If x and y are parallel, then ha = b?

[julypraise]

I defined the angle $\theta$ between $x \in \mathbb{R}^k$ and $y \in \mathbb{R}^k$ in the way that:

$\theta = \cos ^{-1}\frac{<x,y>}{||x|| \cdot ||y||}$

where $\cos ^{-1}:[-1,1] \to [0,\pi]$ is bijective.

And, therefore, $x$ and $y$ are parallel in the same direction if $\theta = 0$ and in the opposite direction if $\theta = \pi$.

Now I can't prove the following theorem within this context:

Theorem: If $x$ and $y$ are parallel in the same direction, then $x = hy$ where $h>0$, and if in the opposite direction, then $x = hy$ with $h<0$.

I feel very grateful if someone post the proof. Thanks.

(Please note that $<x,y> = x_1 y_1 + \cdots + x_k y_k$.)


Actually, what I've done so far for the proof is this:

$<x,y>^2 = (||x|| \cdot ||y||)^2$

$\displaystyle{ x^2 _1 y^2 _1 + \cdots + x^2 _k y^2 _k + 2\sum_{i \neq j} x_i y_i x_j y_j = x^2 _1 y^2 _1 + \cdots + x^2 _k y^2 _k + \sum_{i \neq j} x^2 _i y^2 _j }$

$\displaystyle{ 2\sum_{i \neq j} x_i y_i x_j y_j = \sum_{i \neq j} x^2 _i y^2 _j }$.

But from here I cannot progress further.

 

[zhentil]

Try computing the norm of (||y||/||x||)x-y.