Let $H$ be a complex Hilbert space. Let $T\in B(H)$ be such that $\langle T x,x\rangle = 0$ for all $x\in H$. Prove that $T=0$. Here $\langle\cdot,\cdot\rangle$ is the inner product, and $B(H)$ is the space of bounded linear operators.
How do I show $T$ that is the $0$ operator? I thought of considering $x=au+bv$ in $H$ and then $$T(x)=T(au+bv)=aT(u)+bT(v)$$
Hints would be great.
We assume that $\langle Tx, y \rangle = 0$ for all $x,y \in H$ implies $T=0$.
Consider $$ \begin{aligned} 0 &= \langle T(\alpha x + y), \alpha x + y\rangle \\ &= |\alpha|^2 \langle Tx, x\rangle + \alpha \langle T x, y \rangle + \alpha^* \langle T y, x \rangle + \langle T y, y \rangle \\ &= \alpha \langle T x, y\rangle + \alpha^* \langle T y, x \rangle, \end{aligned} $$ which holds for all $\alpha \in \mathbb{C}$. Now together the cases $\alpha = 1$ and $\alpha = i$ imply that $\langle T x, y \rangle = 0$ for all $x,y \in H$.
Note that we require the Hilbert space to be over $\mathbb{C}$ for this to hold however. For a counterexample consider the Hilbert space $\mathbb{R}^2$ with the standard Euclidean inner product and take $$ T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$ then for $x = (x_1,x_2)$ we have $Tx = (-x_2,x_1)$ and so $\langle Tx, x \rangle = -x_2 x_1 + x_1 x_2 = 0$ but $T \neq 0$.
