Proving that if $\langle Ax,x\rangle =0$ for every $x$, then $A$ is the zero operator I feel kind of dumb but I needed this little claim as a part of a proof I'm writing, and I figured out that I'd better just ask, since I could not find the correct algebraic manipulation needed in order to prove it.
So I want to prove this: Suppose $H$ is an Hilbert space, and $A: H\to H$ is a linear operator (if that matters, which I believe it does not, we can assume that $A$ is a projection).
Suppose that for every $x\in H$, $(Ax,x)=0$. Then $A=0$.
What I tried to do is a little bit of Cauchy-Schwarz and some algebra but nothing worked out.
Thanks!
 A: For general operators $A$, the condition
$$\bigl(\forall x\in H\bigr)\bigl(\langle Ax,x\rangle = 0\bigr)$$
only implies $A = 0$ if $H$ is a $\mathbb{C}$ vector space, not if $H$ is a $\mathbb{R}$ vector space.
For projections however (or for self-adjoint operators), the implication also holds for real spaces: Pick an $x \in \operatorname{im} A$, say $x = Ay$. Then
$$0 = \langle Ax,x\rangle = \langle A^2y,Ay\rangle = \langle Ay,Ay\rangle = \lVert Ay\rVert^2 = \lVert x\rVert^2$$
shows $\operatorname{im} A = \{0\}$.
For self-adjoint operators on real inner product spaces or general operators on complex spaces, the proof uses the respective polarisation identity. For self-adjoint operators on real spaces, we have for any $x,y\in H$
$$0 = \langle A(x+y),x+y\rangle - \langle A(x-y),x-y\rangle = 2\langle Ax,y\rangle + 2\langle Ay,x\rangle \stackrel{\text{s.a.}}{=}2\langle Ax,y\rangle + 2\langle y,Ax\rangle = 4\langle Ax,y\rangle$$
and hence $Ax = 0$ for all $x$.
In the complex case, for all $x,y\in H$:
\begin{align}
0 &= \langle A(x+y),x+y\rangle - \langle A(x-y),x-y\rangle + i\langle A(x+iy),x+iy\rangle - i\langle A(x-iy),x-iy\rangle\\
&=\quad \langle Ax,x\rangle + \langle Ax,y\rangle + \langle Ay,x\rangle + \langle Ay,y\rangle\\
&\quad -\langle Ax,x\rangle + \langle Ax,y\rangle + \langle Ay,x\rangle - \langle Ay,y\rangle\\
&\quad +i\langle Ax,x\rangle + i\langle Ax,iy\rangle + i\langle Aiy,x\rangle + i\langle Aiy,iy\rangle\\
&\quad - i\langle Ax,x\rangle +i\langle Ax,iy\rangle +i\langle Aiy,x\rangle -i\langle Aiy,iy\rangle\\
&= 2 \langle Ax,y\rangle + 2i\langle Ax,iy\rangle + 2\langle Ay,x\rangle + 2i\langle Aiy,x\rangle\\
&= 2\bigl(\langle Ax,y\rangle + i(-i)\langle Ax,y\rangle\bigr) + 2\bigl(\langle Ay,x\rangle + i\cdot i\langle Ay,x\rangle\bigr)\\
&= 4\langle Ax,y\rangle,
\end{align}
again showing that $Ax = 0$ for all $x\in H$. (Note: I used the convention that the inner product is linear in the first argument and antilinear in the second; for the other convention, in the penultimate line the signs of the scalar pulled out of the inner product would be opposite, and the end result would be $4\langle Ay,x\rangle$, which evidently gives the same conclusion.)
