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