Let $A\in\mathbb{R}^{n\times n}$ be a symmetric positive semi-definite matrix, and $B\in\mathbb{R}^{n\times n}$ is symmetric negative semi-definite.
How to prove the eigenvalue of $AB$ is either zero or negative?
Let $A\in\mathbb{R}^{n\times n}$ be a symmetric positive semi-definite matrix, and $B\in\mathbb{R}^{n\times n}$ is symmetric negative semi-definite.
How to prove the eigenvalue of $AB$ is either zero or negative?
$-B$ is again nonnegative definite.
So it suffices to show the eigenvalues of $AB$ are nonnegative for $A, B$ nonnegative definite. We know the eigenvalues of $AB$ are the same as that of $A^{1/2}BA^{1/2}$, where $A^{1/2}$ is the unique square root of $A$. But $A^{1/2}BA^{1/2}$ is nonnegative definite....