I see on Wikipedia that the product of two commuting symmetric positive definite matrices is also positive definite. Does the same result hold for the product of two positive semidefinite matrices?

My proof of the positive definite case falls apart for the semidefinite case because of the possibility of division by zero…

**Answer**

You have to be careful about what you mean by “positive (semi-)definite” in the case of non-Hermitian matrices. In this case I think what you mean is that all eigenvalues are

positive (or nonnegative). Your statement isn’t true if “A is positive definite” means xTAx>0 for all nonzero real vectors x (or equivalently A+AT is positive definite). For example, consider

A=(1225), B=(1−1−12), AB=(−13−38), (1 0)AB(10)=−1

Let A and B be positive semidefinite real symmetric matrices. Then A has a positive semidefinite square root, which I’ll write as A1/2. Now A1/2BA1/2 is symmetric and positive semidefinite, and AB=A1/2(A1/2B) and A1/2BA1/2 have the same nonzero eigenvalues.

**Attribution***Source : Link , Question Author : nullUser , Answer Author : Robert Israel*