How to prove the existence of such a non-zero operator I am self studying  a 2nd course in Functional Analysis using Functional analysis by Rudin and a set of exercises.
I was not able to solve this particular problem and I need help.

Question: Let H be a Hilbert space of dimension $\geq 2$. Show that there exists a non -zero operator $T\in L(H)$ st $\langle Tx,x\rangle  =0   $ for every $x\in H$.

On the converse, let if there exists  $ T\in L(H)$ such that $\langle Tx,x\rangle =0$ for every $x\in H$ then $T \equiv 0$.
But what can be contradiction here?
Kindly help.
 A: I'm not looking at Rudin right now, but the difference is whether you consider a real or a complex Hilbert space.
In a real space, you use the idea that the matrix $A=\begin{bmatrix} 0&1\\-1&0\end{bmatrix}$ satisfies $\langle Ax,x\rangle=0$ for all $x\in \mathbb R^2$. So, given an orthonormal basis $\{e_n\}$, define $T$ as the linear operator that does
$$
Te_1=e_2,\qquad Te_2=-e_1,\qquad Te_{k+2}=0,\qquad\qquad k\in\mathbb N. 
$$
Then, for any $x$
$$
\langle Tx,x\rangle=x_2x_1-x_1x_2=0.
$$
In a complex Hilbert space, the above doesn't work. If $\langle Tx,x\rangle=0$ for all $x$, then $\langle Tx,y\rangle=0$ for all $x,y$ by polarization, and so $T=0$.

Edit: the Polarization Identity is the equality
$$\tag1
\langle Tx,y\rangle=\frac14\,\sum_{k=0}^3i^k\langle T(x+i^ky),x+i^k y\rangle. 
$$
A: On a complex preHilbert space, you are right: if for all vectors $v\in H$, $\langle Tv,v\rangle=0,$ then $T=0.$
But on a real Hilbert space of dimension $\ge2$, take a $\pi/2$-rotation on some plane $P$, and complete by $0$ on $P^\perp.$
