If $A,B$ are positive semidefinite matrices show that all eigen values of $AB$ are non-negative.

Question

If $A,B$ are positive semidefinite matrices show that all eigen values of $AB$ are non-negative.

Since $A,B$ are positive semidefinite then $x^TAx,x^TBx\ge 0\forall x.$ Also all eigen values of both $A,B$ are non-negative.

How to show from here that $AB$ has all eigen values non-negative?

Please help.

Answer 1

Recall for any $n\times n$ matrices $X,Y$, the products $XY$ and $YX$ have same characteristic polynomial. i.e.

$$\det(\lambda I_n - XY) = \det(\lambda I_n - YX)$$ This means $XY$ and $YX$ share same set of eigenvalues (with same multiplicity).

Since $B$ is positive semi-definite, it has a positive semi-definite square root $\sqrt{B}$. A consequence of this is $AB = A\sqrt{B}^2$ and $\sqrt{B}A\sqrt{B}$ share same set of eigenvalues. Notice for any $x \in \mathbb{R}^n$, $$x^T \sqrt{B}A\sqrt{B} x = x^T \sqrt{B}^T A \sqrt{B} x = (\sqrt{B} x)^T A \sqrt{B}x \ge 0$$

This implies $\sqrt{B}A\sqrt{B}$ is positive semi-definite. As a result all eigenvalues of $\sqrt{B}A\sqrt{B}$ and hence all eigenvalues of $AB$ are non-negative.

Answer 2

If we can prove $AB$ is positive semi-definite, then we're done.

Suppose $A,B$ are positive semi-definite matrices. Then $x^TAx,x^TBx\ge 0$ for all real vectors $x$. $\sqrt{A}$ is also positive semi-definite. Write $AB=\sqrt{A}\sqrt{A}B$. Note $Q=v^T\sqrt{A}B\sqrt{A}v = (\sqrt{A}v)^T B (\sqrt{A}v)$, since $\sqrt{A}$ is symmetric. $Q=w^TBw\geq0$, where $w=\sqrt{A}v$, because by assumption $B$ is positive semi-definite. One can conclude that $AB$ and $\sqrt{A}B\sqrt{A}$ (proven to be positive semi-definite) have same non-negative eigenvalues. It follows that $AB$ is positive semi-definite and eigenvalues of $AB$ are non-negative.

Another way to approach this, is by Eigendecomposition. (Suppose further $AB $ is diagonalisable.)