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Let $A$ be a Hermitian Matrix with real entries and $B$ be an anti Hermitian matrix with real entries. i.e., $A^{\dagger}=A$ and $B^{\dagger}=-B$. In case of real entries $\dagger$ is replaced by matrix transpose operation $T$. I am interested in finding the eigenvalues of the matrix $(A-B)(A+B)$ given by the eigenvalue equation

\begin{equation} (A-B)(A+B)\phi^{T}=\Lambda^{2}\phi^{T} \end{equation} Are the Eigenvalues $\Lambda$ always positive? In other words are the matrices $(A-B)(A+B)$ (semi)positive definite?

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The eigenvalues are always real and nonnegative, because $(A-B)(A+B) = (A+B)^\dagger (A+B)$. Any matrix of the form $C^\dagger C$ is positive semidefinite.

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