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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.2012-05-18
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    Exactly what I was writing about 2 minutes ago: the OP *already* fixed the OP so you couldn't now see what I did2012-05-18
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    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.2012-05-18
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    @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 I'll just delete my comment (and you could delete your comment as well, if it is no longer applicable, to avoid these confusions).2012-05-18
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    What about if we define $F(u)=proj_{(v^\perp)}u=k{v^\perp}$ for some $k\in R$. Then $\left=0$ but $F(u)$ is not zero unless $v =0$. Or am I being daft?2012-05-18
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    @azdahak: There is one fixed $F$ once and for all, not a different $F$ for each $v$. Arturo's answer shows why $F(u)$ must be $0$ for all $u$.2012-05-18
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    Perhaps more interesting: If $\langle F(u),u\rangle=0$ for all $u$, must $F$ be $0$? The answer is no in general for real inner product spaces, but yes for complex inner product spaces.2012-05-18
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    @Jonas: Are you assuming $F$ is linear? While it's reasonable, note that the question says "function", not "linear transformation". But you are absolutely right for linear transformations, of course (rotation of $90^{\circ}$ in $\mathbb{R}^2$, and find an eigenvector associated to $\lambda\neq 0$ for the complex case).2012-05-18
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    @ArturoMagidin: Yes, I was thinking of a different problem and should have said that $F$ is linear. Thanks for the correction. (Of course if $F$ could be arbitrary, the answer to that question would always be no unless $V=\{0\}$, e.g. $F(0)\neq 0 $ and $F(u)=0$ for all $u\neq 0$.) Your suggestion for the complex linear case works if there is a nonzero eigenvalue, but it is true in general, as can be seen for example by expanding $\langle F(u+iv),u+iv\rangle$ and $\langle F(u+v),u+v\rangle$.2012-05-18
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    @Jonas: Good point... forgot about nilpotent matrices for a moment there...2012-05-18
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    @ArturoMagidin: Or worse, on infinite dimensional spaces linear transformations need not have any eigenvalues.2012-05-18
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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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    why I can set v=F(u)?2012-05-18
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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