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

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  • $\begingroup$ 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. $\endgroup$
    – MSMalekan
    Commented May 20, 2018 at 10:43
  • $\begingroup$ I think it is true iff T is positive $\endgroup$
    – Schüler
    Commented May 20, 2018 at 11:26
  • $\begingroup$ If $\sigma(T)=\{0\}$, then, sure, $T$ is positive. $\endgroup$
    – MSMalekan
    Commented May 20, 2018 at 11:36
  • $\begingroup$ Gould you please explain me why? $\endgroup$
    – Schüler
    Commented May 20, 2018 at 12:16
  • $\begingroup$ Because $T$ is positive iff $\sigma(T)\subset \mathbb R_{\geq0}$. $\endgroup$
    – MSMalekan
    Commented May 20, 2018 at 13:45

1 Answer 1

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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 $AB$.

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    $\begingroup$ 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. $\endgroup$
    – Jochen Glueck
    Commented May 12, 2018 at 22:16
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    $\begingroup$ 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) $AB = 0$. $\endgroup$
    – Nik Weaver
    Commented May 12, 2018 at 22:24
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    $\begingroup$ That makes things clearer for me. Thanks a lot! $\endgroup$
    – Jochen Glueck
    Commented May 12, 2018 at 22:48
  • $\begingroup$ Sure, no problem. $\endgroup$
    – Nik Weaver
    Commented May 12, 2018 at 23:02
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    $\begingroup$ Incidentally, this argument also shows that $\sigma(AB)$ is contained in $[0, \infty)$. $\endgroup$
    – Nik Weaver
    Commented May 13, 2018 at 2:02

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