TITLE: To show operator $T=0$ where $T\in B(H)$
QUESTION [0 upvotes]: If $T$ is bounded linear operator $T:H\to H$, and for all $x,y$ in $H$, we have $<Tx,y>=0$, then can I conclude that $T=0$. I am struggling to find mathematical reasoning for this. Any help. Thanks.

REPLY [1 votes]: Assume $T$ is non zero. Then there should exist $x$ such that $Tx\neq 0$. Since the norm is positive define we should have
$$0\neq \|Tx\|^2=\langle Tx,Tx\rangle$$
but since $\langle Tx,y\rangle=0$ for all $y$ this must also hold for $y=Tx$, so the above should be zero.
This is a contraction, thus $T$ must be zero.
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