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I want to show that for $S,T\in B(H)$ bounded operators on Hilbertspace with $S\geq 0,T\geq 0$ and $ST=TS$, we have $S+T\geq 0$, and $ST\geq 0$.

$T\geq 0$ means $(Tx,x)\geq 0$.

To me it seems that $((S+T)x,x) = (Sx,x)+(Tx,x)\geq 0$. But for $ST$ i like some help.

There is a theorem in Rudin (12.32) that says that for every $T\in B(H)$ we have an equivalence:

$T$ is positive $\Leftrightarrow$ $T$ is self adjoint, i.e. $T=T^*$ and $\sigma(T)\subset [0,\infty)$.

Since $S, T$ are self adjoint we know that $ST=TS = S^*T^* = T^*S^*$ ($ST$ is selfadjoint since $S,T$ commute). Also I read somewhere that if $S,T $ commute then $\sigma(ST)\subset \sigma(S)\sigma(T)$, but I am not sure if this is true. Clearly I cant prove it, but if it were true by this ''Theorem'' it would follow that $ST \geq 0$.

Is there a more straightforward way?

DinkyDoe
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I would use $ST = S^{1/2} T S^{1/2}$. But I don't know if you have done square roots of positive operators yet.

Stephen Montgomery-Smith
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  • I have seen them. But, I have no idea how this decomposition with square roots helps? Actualy "every positive operator has a square root" is the next Theorem in Rudin 12.33 :P – DinkyDoe Nov 27 '13 at 23:22
  • But $T$ has a square root too, so is it right that $ST = S^{1/2}T^{1/2}S^{1/2}T^{1/2}$ so they all commute and $ST = (S^{1/2}T^{1/2})^2$. Can we conclude then $ST \geq 0$? :P – DinkyDoe Nov 27 '13 at 23:30
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    $T$ does have a square root. So you could use your argument, say that $S^{1/2}T^{1/2}$ is self adjoint, and then say its square is positive. – Stephen Montgomery-Smith Nov 28 '13 at 00:17
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    But what I was thinking was $\langle S^{1/2}TS^{1/2}x,x\rangle = \langle Ty,y\rangle$ where $y = S^{1/2} x$. – Stephen Montgomery-Smith Nov 28 '13 at 00:18
  • Ha nice, that works too :--) – DinkyDoe Nov 28 '13 at 02:00