Im working with the following definition:
Two vectors, $\vec{x}$ and $\vec{y} \in \Bbb R^n$ are parallel iff $|\vec{x} \cdot\vec{y}|=\|\vec{x}\|\|\vec{y}\|$
Then, I must prove that if two vectors $\vec{x}$ and $\vec{y}$ are parallel, one is a scalar multiple of the other. That is, $\vec{x} = \lambda\vec{y}, \lambda \in \Bbb R$
I've tried to prove it directly but its too messy on the algebra, I'm hoping there is a simpler way to prove it.
Thanks!
Best Answer
Elaborating on Mr. Leuuwen's comment, the usual proof for Cauchy-Schwarz suggests some ideas. One possible way to go follows:
(Let $x \cdot y$ stand for inner product of $x$ and $y$.)
Suppose $|x \cdot y | = \|x\| \|y\|.$ Dividing by $\|y\|$, we get $$\|x\| = \frac{|x \cdot y|}{\|y\|},$$ which taking into account the definition ($|y\cdot y| = \|y\| \|y\| $) equals $$\frac{|x \cdot y|}{|y\cdot y|}\|y\|.$$ This shows that $\|x\| = \lambda_1 \|y\|$ for some scalar $\lambda_1$. To get the result $x = \lambda y$, one could apply the definition of parallel again (to get $x\cdot y = \pm \lambda_1\, y \cdot y$ ) and note the angle between $x$ and $y$ is defined via inner product $x \cdot y$ (see the answer by Mr Rose above), or lend even more from the proof and consider that if we write $x$ as $$x = \lambda y + z$$ where $z$ and $x$ are orthogonal (so $z \cdot x = 0$), then from the other assumptions will follow that $\|z\| = 0$.