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?

  • $\begingroup$ 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

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.


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