Product of two positive compact, self adjoint operators

Question

If we have two positive compact , self adjoint operators; $A$, $B$. Is the product $AB$ a positive operator?

Answer 1

No. Counterexample: $$ A = \pmatrix{ 1 & 2\cr 2 & 5\cr},\ B = \pmatrix{1 & -1\cr -1 & 2\cr}\\ AB = \pmatrix{-1 & 3\cr -3 & 8\cr}\\ \left\langle AB\pmatrix{1\\0},\pmatrix{1\\0} \right\rangle = \pmatrix{1&0} A B \pmatrix{1\cr 0\cr} = -1 $$ (example taken from here).

Answer 2

I guess that "positive" here means that the quadratic form is positive, right? In this case, I guess that the proposition is false unless $[A, B]=0$. You can find counterexamples even in finite dimensional spaces, so the compactness assumption does not help you much.

Answer 3

The product of positive operators is not even selfadjoint in general. For instance, $$ \begin{bmatrix}1&0\\0&0\end{bmatrix}\,\begin{bmatrix}1&1\\1&1\end{bmatrix}=\begin{bmatrix}1&1\\0&0\end{bmatrix}, $$ while $$ \begin{bmatrix}1&1\\1&1\end{bmatrix}\,\begin{bmatrix}1&0\\0&0\end{bmatrix}=\begin{bmatrix}1&0\\1&0\end{bmatrix}. $$ Note that since $A,B$ positive implies that they are selfadjoint, $AB$ positive implies that $AB=BA$. So the product of any two non-commutative positives will not be positve. In finite dimension this means that two positives commute if and only if they are simultaneously diagonalizable.