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If $A$ is a Hermitian operator on an infinite dimensional Hilbert space such that $\langle Ax|x\rangle=0$ for all vectors $x$, can we prove $A=0$ without the spectral theorem? The proof seems straightforward if we diagonalise $A$ into its eigenbasis, but can we do this without resorting to such a 'powerful' result?

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You can use the trick which lets you recover a scalar product from the associated norm. Let $A$ be as in your question, and $x$, $y$ be any elements of the Hilbert space. Then:

$$0 = \langle A(x+y)|x+y\rangle = \langle Ax|x\rangle + \langle Ay|y\rangle+\langle Ax|y\rangle+\langle Ay|x\rangle = \langle Ax|y\rangle+\langle Ay|x\rangle.$$

Since $A$ is Hermitian, we get $\langle Ax|y\rangle = -\langle Ay|x\rangle=-\langle y|Ax\rangle = -\overline{\langle Ax|y\rangle}$, so $\langle Ax|y\rangle$ is pure imaginary for all $x$ and $y$. But then $\langle Ax|iy\rangle = i \langle Ax|y\rangle$ is also pure imaginary, so $\langle Ax|y\rangle = 0$ for all $x$ and $y$.

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In the real case it is especially easy. Expand the right hand side of the following: $$ 0=\langle A(x+y), x+y\rangle, $$ you get $2\langle Ax, y\rangle=0$ for all $x, y\in H$. In the complex case you have to play a little bit more because you arrive at $2\Re \langle Ax, y\rangle=0$, but that's not a serious issue.

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