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.


  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

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 $f(\lambda) = 0$

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ " 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.