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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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    I deleted my comment since it just inelegantly repeated what came above. :P2012-05-18

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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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    @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.2012-05-18