Suppose that for all $u \in V, \langle T(u), u \rangle =0$. How to prove that T is the 0 operator, that is $T(v)=0$ for all $v \in V$? Hint: Apply information on my last question about polarization identity to obtain that $\langle T(u), v \rangle = 0$ for all $u, v \in V$ and then let $v=T(u)$. this is a homework in the area of linear algebra.
closed as off-topic by Jonas Meyer, N. F. Taussig, Davide Giraudo, hardmath, Aaron Maroja Mar 22 at 17:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
You've left out some important information. It's not enough to guarantee T(v) is the zero operator just to know T(v) orthogonal to v for all v. (Think about a right angle rotation in the plane.)
Of course if T(v) is orthogonal to u for all u, then (in particular) T(v) must be zero since < T(v),T(v)> = 0 implies that.
So look for the missing extra condition(s)...
Edited per Robin Chapman's comments, to incorporate the OP's earlier question:
The polarization identity expresses <T(v),u> for any v,u in a complex inner product space as a linear combination of terms of the form <T(w),w>. Thus if all of the latter are zero, so must <T(v),u> be always zero, and the result follows.