Let \(A,B\) be Hermitian \(n\times n\) complex matrices, and suppose \(A\) is positive definite. Prove that every eigenvalue of \(AB\) is real.
Solution
1. The matrix \(A\) has an invertible Hermitian positive-definite squareroot \(S=A^{1/2}\).
By the spectral theorem, there is a unitary matrix \(U\) and positive real numbers \(\lambda_1,\ldots,\lambda_n\) such that \[ A = U\operatorname{diag}(\lambda_1,\ldots,\lambda_n)U^*. \] Set \[ S \coloneqq U\operatorname{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n})U^*. \] Then \(S=S^*\), every eigenvalue of \(S\) is positive, \(S\) is invertible, and \(S^2=A\).
2. The matrix \(AB\) is similar to \(SBS\).
Since \(A=S^2\) and \(S\) is invertible, \[ S^{-1}(AB)S = S^{-1}S^2BS = SBS. \]
3. The matrix \(SBS\) is Hermitian.
Because both \(S\) and \(B\) are Hermitian, \[ (SBS)^* = S^*B^*S^* = SBS. \]
4. Every eigenvalue of \(AB\) is real.
By step 3 and the spectral theorem, every eigenvalue of the Hermitian matrix \(SBS\) is real. Similar matrices have the same characteristic polynomial and therefore the same eigenvalues. Step 2 now gives \[ \boxed{\text{every eigenvalue of }AB\text{ lies in }\RR}. \]
5. Q.E.D.
Step 4 is the required conclusion.