# Is the composition of positive, self-adjoint operators on a Hilbert space positive?

Let $$A,B$$ be self-adjoint, bounded, linear operators on a Hilbert space $$H$$ over $$\mathbb{R}$$ such that $$\langle Ax,x\rangle\geq 0$$ and $$\langle Bx,x\rangle\geq 0$$ for all $$x\in H$$. Does it also hold that $$\langle ABx,x\rangle\geq 0$$ for all $$x\in H$$?

• The answer is no, but the details depend on whether you're talking about Hilbert spaces over $\Bbb R$ or over $\Bbb C$. Over $\Bbb R$, some care is required in constructing $A,B$ such that $\langle ABx ,x \rangle < 0$. Commented Jan 23 at 15:21
• In the simplest case, just take $H$ to be a two-dimensional space (either over $\mathbb{C}$ or over $\mathbb{R}$), then take $A = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}$. Commented Jan 23 at 19:35
• You're right, thank you. I mean Hilbert spaces over $\mathbb{R}$ and updated the question. I think for $𝐴𝐵$ to be positive more assumptions are needed, for instance, that $𝐴$ and $𝐵$ commute. Does it maybe also work, if $A$ is strictly positive?
– Suim
Commented Jan 24 at 9:54
• @Suim Assuming $A$ is strictly positive does not work, just change $A$ to be $\begin{pmatrix} 2 & 0\\0 & 1\end{pmatrix}$ in the example. Similarly, assuming both $A$ and $B$ are strictly positive does not work either. As Disintegrating By Parts pointed out in their answer, this happens if and only if $A$ and $B$ commute. Commented Jan 25 at 18:59

Assume that $$A,B$$ are positive operators. A positive operator is self-adjoint. So, if $$AB \ge 0$$, then it is necessary that $$(AB)^*=AB$$, or $$BA=AB$$. In that case $$AB \ge 0$$ because, you may write $$A=\sqrt{A}^2$$ where $$\sqrt{A}$$ is the unique positive square root of $$A$$ that commutes with everything that commutes with $$A$$. Then you end up with $$\sqrt{A}B=B\sqrt{A}$$ and $$AB=\sqrt{A}\sqrt{A}B=\sqrt{A}B\sqrt{A} \ge 0.$$