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Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.

Let $$\mathcal{B}(F)^+=\left\{T\in \mathcal{B}(F);\,\langle Tx, x\rangle\geq 0,\;\forall\;x\in F\;\right\}$$

Let $A,B\in\mathcal{B}(F)^+$ such that $A\neq 0$ and $B\neq 0$. It is possible that $AB=0$ or $BA=0$?

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Yes. Suppose $Y$ is a non-trivial closed subspace of $F$ and recall that we have $F = Y \oplus Y^\perp$. Further the orthogonal projections $P_Y$ and $P_{Y^\perp}$ onto $Y$ and $Y^\perp$ respectively are bounded, positive operators and since $Y \subset \ker P_{Y^\perp}$ and $Y^\perp \subset \ker P_Y$ we have $P_Y P_{Y^\perp} = 0 = P_{Y^\perp} P_Y$.

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