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I need some help to show that:

If $| \langle u,v \rangle | = \|u\| \|v\|$ then $u=\lambda v$ for some scalar $\lambda$.

We have to consider this over an arbitrary field $\mathbb{F}$.

I appreciate your suggestions.

Thanks.

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  • $\begingroup$ This question is a possible duplicate of this MSE post. $\endgroup$ – Jose Arnaldo Bebita-Dris Sep 4 '14 at 12:50
  • $\begingroup$ How do you define $\|u\|$ over a finite field? $\endgroup$ – Gerry Myerson Sep 4 '14 at 12:54
  • $\begingroup$ We define $\|u\| = \sqrt{\langle u , u \rangle}$ $\endgroup$ – JimmyJP Sep 4 '14 at 13:07
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    $\begingroup$ The result you are trying to prove is not necessarily true. Depending on what $u$ and $v$ are, what might be provable is the result that there is a positive real number $\lambda$ such that $\|u-\lambda v\| = 0$ which includes as a special case the result that $u = \lambda v$. See the Appendix of this document for a proof. $\endgroup$ – Dilip Sarwate Sep 4 '14 at 13:08
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    $\begingroup$ @DilipSarwate: It is one of the axioms of an inner product space that $\|u\|^{2} = 0$ if and only if $u = 0.$ $\endgroup$ – Geoff Robinson Sep 4 '14 at 13:42
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Take $f(\lambda) = \|u - \lambda v\|^2 = \langle u - \lambda v, u - \lambda v\rangle = \|v\|^2\lambda^2 - 2\langle u, v \rangle\lambda + \|u\|^2$, which is a quadratic function of $\lambda$ and your given condition implies there exists $\lambda$ such that $f(\lambda) = 0$

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    $\begingroup$ No, you have shown that there exists a $\lambda$ such that $f(\lambda) = \|u-\lambda v\|^2 = 0$, which does not necessarily imply that $u = \lambda v$; the two functions might differ on a set of measure $0$. This possibility does not occur in finite-dimensional vector spaces but must be considered in more general spaces. $\endgroup$ – Dilip Sarwate Sep 4 '14 at 13:14
  • $\begingroup$ @DilipSarwate In this kind of situation, such as $L^2$ function, we usually consider class of equivalent functions as element in this space. $\endgroup$ – Petite Etincelle Sep 4 '14 at 14:13
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    $\begingroup$ " we usually consider class of equivalent functions as element in this space." Exactly my point. To me, a general statement such as $u = \lambda v$ means that $u(x) = \lambda v(x)$ for all $x$. The equality means different things in finite-dimensional spaces versus more general spaces. $\endgroup$ – Dilip Sarwate Sep 4 '14 at 14:36

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