0
$\begingroup$

Let $V$ be an inner product space over a field $F$. If $u,v \in V$ with $\|u\|=\|v\|=1$ and $\langle u,v \rangle=1$, then prove that $u=\alpha v$ for some $\alpha \in F$.

Can I say that here the equality in Cauchy Schwarz holds so they must be linearly dependent?

because $|\langle u,v\rangle| = \|u\|\|v\|$ here

so $u=\dfrac{\langle u,v\rangle}{\|v\|^2} v$? so $\alpha =1?$

or we can also compute, $\|u-v\|^2$ directly which we get as $0$.

What is the geometric significance here? Can I imagine this result in an intuitive way?

$\endgroup$
2
  • $\begingroup$ You must mean $\alpha$ in the underlying field, not $\alpha\in V$? $\endgroup$ Oct 18 '19 at 22:21
  • $\begingroup$ Yes, corrected it. Thanks. $\endgroup$
    – Abhay
    Oct 18 '19 at 22:23
1
$\begingroup$

Your insight is good:

$$\left\|u-v\right\|^2=\langle u-v,u-v\rangle=\left\|u\right\|^2-\langle u,v\rangle-\langle v,u\rangle+\left\|v\right\|^2=0\implies u-v=0$$

Remember that $\;1=\langle u,v\rangle=\left\|u\right\|\left\|v\right\|\cos\theta=\cos\theta\;$ , with $\;\theta=\;$ angle between $\;u,\,v\;$ , so the above means $\;\theta=0\;$ ( or any other integer multiple of $\;2\pi\;$), so $\;u,\,v\;$ are vectors of the same length and in the same direction $\;\implies\;$ they're equal.

$\endgroup$

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.