Complex Operator that sends every vector to an orthogonal image. If $T$ is an operator on a complex vector space $V$ such that, for all vectors $v\in V$, $\langle Tv, v\rangle=0$, then one can show that $T$ must be the zero operator by using a complicated formula that expresses $\langle Tu, w\rangle$ in terms of inner products of the form $\langle Tv, v\rangle$. Is there a proof of this statement that is more elegant/easier to come up with on your own?
 A: Here is a proof that avoids thinking about the real and imaginary parts of the operator. 
For any $t\in\mathbb C$ and $v,w\in V$,
$$
0=\langle T(v+tw),v+tw\rangle=\langle Tv,v\rangle+|t|^2\langle Tw,w\rangle+t\langle Tw,v\rangle+\bar t\langle Tv,w\rangle\\
=t\langle Tw,v\rangle+\bar t\langle Tv,w\rangle.
$$
Choosing $t=1$ and $t=i$, we get $\langle Tw,v\rangle\pm\langle Tv,w\rangle=0$, so both terms are zero. From $\langle Tv,w\rangle$ for all $v,w$ we get $T=0$. 
A: Suppose $\langle Tv,v\rangle=0$ for every $v$. Decompose $T=T_1+iT_2$ with $T_i$ self-adjoint, that is, $T_i=T_i^*$. For every $v\in V$,
$$0=\langle(T_1+iT_2)v,v\rangle=\langle T_1v,v\rangle+i\langle T_2v,v\rangle,$$
so $0=\langle T_iv, v\rangle$. Therefore, it suffices to show that $T_i=0$, so we can assume that $T=T^*$.
For every $x,y\in V$,
$$0=\langle T(u+w),u+w\rangle=\langle Tu,u\rangle+\langle Tu, w\rangle+\langle Tw,u\rangle+\langle Tw,w\rangle=\langle Tu,w\rangle+\langle Tw, u\rangle;$$
Since $T=T^*$, we obtain $0=\langle Tu,w\rangle+\langle w,Tu\rangle=\langle Tu,w\rangle+\overline{\langle Tu,w\rangle}=2\operatorname{Re}\langle Tu,w\rangle$.
Therefore, if $w\in V$ is fixed, the linear function $u\mapsto \langle Tu,w\rangle$ has image contained in $i\mathbb{R}$, so its image, being a (complex) linear subspace of $\mathbb{C}$, must equal $\left\{0\right\}$. That means that $\langle Tu,w\rangle=0$ for every $u,w\in V$, thus $T=0$.
PS.: The last argument works only for complex vector spaces. In the real case, we can also show that every self-adjoint operator $T$ in a real inner-product space that satisfies $\langle Tv,v\rangle=0$ for every $v$ must be $0$. For this, simply notice that $0=\langle T(x+y),x+y\rangle=2\langle Tx,y\rangle$. The complex case also follows from the real one.
A: First, observe that this only works for complex vector spaces. In $\mathbb{R}^2$, for example, you can let $T$ be a rotation by $90^\circ$, then $\langle Tv,v \rangle = 0$ yet $T \neq 0$.
For finite-dimensional complex vector space, i.e. for $\mathbb{C}^n$, you can argue in terms of the jordan normal form of $T$. 
First, all the eigenvalues of $T$ must be zero, because if there was a $\lambda \neq 0$ with $Tv = \lambda v$, then $\langle Tv,v\rangle = \lambda\|v\|^2 \neq 0$. If all Jordan blocks in the JNF of $T$ have size $1\times 1$ we're done (since their only entry is then an eigenvalue, and they are all zero). Assume thus that there is some $k\times k$ Jordan block, and let $b_1,\ldots,b_k$ be the corresponding basis vectors. Then $Tb_1 = 0$ and $Tb_{i+1} = b_i$, and therefore $$
  \langle T(\mu b_1 + b_2), \mu b_1 + b_2 \rangle
  = \langle b_1, \mu_1 b_1 + b_2 \rangle
  = \mu\|b_1\|^2 + \langle b_1,b_2 \rangle \text{.}
$$
But the last expression is surely non-zero for some $\mu$, and this thus contradics that $\langle Tv,v \rangle = 0$ for all $v$. Thus, all Jordan block have indeen size $1\times 1$, and we've already shown that $T = 0$ in that case
