Parallel vectors in $\mathbb{R}^n$

Question

Definition: We say that $\vec{x},\vec{y}\in\mathbb{R}^n$ are parallel vectors if $|\vec{x}\cdot \vec{y}| = \|\vec{x}\|\|\vec{y}\|$. (i.e equality holds in Cauchy–Schwarz inequality)

I'm having some problems showing that if $\vec{x},\vec{y}$ are parallel vectors in $\mathbb{R}^n$, then one is a scalar multiple of the other.

Answer 1

First of all, we usually restrict the notion of parallel vectors to non-zero vectors.

Recall that for $x, y \in \mathbb{R}^n\setminus\{0\}$, $|x\cdot y| = \|x\|\|y\|\cos\theta$ where $\theta$ is the angle between $x$ and $y$. If $x$ and $y$ are parallel, then $|x\cdot y| = \|x\|\|y\|$, so $\cos\theta = 0$. Therefore $\theta = 0$, in which case $x$ are $y$ are pointing in the same direction, or $\theta = \pi$, in which case they are pointing in opposite directions. Can you take it from here?

Answer 2

Hint: Assume $x\neq0$. A quick calculation reveals that the length of the vector $$y-\frac{x\cdot y}{\|x\|^2}x$$ equals $0$ provided that $|x\cdot y|=\|x\|\|y\|$.

Answer 3

Let $x = \|x\|n$ where $n$ is the unit vector in the direction of $x$. Now if $y$ is parallel to $x$ they must have same support, that is $y = \pm \|y\|n$ sub unit vector we get $y = (\text{scalar})\,x$.