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!

Solution 1:

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.)