Let $|\langle u,v\rangle|=\|u\| \cdot \|v\|.$ How to show that $u,v$ are linearly dependent?

Question

Let $|\langle u,v\rangle|=\|u\|\cdot \|v\|.$ How to show that $u,v$ are linearly dependent?

Without loss of generality let $u\ne0,v\ne0.$

Then, $\displaystyle\left\langle\frac{u}{\|u\|},\frac{v}{\|v\|}\right\rangle=1=\left\langle\frac{u}{\|u\|},\frac{u}{\|u\|}\right\rangle\\\implies\displaystyle\left\langle\frac{u}{\|u\|},\frac{u}{\|u\|}-\frac{v}{\|v\|}\right\rangle=0$

I don't know what to do next.

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Added: A thought that just occurred to me:

$\displaystyle\left\langle\frac{u}{\|u\|},\frac{u}{\|u\|}-\frac{v}{\|v\|}\right\rangle=0=\left\langle \frac{v}{\|v\|},0\right\rangle\\\implies \displaystyle\left\langle\frac{u}{\|u\|}-\frac{v}{\|v\|},\frac{u}{\|u\|}-\frac{v}{\|v\|}\right\rangle=0\\\implies \dfrac{u}{\|u\|}-\dfrac{v}{\|v\|}=0$

Does it work?

Answer 1

Write $\lambda=\langle u,v\rangle/\Vert v\Vert^2$. Let $$ w=u-\lambda v. $$ Then $$ \begin{aligned} \langle w,w\rangle&=\langle u,u\rangle-\overline{\lambda}\langle u,v\rangle-\lambda\langle v, u\rangle+|\lambda|^2\langle v,v\rangle\\ &=\Vert u\Vert^2-2\frac{|\langle u,v\rangle|^2}{\Vert v\Vert^2}+\frac{|\langle u,v\rangle|^2}{\Vert v\Vert^4}\Vert v\Vert^2\\ &=\Vert u\Vert^2-2\frac{\Vert u\Vert^2\cdot\Vert v\Vert^2}{\Vert v\Vert^2}+ \frac{\Vert u\Vert^2\cdot\Vert v\Vert^2}{\Vert v\Vert^2}\\ &=\Vert u\Vert^2(1-2+1)=0. \end{aligned} $$ So $w=0$ and $u$ and $v$ must be dependent.

Answer 2

Suppose $u=kv$ for some $k\neq0$ what will you get?

Answer 3

Suppose $u\neq 0$ and $v\neq 0$ are linearly dependent, i.e. exist coefficients $ \lambda, \gamma$ not both equal to $0$ such that $w:=\lambda u+\gamma v=0$.

But then

$$0=|\langle u,w\rangle|=\|u\|(|\lambda|\|u\|+|\gamma|\|v\|).$$

What can you conclude ($\|u\|\neq 0$)? You need to consider 2 different cases.

Answer 4

An idea: let $\,x\in\Bbb R\,$ be a parameter, then

$$ \langle u+xv,u+xv\rangle=||u||^2+2\text{Re}\langle u,v\rangle x+x^2||v||^2\;\;\;\;(**)$$

The above real quadratic in $\,x\,$ has a unique root iff

$$\Delta:=b^2-4ac=4\left(\text{Re}\langle u,v\rangle\right)^2-4||u||^2\,||v||^2=0\iff\left|\text{Re}\langle u,v\rangle\right|=||u||\,||v||$$

In this case the quadratic is

$$||v||^2x^2\pm 2||u||\,||v||\,x+||u||^2 =\left(||v||x\pm||u||\right)^2$$

and the unique (double root) is

$$x=\pm \frac{||u||}{||v||}$$

and thus (**) becomes

$$0=\langle u+xv,u+xv\rangle\iff u=-xv\iff\;u\;,\;v\;\;\text{are linearly dependent}$$

Answer 5

If we are talking about a real inner product, $$ \langle\,u,v\,\rangle=\pm\|u\|\|v\| $$ implies $$ \langle\,u\|v\|\mp v\|u\|,u\|v\|\mp v\|u\|\,\rangle=0 $$ Therefore, $u\|v\|\mp v\|u\|=0$.

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If we are talking about a complex inner product, then $$ |\langle\,u,v\,\rangle|=\|u\|\|v\| $$ We can find a $z\in\mathbb{C}$ so that $|z|=1$ and $\langle\,zu,v\,\rangle\ge0$ (that is, it is a non-negative real). $$ \begin{align} \langle\,zu\|v\|-v\|u\|,zu\|v\|-v\|u\|\,\rangle &=2\|u\|^2\|v\|^2-2\mathrm{Re(}\langle\,zu,v\,\rangle)\|u\|\|v\|\\[6pt] &=0 \end{align} $$