In a complex vector space, $\langle Tx,x \rangle=0 \implies T = 0$ Suppose $T$ is a linear operator on a complex inner product space. Is it a theorem that if $\langle Tx,x\rangle=0$ for all $x$ in the space then $T=0$. The theorem fails in the real case, as seen for instance by rotation by $\pi/2$ on $\mathbb{R}^2$.
Is there anything deeper behind this fact, or can it mostly be looked at as a quirk of the conjugate-linearity of the complex inner product?
If anyone is interested in looking up this proof, it is theorem 9.2 in Roman's Linear Algebra, third edition.
 A: The key is to see how $T$ acts on an arbitrary basis. Suppose $\langle Tx,x \rangle = 0$ for all $x \in V$, where $V$ is a (real or complex) vector space.
If $T$ acts on a real space with (Hamel) basis $\{e_j\}_{j \in \alpha}$, then we have


*

*$\langle e_j, Te_j \rangle = 0$

*$
\langle e_j + e_k, T(e_j + e_k) \rangle = 0 \implies
\langle e_j, T e_k \rangle = -\langle Te_k, e_j \rangle
$


This is enough to deduce that $T = -T^*$.  If $T$ acts on a complex space, we have the additional constraint


*

*$
\langle e_j + ie_k, T(e_j + ie_k) \rangle = 0 \implies
\langle e_j, T e_k \rangle = \langle Te_k, e_j \rangle
$


This, together with the other two properties, allows us to deduce that $T = 0$.
We deduce that on a real space, $\langle Tx,x \rangle = 0$ for all $x \in V$ $\iff T^* = -T$, and on a complex space, $\langle Tx,x \rangle = 0$ for all $x \in V$ $\iff T = 0$.
Note: I haven't explicitly proved the converse in either case.  I think that you'll find that, in each case, the proof is straightforward.

While I can't say whether the result is deep, I can say that this shows that the inner product becomes much more powerful over complex spaces.
A consequence of this quirk is that when one defines positive definite operators over a real inner product space, it is significant whether one specifies that the operator must also be self-adjoint.  As real bilinear forms, matrices act the same up to their self-adjoint part.  That is, we have $A + A^*= B + B^* \iff \langle x,Ax \rangle = \langle x,Bx \rangle$ for all $x$.
For complex inner-product spaces, the additional specification of self-adjointness is redundant, and we have $A = B \iff \langle x,Ax \rangle = \langle x,Bx \rangle$ for all $x$.

Another interesting quirk: the statement
$$
\|x+y\|^2 = \|x\|^2 + \|y\|^2 \iff \langle x,y \rangle = 0
$$
is only true for real inner-product spaces.
A: I believe this fact follows easily for self-adjoint operators (in finite dimension) and since every $T=R+iS$, where $R,S$ are self-adjoint operators, the theorem follows for every $T$. I wrote the proof bellow.
The condition $\langle Tx,x\rangle=0$, for every $x$, implies $0=\langle Rx,x\rangle+i\langle Sx,x\rangle$. 
Now, $\langle Rx,x\rangle, \langle Sx,x\rangle\in\mathbb{R}$ for every $x$, since $R$ and $S$ are self-adjoints. Thus, $\langle Rx,x\rangle=\langle Sx,x\rangle=0$, for every $x$.
We know by spectral theorem (in finite dimension) that $R$ is diagonalizable. If $x$ is an eigenvector associated to an eigenvalue $a$ then $0=\langle Rx,x\rangle=a\langle x,x\rangle$ and $a=0$. Thus any eigenvalue is zero and since $R$ is diagonalizable then $R=0$. Of course the same occurs with $S$. Thus, $T=0$. $\square$
