TITLE: $\langle Tu\;|\;u\rangle=0,\;\forall u\in E \Longrightarrow T=0$?
QUESTION [10 upvotes]: Let $E$ be  a complex Hilbert space. Let $T\in \mathcal{L}(E)$.
I have two questions:

Why it is not true that for an arbitrary operator $T\in \mathcal{L}(E)$, we have $\langle Tu\;|\;u\rangle=0,\;\forall u\in E \Longrightarrow T=0$? And is  this property true for normal operators?

I think it is true for self adjoint operators because the norm of a self adjoint operators is given by
$$\left\|T\right\|= \sup\big\{\;\left|\langle Tu\;|\;u\rangle \right|,\;\;u \in E\;, \left\| u \right\| = 1\;\big\}$$
Thank you.

REPLY [5 votes]: On a complex Hilbert space $H$ the following proposition holds:

Let $T : H \to H$ be a bounded linear map such that $\langle Tx, x\rangle \in \mathbb{R}$ for all $x \in H$. 
Then $T^* = T$, i.e. $T$ is self-adjoint.

Proof:
We have $\langle Tx, x\rangle = \overline{\langle x, Tx\rangle} = \langle x, Tx\rangle$ for all $x\in H.$ 
\begin{align}
4\langle Tx, y\rangle &= \langle T(x+y), x+y\rangle - \langle T(x-y), x-y\rangle + i\langle T(x+iy), x+iy\rangle - i\langle T(x-iy), x-iy\rangle\\
&= \langle x+y, T(x+y)\rangle - \langle x-y, T(x-y)\rangle + i\langle x+iy, T(x+iy)\rangle - i\langle x-iy, T(x-iy)\rangle\\
&= 4\langle x, Ty\rangle
\end{align}
Hence $\langle Tx, y\rangle = \langle x, Ty\rangle$ for all $x, y \in H$ so $T^* = T$.
Now in your case we have $\langle Tx, x\rangle = 0 \in \mathbb{R}$ so in particular $T^* = T$.
Using the formula you gave for the norm of a self-adjoint operator we obtain:
$$\|T\| = \sup_{\|x\| = 1} |\langle Tx, x\rangle| = 0 \implies T = 0$$