Multiplication of a symmetric matrices [closed]

Question

I'm wonder if the next claim is true or not:
If A,B is a symmetric matrices over the real numbers, and A is PSD , B is PD implies than AB is PSD. PD - positive definite
PSD - positive semidefinite
If it is true , how can i prove it? and if it wrong , I wonder if there is some examples that shows that.

Answer 1

It is unclear whether you claim, or not, that $AB$ is symmetric. In general, the product of symmetric matrices is not symmetric.

But the following is true and classical (see for instance my book Matrices, Springer GTM216).

If $A$ is symmetric positive definite and $B$ is symmetric, then the product $AB$ (even if not symmetric) is diagonalizable with real eigenvalues $\lambda_1\le\cdots\le\lambda_n$. The distribution of the signs ($-,0,+$) of the $\lambda_j$'s is the same as that of the signs of the eigenvalues $b_1\le\cdots\le b_n$ of $B$.

In particular, if $B$ is positive semi-definite, then the eigenvalues of $AB$ are non-negative. By continuity, if bot $A,B$ are PSD, then the eigenvalues of $AB$ are non-negative, though it might not be diagonalizable.

The proof uses the fact that $AB$ is similar to $A^{1/2}BA^{1/2}$. The latter is symmetric, hence diagonalisable, and represents the same quadratic form as $B$, though in a different basis.