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Let $H$ be the Hilbert space and $T$ be bounded linear operator on $H$. If $$\langle T^2x,x\rangle =0, \forall x \in H \quad \text{and} \quad \langle Tx,x\rangle =0, \forall x \in H, $$ then $T=0$.

I thought so much about this problem but could not get any clue to tackle this problem. Someone give the hint to solve this one thank you..!!

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  • $\begingroup$ Is your Hilbert space over $\mathbb R$ or $\mathbb C$? I ask because in the $\mathbb C$ case you don't need the assumption that $\langle T^2x,x\rangle=0$ for all $x\in H$. $\endgroup$ – Aweygan Feb 9 '18 at 16:15
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    $\begingroup$ Note that $\langle T x , T^2x \rangle = 0$. Then $\forall$ $x$ $\in$ $H$ $\langle T x , Tx \rangle = \langle Tx, Tx + T^2 x \rangle = \langle x +Tx, Tx + T^2 x \rangle = \langle (x+ Tx), T (x + T x) \rangle = 0 $ $\Rightarrow T = 0$. What I don't understand is why $T$ need to be bounded. $\endgroup$ – Matheus Manzatto Feb 9 '18 at 16:56
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Hint : It is enough to prove that $$\langle Tx,Tx\rangle=0.$$ To show this, you can add suitable terms that are orthogonal to $Tx$ into the inner product to obtain a term of the form $\langle Ty,y\rangle$ for some $y$.

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  • $\begingroup$ Thanks Arnaud.. :) $\endgroup$ – Chiranjeev_Kumar Feb 10 '18 at 4:45

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