= Klein's inequality
{c}
{title2=$\operatorname{Tr}[A(\ln A-\ln B)]\geq\operatorname{Tr}(A-B)$}
For positive-definite <Hermitian matrices> $A,B$, Klein's inequality gives
$$
\operatorname{Tr}[A(\ln A-\ln B)]\geq\operatorname{Tr}(A-B),
$$
with equality exactly when $A=B$. To prove it, let $a_i,b_j$ be their <eigenvalues> and $u_i,v_j$ their orthonormal eigenvectors. The weights $w_{ij}=|\langle u_i,v_j\rangle|^2$ have row and column sums one. The difference between the two sides is
$$
\sum_{i,j}w_{ij}\left[a_i\ln\frac{a_i}{b_j}-a_i+b_j\right]\geq0
$$
by the scalar <logarithm inequality> $\ln t\leq t-1$. Equality requires $a_i=b_j$ whenever $w_{ij}>0$, implying $A=B$. Limits extend the result to positive semidefinite matrices with the appropriate support condition. Applying it to trace-one matrices proves <nonnegativity of quantum relative entropy>.
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