In my reference, Box 11.2, Page 512, Chapter 11, Entropy and Information, Quantum Computation and Quantum Information by Nielsen and Chuang, proof of the Fannes' inequality contains
here $\rho$ and $\sigma$ are positive semidefinite(hermitian with nonzero eigenvalues) with eigenvalues lies in the interval $[0,1]$(constitute a probability distribution).
In another document, Theorem 10.2, Lecture 10: Continuity of von Neumann entropy;
quantum relative entropy, John Watrous' Lecture notes, it is given that
How can we prove that any hermitian operator can be decomposed in terms of the difference between two positive semidefinite operators?
Here, $A=VD_AV^\dagger$ and $B=UD_BU^\dagger$ need not commute (not simultaneously diagonalized).
The difference between two hermitian matrices is hermitian, ie., $H=A-B$ is hermitian.
$H=A-B=WD_HW^\dagger$, and we are free to choose diagonal matrices $D_p$ and $D_q$ such that $D_H=D_p-D_q$, i.e., $H=WD_HW^\dagger=W(D_p-D_q)W^\dagger=WD_pW^\dagger-WD_qW^\dagger=P-Q$
It would be helpful if one could point in the right direction.
Best Answer
Since $\rho - \sigma$ is Hermitian, you can diagonalize it and all its eigenvalues fall between $[-2,2]$ by the triangle inequality. Let $Q$ be the diagonal matrix whose eigenvalues are exactly the non-negative eigenvalues of $\rho - \sigma$ and let $R$ be the diagonal matrix whose eigenvalues are the negatives of the negative eigenvalues of $\rho - \sigma$. So, both $Q$ and $R$ are positive and if you've ordered the eigenvalues right, $Q-R = \rho-\sigma$ after a suitable unitary change of basis. Note that $Q$ and $R$ will have orthogonal support since $Q$ only supports the positive eigenspace and $R$ only supports the negative eigenspace, and these are obviously orthogonal (no eigenvalue is both negative and positive).