# In a complex vector space with inner product: If $\langle T(v),v \rangle=0$ for any $v \in V$ then $T=0$

I'm having a hard time trying to prove the proposition of the title. In the accepted answer to this post written by Christopher A. Wong he mentions the following variation of the polarization identity:

$$\langle T(x), y \rangle = \frac{1}{4} \left( \langle T(x + y), x + y \rangle + \langle T(x - y), x -y \rangle + i \langle T(x + iy), x + iy \rangle + i \langle T(x - iy), x - iy \rangle \right)$$ but I can't figure out how to get it from the standard version of polarization identity. I'm asking for some hint on how to prove the proposition of the title without using this identity, or alternatively some hint to prove this version of polarization identity.

Thanks

Added: I've tryed to prove the proposition by plugging in $x+y$ and $x+iy$ instead of $v$ and noticing $T$ is self-adjoint, and after some manipulations I got: $$Re(\langle T(y),x \rangle) + Im(\langle T(y),x \rangle)=0$$ but this doesn't seem to imply $\langle T(y),x \rangle=0$ so I don't know ho to proceed from here

That statement of the polarization identity is not correct. It should be $$\langle T(x), y \rangle = \frac{1}{4} \left( \langle T(x + y), x + y \rangle \color{red}{-} \langle T(x - y), x -y \rangle + i \langle T(x + iy), x + iy \rangle \color{red}{-} i \langle T(x - iy), x - iy \rangle \right).$$ With this correction, the proof is basically trivial: just expand out the right-hand side using sesquilinearity of the inner product. You will find that everything cancels except for the $\langle T(x),y\rangle$ terms, of which there are $4$ to cancel the $\frac{1}{4}$ factor.