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



> 
> Let $A,B\in \mathcal{B}(F)^+:=\left\{T\in \mathcal{B}(F);\,\langle Tx, x\rangle\geq 0,\;\forall\;x\in F\;\right\}$, be such that $AB\neq 0$. I want to show that 
>  $$\sigma(AB)\neq\{0\}\;.$$
> 
> 
> 



Spectrum of the product of operators
Use the fact that $\sigma(ST) \cup \{0\} = \sigma(TS) \cup \{0\}$. So if $A$ and $B$ are positive then, except possibly for the point $0$, $\sigma(AB)$ equals $\sigma(A^{1/2}BA^{1/2})$. If $AB \neq 0$ then the latter is a nonzero positive operator, and hence it has a nonzero element in its spectrum. Therefore so does $AB$.


