# If $u,v \in V$ with $\|u\|=\|v\|=1$ and $\langle u,v\rangle=1$, prove that $u=\alpha v$ for some $\alpha \in F$

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?

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

## 1 Answer

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.