by expanding the diagonal sums. Since $X$ commutes with $R_s$, it rewrites the integral as $\mathrm{tr}(H{\int }_0^{\mathrm{\infty }}X{R}_s^{2}ds)$. Diagonalizing $X$ and using ${\int }_0^{\mathrm{\infty }}x(x+s{)}^{-2}ds=1$ for each eigenvalue $x>0$ makes the matrix integral $I$. Thus $D\mathrm{tr}(X\log X)[H]=\mathrm{tr}(H(\log X+I))$.

Apply step 3.1 first with $X=D$ and direction $H$, then with $X=I-D$ and direction $-H$. The identity terms cancel, leaving $D\mathcal{S}(D)[H]=\mathrm{tr}(H(\log D-\log (I-D)))$.