The answer is yes! If H and K are two hermitian matrices that are PSD, then HK can be diagonalized and has positive eigenvalues. This follows from the following proposition by Denis Serre (Matrices Theory and Applications, GTM 216, Proposition 6.1): Prop: If H and K are two Hermitian matrices with H positive symmetric definite, then HK is diagonalizable with real eigenvalues. Furthermore, the number of positive eigenvalues of HK is equal to that of K. Proof. Since H is hermitian, it can be diagonalized w.r.t a unitary matrix U and its eigenvalues are real: H = U^* diag(mu_1, ..., mu_n) U. Because H is PSD, mu_i > 0 so if we let S = U^* diag(sqrt(mu_1), ..., sqrt(mu_n)) U, we have S^2 = H and S is also Hermitian PSD. Observe that HK = SS K SS^{-1} = S(SK S) S^{-1} = S(S^* K S) S^{-1}. Therefore, HK and S^*KS have the same eigenvalues and since S^* K S is hermitian, it can be diagonalized and it has real eigenvalues, and so the same holds for HK. The number of positive eigenvalues of S^*KS is the largest dimension of a subspace of complex^n on which the hermitian form x |-> x^* S^* K S x is positive definite, which means that S\complex^n is the subspace on which the hermitian form y |-> y^*Ky is positive definite. Since S is nonsingular, this max dimension is the same. Best, -- Jean