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I just wanted to be sure about something. The implication $\langle Tx,x\rangle =0$ , then T is zero , holds only if $T$ is self-adjoint right? If $T$ is an arbitrary operator, we need to have $\langle Tx,Tx \rangle $ or $\langle Tx,y\rangle$ being zero for all $x,y$ in order to conclude this, right?

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    For real spaces, $\langle Tx,x\rangle = 0$ for all $x$ is not sufficient to conclude $T = 0$ unless $T$ is self-adjoint. For complex spaces, the self-adjointness is not necessary. – Daniel Fischer Oct 18 '14 at 19:09
  • why is this sufficient in complex spaces, is it polarization? –  Oct 18 '14 at 19:10
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    Yes, it's polarisation. Try to see yourself that $\langle Tx, x\rangle = 0$ for all $x$ implies $\langle Tx,y\rangle = 0$ for all $x,y$ in the complex case. – Daniel Fischer Oct 18 '14 at 19:11
  • Related: https://math.stackexchange.com/questions/1307747/prove-that-if-langle-tx-x-rangle-0-for-all-x-in-x-then-t-0?rq=1 – Nate Eldredge Apr 20 '21 at 22:48

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Let $\langle \cdot, \cdot \rangle$ be the dot product and $T$ be the operator taking $e_1$ to $e_2$ and $e_2$ to $-e_1$, i.e.

$$T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$

So we have $$\left\langle T\begin{pmatrix}\alpha \\ \beta \end{pmatrix}, \begin{pmatrix}\alpha \\ \beta \end{pmatrix}\right\rangle = \left\langle \begin{pmatrix}-\beta \\ \alpha \end{pmatrix}, \begin{pmatrix}\alpha \\ \beta \end{pmatrix}\right\rangle = - \alpha \beta + \alpha \beta = 0$$

But clearly $T$ is not zero.

So what's going on here? Well, the eigenvectors of this matrix aren't real, so we don't see them then we take $\langle T v, v\rangle$ for all $v \in \mathbb{R}^2$. If we extend to $\mathbb{C}^2$, then we can see the eigenvectors, and it's no longer the case that $\langle T v, v \rangle = 0$ for all $v \in \mathbb{C}^2$.

For a self-adjoint real operator, all the eigenvectors are real, so we don't run into this problem.

Daniel McLaury
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