If the expectation $\langle v,Mv \rangle$ of an operator is $0$ for all $v$ is the operator $0$? I ran into this a while back and convinced my self that it was true for all finite dimensional vector spaces with complex coefficients. My question is to what extent could I trust this result in the infinite dimensional case. If there is a circumstance where it is not true then I would like the corresponding counter example.
The statement is $$\langle v| \textbf{M} v \rangle = 0 \quad \forall \mid v\rangle \in \mathbb{V} \Rightarrow \textbf{M}=0$$


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*$\textbf{M}$ is a linear operator on $\mathbb{V}$. 

*$\mathbb{V}$ is a vector space over the field $\mathbb{C}$. 


In the case of finite dimensions we can see that this is true by going to the eigenbasis of $\textbf{M}$. In that case if $\textbf{M} \mid \lambda_i \rangle  = \lambda_i \mid \lambda_i \rangle $ then $\lambda_i \langle \lambda_i \mid \lambda_i \rangle= \langle \lambda_i \mid \textbf{M} \lambda_i \rangle = 0 \quad \forall i$. This means the diagonal form of $\textbf{M}$ is the zero matrix. This is assuming that the eigen vectors of $\textbf{M}$ can form a basis for the space which I believe is always true in finite dimensional vector spaces with complex coefficients because of the spectral theorem.
Thanks ahead of time.

This is true in general for all vector spaces over $\mathbb C$. Indeed, one can consider $v = x-y$ and $v=x-iy$ to come up with the identity $\langle x, My \rangle=0$ for all $x, y$.
