Extend the <quantum conditional entropy> from normalized states to positive operators by
$$
F(X_{AB})
=-\operatorname{Tr}(X_{AB}\log X_{AB})
+\operatorname{Tr}(X_B\log X_B).
$$
Since $\operatorname{Tr}X_{AB}=\operatorname{Tr}X_B$, the two terms involving $\log t$ cancel under $X\mapsto tX$, so $F(tX)=tF(X)$. Thus $F$ is <positively homogeneous>, and part (b) extends its <concavity> from states to the positive cone.

Apply part (c) with $X=\sigma_{AB}$ and $Y=\rho_{AB}$. Differentiating the <matrix logarithm> under the trace gives
$$
\left.\frac d{dt}\right|_{t=0}F(\sigma+t\rho)
=-\operatorname{Tr}(\rho\log\sigma)
+\operatorname{Tr}(\rho_B\log\sigma_B).
$$
The inequality from part (c), after moving $F(\rho)$ to the left, becomes
$$
\boxed{
D(\rho_{AB}\|\sigma_{AB})
\geq D(\rho_B\|\sigma_B)}.
$$
This is the <data-processing inequality for quantum relative entropy> under <partial trace>. Tensoring each output with the appropriate maximally mixed state does not change either side, so it also proves data processing under normalized partial traces. Singular $\sigma$ follows by approximation on its support.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 323 / 3 / ii / d / Solution

{title2=$\log A$}
{wiki}

For a positive-definite <Hermitian matrix> with <spectral decomposition> $A=U\operatorname{diag}(\lambda_j)U^*$, the matrix logarithm is $\log A=U\operatorname{diag}(\log\lambda_j)U^*$. More generally it is a branch of the inverse of the <matrix exponential> when such a branch exists.

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 Incoming links: Matrix logarithm