Prove that the only operator on $\mathbb{C}$ for which his inner product is zero is zero

Question

How do I prove this statement: Let $V$ be an unitary vector space over $\Bbb C$, $(,)$ be an inner product on $V$ and $\Bbb A$ operator $V\rightarrow V$. Then $(\Bbb Av,v)=0$ if and only if $\Bbb A=0$.

I know that on $\Bbb R$ this does not apply, since we are searching for an operator that assigns to my vector a vector that is orthogonal and so every orthogonal projection on a subspace of $\Bbb R$ can be this operator. But how is this possible on $\Bbb C$? Thank you.

@Git Gud: C should be the set of complex numbers, I don't know how to format it. Resp. it should be V, an unitary subspace on complex numbers — user74200, Apr 25 '13 at 07:41
You probaby want to type $A$ instead of $\Bbb A$, in order to get $A$ instead of $\Bbb A$. — Git Gud, Apr 25 '13 at 07:47
An orthogonal projection on $\Bbb R$ isn't a counterexample (look at vectors in the image). A $90^\circ$ rotation in the plane is. — PseudoNeo, Apr 25 '13 at 07:50
@Git Gud in my textbook operators are also double letters, I am just quoting — user74200, Apr 25 '13 at 08:13

score 11 · Accepted Answer · answered Apr 25 '13 at 07:44

This proof is (sort of) from Advanced Linear Algebra by Steven Roman.

Let $V$ be a complex inner product space and let $A$ be a linear operator on $V$ with the given property. For all $v\in V$ and $r\in\mathbb{C}$ we have \begin{align} 0 &= \langle A(rv+Av),rv+Av \rangle \\ &= |r|^2\langle Av,v\rangle + \langle A^2v,Av\rangle + r\langle Av, Av\rangle + \overline{r}\langle A^2 v, v\rangle \\ &= r\langle Av, Av\rangle + \overline{r}\langle A^2 v, v\rangle. \end{align} Set $r=1$ to get $$\langle Av, Av\rangle + \langle A^2 v, v\rangle=0,$$ and set $r=i$ to get $$\langle Av, Av\rangle - \langle A^2 v, v\rangle=0.$$ Therefore $\Vert Av \Vert^2 = 0$.

Prove that the only operator on $\mathbb{C}$ for which his inner product is zero is zero

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