# Norm of vector in $\mathbb{R}^3$ with multiple

If we have a vector in $\mathbb{R}^3$ (or any Euclidian space I suppose), say $v = (-3,-6,-9)$, then:

1. May I always "factor" out a constant from a vector, as in this example like $(-3,-6,-9) = -3(1,2,3) \implies (1,2,3)$ or does the constant always go along with the vector?
2. If yes on question 1, then if I want to compute the norm, is the correct computation the following: $||v|| = |-3|\sqrt{14} = 3\sqrt{14}$ ? If so, is the only reason that we take the absolute value of -3 because we don't want a negative length?

I'm sorry if things are obvious but I just want to make sure I actually get this correctly.

Best regards

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The answer is yes to both.

1. If you have a vector $(v_{1},...,v_{n})\in\mathbb{R}^{n}$ and a real number $a\in \mathbb{R}$, then $(av_{1},...,av_{n})=a(v_{1},...,v_{n})$. This is the definition for multiplication of a vector by a real number.

2. If $v\in\mathbb{R}^{n}$ and $a\in \mathbb{R}$, then it always holds that $\|av\|=|a|\cdot \|v\|$. In your case of $\mathbb{R}^{3}$ and Euclidean norm, you can see it for example as following. If $v=(v_{1},v_{2},v_{3})\in\mathbb{R}^{3}$ and $a\in\mathbb{R}$, then \begin{equation*} \|av\|=\sqrt{(av_{1})^{2}+(av_{2})^{2}+(av_{3})^{2}}=\sqrt{a^{2}(v_{1}^{2}+v_{2}^{2}+v_{3}^{2})}=|a|\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}}=|a|\cdot\|v\|. \end{equation*}

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As a rule, this all falls from the distributive rule. If $v=(ax,ay,az)$, then \begin{align}||v|| &= \sqrt{(ax)^2 + (ay)^2+(az)^2} = \sqrt{a^2(x^2+y^2+z^2)}\\ &=\sqrt{a^2}\sqrt{x^2+y^2+z^2}\end{align}

And $\sqrt{a^2}=|a|$

And yes, ultimately, this is all because we want distances to be positive, but we also want the norm to be well-defined.If $w=(1,-2,3)$, is $|w|=\sqrt{14}$ or $-\sqrt{14}$? There is no guidance here to choose one or the other.

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The answer is yes to both questions. In any normed vector vector space, $\|\alpha v\| = |\alpha|\|v\|$ for any vector $v$ and scalar $\alpha$.

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Starting with your question 2., the general relation $\|\lambda v\| = |\lambda| \|v\|$ always holds. From that you can answer positively to your first question (even if $\lambda$ is real)

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You are perfectly entitled to "factorise" a vector, as you say $(-3,-6,-9) = -3(1,2,3).$ The important thing here is that this factorisation shows that the vectors $(-3,-6,-9)$ and $(1,2,3)$ are linearly dependent. In the case of two vectors, this means that they are parallel vectors.

In general, the vectors $(\lambda x, \lambda y, \lambda z)$ and $\lambda(x,y,z)$ are identical. The represent the same vector. Moreover, the vector $(\lambda x, \lambda y, \lambda z)$ is exactly $\lambda$-times the vector $(x,y,z)$.

This is a nice way to simplify the calculations because:

$$||(\lambda x, \lambda y, \lambda z)|| = |\lambda| \cdot ||(x,y,z)|| \, .$$

This question hints towards a very interesting topic: projective space. The projective plane, denoted by $\mathbb{RP}^2$, can be defined using an equivalence relation: $\mathbb{RP}^2 = \mathbb{R}^3\backslash \sim$ where

$$(x_1,y_1,z_1) \sim (x_2,y_2,z_2) \iff \exists \ \lambda \neq 0 : (x_1,y_1,z_1) = \lambda(x_1,y_2,z_2) \, .$$

The equivalence classes are denoted by by homogeneous coordinates $(x_1:y_1:z_1)$.

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