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Let $T$ be a self-adjoint operator on a Hilbert space $H$. If for all $x\in H$, $\langle Tx,x\rangle=0$, is $T=0$?

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  • $\begingroup$ Yes. For a self-adjoint operator, we even have $\|T\|=\sup_{\|x\|\leq 1}|(Tx,x)|$. This follows from polarization tricks. Note that your statement is true in the complex case without even assuming $T$ is self-adjoint. $\endgroup$ – Julien May 3 '13 at 15:01
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If the ground field is $\mathbb{R}$, then note that $$ \langle Tx,y\rangle=2^{-2}\sum\limits_{k=0}^1 i^{2k}\langle T(x+i^{2k}y),x+i^{2k}y\rangle $$ If the ground field is $\mathbb{C}$, then note that $$ \langle Tx,y\rangle=2^{-2}\sum\limits_{k=0}^3 i^{k}\langle T(x+i^{k}y),x+i^{k}y\rangle $$ In both cases you get $\langle Tx,y\rangle=0$ for all $x,y\in H$. This implies $T=0$.

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