Clarification about positive semidefiniteness

Question

I am trying to understand whether or not the product of two positive semidefinite matrices is also positive semidefinite. This topic has already been discussed in the past here. For me $A$ is positive definite" means $x^T A x > 0$ for all nonzero real vectors $x$, in this case @RobertIsrael gives a counterexample: $$ A = \pmatrix{ 1 & 2\cr 2 & 5\cr},\ B = \pmatrix{1 & -1\cr -1 & 2\cr},\ AB = \pmatrix{-1 & 3\cr -3 & 8\cr},\ (1\ 0) A B \pmatrix{1\cr 0\cr} = -1$$ However, then proceeds to prove that for $A$, $B$, positive semidefinite real symmetric matrices the result holds.

The proof is very short, quoting the answer: "Then $A$ has a positive semidefinite square root, which I'll write as $A^{1/2}$. Now $A^{1/2} B A^{1/2}$ is symmetric and positive semidefinite, and $AB = A^{1/2} (A^{1/2} B)$ and $A^{1/2} B A^{1/2}$ have the same nonzero eigenvalues."

So then the question is: what does it mean to be positive semidefinite real symmetric and why are $A$, $B$ not of this type?

Answer 1

The problem is not in $A$ and $B$, but in their product $AB$: the product is not symmetric; hence, there is no clear definition of positive definiteness.

In standard parlance, a Hermitian (or symmetric) matrix $M$ is positive definite if $x^T M x > 0$ for all $x$ (and this corresponds to $M$ having only positive eigenvalues). If $M$ is not symmetric, then $x^T M x$ may be zero or negative, even tho $M$ has only positive eigenvalues (as in the example by RobertIsrael).

So, for positive definiteness to hold, you need that $M = AB$ is Hermitian.

Answer 2

A real matrix $A$ is positive-semidefinite if $A$ is symmetric and $x^TAx \ge 0$ for all $x \in \mathbb{R}^n$.

Product of two positive-semidefinite matrices $A,B$ is again a positive-semidefinite matrix if and only if $AB = BA$.

Proof.

Assume $AB = BA$.

Then $(AB)^T = B^TA^T = BA = AB$ so $AB$ is symmetric.

$A$ has a (unique) positive-semidefinite square root $A^{1/2}$. Furthermore, $AB = BA$ implies $A^{1/2}B = BA^{1/2}$.

We have

$$x^TABx = x^TA^{1/2}A^{1/2}Bx = \left(A^{1/2}x\right)^TA^{1/2}Bx = \left(A^{1/2}x\right)^T B \left(A^{1/2}x\right) \ge 0$$

Therefore $AB$ is positive-semidefinite.

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Conversely, assume that $AB$ is positive-semidefinite. In particular, $AB$ is symmetric so

$$AB = (AB)^T = B^TA^T = BA$$

Hence $AB = BA$.