TITLE: In a complex vector space, $\langle Tx,x \rangle=0 \implies T = 0$
QUESTION [9 upvotes]: 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.

REPLY [6 votes]: 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.