In Theorem 3 in this paper, https://core.ac.uk/download/pdf/82822897.pdf, ``On a product of positive semidefinite matrices, A.R. Meenakshi, C. Rajian, Linear Algebra and its Applications, Volume 295, Issues 1–3, 1 July 1999, Pages 3–6." the following has been proven,
Theorem : If $A$ and $B$ are PSD, $AB$ is PSD iff $AB$ is normal, ie, $(AB)^T AB = AB(AB)^T$.
But we need to note : that in this paper, a real square matrix $A$ is said to be PSD if $\exists$ another matrix $Q$ s.t $A= QQ^T$ i.e their PSD matrices are somewhat special - as in their PSD matrix $A$ not only have all non-negative eigenvalues and but its also true that for all vectors $\vec{v}$ we would have, $\vec{v}^TA\vec{v} \geq 0$.
My question is 3 fold,
Does it follow from above (or anything else in the paper!) that ``The product of two symmetric PSD matrices is PSD, iff the product is also symmetric - if PSD is defined as real matrices $A$ s.t for all vectors $v$, $v^TAv \geq 0$" (...Or does this need an independent proof? If yes, what?...)
But we do know that there exists symmetric PSD matrices like $A = [[1,2],[2,5]]$ and $B=[[1,-1],[-1,2]]$ s.t $AB = [[-1,3],[-3,8]]$ is PSD (but not symmetric) in the sense of having only non-negative eigenvalues but $AB$ is not PSD in the sense that for $\vec{v} = [1,0]^T$, $\vec{v}^T (AB) \vec{v} = -1$. So this is an example where in the non-negative eigenvalue sense a product of symmetric PSD matrices can be PSD while the product being assymetric.
Is there a more general theorem than the quoted paper which encompasses the above kind of example?
Let $A$ and $B$ be two PSD matrices where B is assymmetric but feel free to assume that A is symmetric if you need to. Now under what combinations of (definitions of PSD + conditions on A,B) can one say that AB is also PSD?
