Spectrum of the product of operators

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 $$\sigma(AB)\neq\{0\}\;.$$

• For an element $T\in B(F)$, $\sigma(T)=\{0\}$ iff $T=0$, this is because the identity function on $\{0\}$ is the zero function. – Meisam Soleimani Malekan May 20 '18 at 10:43
• I think it is true iff T is positive – Schüler May 20 '18 at 11:26
• If $\sigma(T)=\{0\}$, then, sure, $T$ is positive. – Meisam Soleimani Malekan May 20 '18 at 11:36
• Gould you please explain me why? – Schüler May 20 '18 at 12:16
• Because $T$ is positive iff $\sigma(T)\subset \mathbb R_{\geq0}$. – Meisam Soleimani Malekan May 20 '18 at 13:45

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 $AB$.
• Could you explain why $AB \not= 0$ implies $A^{1/2}BA^{1/2} \not= 0$? I assume it has something to do with the positive semi-definiteness of $B$, since this is not correct if $B$ is merely self-adjoint. – Jochen Glueck May 12 '18 at 22:16
• Sure. We have $(A^{1/2}B^{1/2})(A^{1/2}B^{1/2})^* = A^{1/2}BA^{1/2}$. So $A^{1/2}BA^{1/2} = 0$ implies $A^{1/2}B^{1/2} = 0$ implies (multiplying by $A^{1/2}$ on the left and $B^{1/2}$ on the right) $AB = 0$. – Nik Weaver May 12 '18 at 22:24
• Incidentally, this argument also shows that $\sigma(AB)$ is contained in $[0, \infty)$. – Nik Weaver May 13 '18 at 2:02