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Prove/Disprove that if $X$ is a complex inner product space and $Q:X\to X$ satisfies $\langle Qx,x\rangle =0\forall x\in X$ then $Q=0$.

I think the above is false because if we take $Q$ to be a rotation by an angle $90^\circ$ then $\langle Qx,x\rangle =0\forall x\in X$ but $Q\neq 0$

Is it true?Please help.

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The statement is true. I'm assuming that $Q$ is linear. The following polarization identity is key to the proof that it is true:

Let $x$, $y \in X$ then

$\langle Qx,y\rangle=\frac{\langle Q(x+y),x+y\rangle-\langle Q(x-y),x-y\rangle}{4} +i\frac{\langle Q(x+iy),x+iy\rangle-\langle Q(x-iy),x-iy\rangle}{4}$. Do you know how to finish from there?

To demonstrate that your idea was incorrect let $X=\mathbb{C}^2$ and let $Q = \left[\begin{smallmatrix} 0 & -1 \\ 1 & 0\end{smallmatrix} \right]$ be the rotation by $90$ degrees. Recall that $\mathbb{C}^2$ is an inner product space with the inner product given by $\left = \bar{x}^{T}y$ Take, e.g., $x=\left[\begin{smallmatrix} 1+i\\ 1-i\end{smallmatrix} \right]$ then $$Qx = \left[\begin{smallmatrix} i-1\\ i+1\end{smallmatrix} \right]$$ and so $$ \left = [-1-i, 1-i] \left[\begin{smallmatrix} 1+i\\ 1-i\end{smallmatrix} \right]=-4i.$$

Your idea with the rotation would be correct if $X$ was a real inner product space.

  • 0
    I want to prove it geometrically,How can it be done? – 2017-02-23
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    What do you mean? You don't want to use this polarisation identity, but why? – 2017-02-23
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    No you got it wrong. I mean what does rotation in $\Bbb C$ mean ?The same problem if done in $\Bbb R$ gives different result.Difference between rotation in $\Bbb R$ and $\Bbb C$? – 2017-02-23