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I really don't even understand this question ( I guess it just a simple one but I don't understand this function given)

Given $V$, an inner product space and function $F\colon V\to V$ such that for every $u,v$ vectors in $V$, $\langle F(u),v\rangle =0$. I need to prove that $F(u)=0$ for each $v\in V$.

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You seem to have ommited some stuff from the question since as it is written now it makes no sense. Check this. – DonAntonio May 18 '12 at 16:37
Exactly what I was writing about 2 minutes ago: the OP already fixed the OP so you couldn't now see what I did – DonAntonio May 18 '12 at 16:48
Imo, it should be forbidden, or even better: impossible to fix a post which has already been on the board for several minutes (say, 5 or so), as otherwise these misunderstandings can pop up. – DonAntonio May 18 '12 at 16:49
@DonAntonio: Not at all! I'll just delete my comment (and you could delete your comment as well, if it is no longer applicable, to avoid these confusions). – Arturo Magidin May 18 '12 at 16:54
What about if we define $F(u)=proj_{(v^\perp)}u=k{v^\perp}$ for some $k\in R$. Then $\left<F(u),v\right>=0$ but $F(u)$ is not zero unless $v =0$. Or am I being daft? – azdahak May 18 '12 at 17:16
up vote 3 down vote accepted

Hint. Let $u\in V$. Set $v=F(u)$. The condition $0=\langle F(u),v\rangle$ tells you what about $F(u)$?

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why I can set v=F(u)? – Mary May 18 '12 at 16:50
@Nusha: Because your condition says that it holds for every vector $u$ and $v$; in particular, it holds for $v=F(u)$, because that's a vector. – Arturo Magidin May 18 '12 at 16:54
Thank you very much! – Mary May 18 '12 at 16:58

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