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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!

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    $\begingroup$ Look up a proof of the Cauchy-Schwarz inequality, which in part says just this. $\endgroup$ – Marc van Leeuwen Sep 14 '14 at 4:57
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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 $\|z\| = 0$.

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I have tried this: if y is not a multiple of x, then x and y cannot be parallel. Assume $y\neq \lambda x$ for all $\lambda\in F$, where $F$ is the field. Then the inner product $(y-\lambda x,y-\lambda x)>0$ for all $\lambda\in F$. In this case you get a quadratic polynomial in $\lambda$ which is always positive. So, discriminant is negative, which follows that $$ (x,y)<\|x\|\cdot\|y\| \Longrightarrow \cos(\theta)=\frac{(x,y)}{\|x\|\cdot\|y\|}\neq 1 $$ and so $x$ and $y$ are not parallel. Result: If $x$ and $y$ parallel then one is a multiple of the other.

Note1: $x$ and $y$ parallel means the angle between them is zero.

Note2: Proof of the converse is very easy.

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One possible solution is to note that $$ x \cdot y = \|x\|\|y\| \cos\theta $$ where $\theta$ is the angle between the two vectors. Using your definition of parallel, this tells us that $|\cos\theta| = 1$, which (suitably interpreted) should get you what you want.

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