T is a linear operator on V, with an adjoint. V is a complex inner product space, not necessarily of finite dimension. How do I prove $$\left \langle T(x),x \right \rangle=0,\forall x\in V\Rightarrow T=T_0$$ I at first thought I just had to state that $T(x)\perp x,\forall x\in V$ so $T=T_0$ but that obviously isn't necessarily true. I'm aware that there is a similar problem already on stack exchange, but the question hint asks me to do it by substituting $x$ with $x+y$ and then $x+iy$. After expanding the inner product to $$\left \langle T(x),x \right \rangle+\left \langle T(x),y \right \rangle+\left \langle T(y),x \right \rangle+\left \langle T(y),y \right \rangle=0$$I'm still not sure how to prove $T=T_0$. I came up with some ideas like $T(x)\perp y$ for any arbitrary x and y, but I'm not sure if I can state that definitevely just from what I have above.
Edit: $T_0=0,\forall x$
Follow the hint. The equation you get after substituting $x + y$ in place of $x$ can be written as $$ \langle T(x),y\rangle + \langle T(y),x\rangle = 0 $$ since $\langle T(x),x\rangle = \langle T(y),y\rangle = 0$ (given). Similarly, second substitution gives, $$ - \langle T(x),y\rangle + \langle T(y),x\rangle = 0 $$
Thus, $\langle T(x),y\rangle = 0$ for any $x$ and $y$. Fix $x$, let $y = T(x)$, then $\langle T(x), T(x)\rangle = 0 \implies T(x) = 0$. Since $x$ is arbitrary, $T(x) = 0$ for all $x$, hence $T = T_0$.