On the existence of scalar multiplication of free vectors?

Question

When we study analytical geometry at the undergraduate level we define free vectors as oriented line segments. This raises a problem when we try to formalize the multiplication of a free vector by a scalar. In order to elaborate on that, denote by $\mathbb V^3$ the set of free vectors. How can we show there exists a unique map

$$\cdot : \mathbb R\times \mathbb V^3\rightarrow \mathbb V^3, (\lambda, \overrightarrow{v})\mapsto \lambda \overrightarrow{v}$$

such that:

(1) $0\cdot \overrightarrow{v}=\overrightarrow{0}$

(2) $\lambda \overrightarrow{v}$ has the same direction as $\overrightarrow{v}$

(3) $\lambda \overrightarrow{v}$ has the same orientation as $\overrightarrow{v}$ if $\lambda>0$ and opposite orientation when $\lambda<0$

(4) $|\lambda \overrightarrow{v}|=|\lambda||\overrightarrow{v}|.$

Back in the past I found a book that proved this but I ended up forgetting which book was that. Do you know how to prove the existence of such a map or where I can find the proof?

Thanks.

Answer 1

You mention in a comment that you want to use the abstract Euclidean space $E$ rather than the coordinate space $\mathbb R^3$. The space $E$ has the nice property that every point in the space has equal status, unlike the space $\mathbb R^3$ in which one point, $(0,0,0),$ is singled out as the origin.

But this presents you with a special difficulty when you try to define your scalar product. The definition of the scalar product requires the existence of a unique vector $\vec 0$ such that for any vector $\vec v,$ no matter which you choose,

$$ 0 (\vec v) = \vec 0.$$

The question then is: which vector in $\mathbb V^3$ is $\vec 0$?

There can only be one zero vector, hence you must single out just one of your "free vectors" to be the zero vector. As soon as you do so, you have established a special point in your space $E,$ the same way that the origin is a special point in $R^3.$ In order to define a scalar product on your "free vectors," you are going to have to give up one of the most significant differences between your space and $R^3.$

Note that once you find $\vec 0$ such that $0 (\vec v) = \vec 0,$ the property $|\lambda \vec{v}|=|\lambda||\vec{v}|$ implies that $|\vec 0|=|0\vec v|=|0||\vec v| = 0.$ That is, both endpoints of $\vec 0$ must be the same point.

As long as you are willing to identify $\vec 0$ as a particular point in $E,$ you might as well call that point the origin of $E.$ You can then define $\lambda \vec v$ for any free vector $\vec v$ as follows:

For any point $p \in E,$ if $p = \vec 0$ then $\lambda p = \vec 0$; otherwise define $\lambda p$ as a point on the line through $\vec 0$ and $p$ such that the distance from $\vec 0$ to $\lambda p$ is $|\lambda|$ times the distance from $\vec 0$ to $p$ and such that $\vec 0$ is strictly between $p$ and $\lambda p$ if and only if $\lambda < 0.$

Then for any free vector $\vec v$, if $\vec v$ is the directed line segment from $p \in E$ to $q \in E,$ let $\lambda \vec v$ be the directed line segment from $\lambda p$ to $\lambda q.$

You can check that this satisfies all four of your properties.

Answer 2

I'll write my answer under the presumption that by the superscript $3$ on $\mathbb V^3$ you are working in $\mathbb R^3$. Also, given $p,q \in \mathbb R^3$ I'll use $\overline{pq}$ or $\overline{p,q}$ as the notation for the free vector with initial endpoint $p$ and terminal endpoint $q$.

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Regarding existence, one construction goes like this.

For each $p \in \mathbb R^3$ let $$\mathbb V^3_p = \{\overline{pq} \mid q \in \mathbb R^3\} $$ Also, let $T_p$ be the translation of $\mathbb R^3$ which takes the origin to $p$, defined by $$T_p(x)=x+p $$ Here, of course, I am using ordinary vector addition on $\mathbb R^3$.

Then scalar multiplication can be defined on $\mathbb V^3$ by defining it separately on $\mathbb V^3_p$ for each $p$:

• If $p=0$ then for each $\vec v = \overline{0x} \in \mathbb V^3_0$ define $r \vec v = \overline{0,rx}$. This is essentially ordinary scalar multiplication, carried out on the terminal endpoint of free vectors with initial endpoint $0$.
• If $p \ne 0$ then for each $\vec v = \overline{p,x} \in \mathbb V^3_0$ define $$r \vec v = T_p ( r T_p^{-1}(\vec v)) = \overline{p,p + r (x-p)} $$ Think of this as changing coordinates so that $p$ becomes the origin, and then in those new coordinates using ordinary scalar multiplication to define scalar multiplication as usual.

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However, uniqueness is false. For example, having first defined the scalar multiplication above, which I'll denote $r \cdot \vec v$, I could do something silly to get another scalar multiplication $r * \vec v$. For example, I could pick two points $p_1,p_2$, define $S$ to be the translation of $\mathbb R^3$ that takes $p_1$ to $p_2$, and then:

• If $p \not\in \{p_1,p_2\}$ then $r * \vec v = r \cdot \vec v$
• If $p=p_1$ then $r * \vec v = S(r \cdot \vec v)$
• If $p=p_2$ then $r * \vec v = S^{-1}(r \cdot \vec v)$

If you want something unique, you're going to have to impose an additional condition. One natural condition to impose would be that for each $p \in \mathbb R^3$, if $\vec v \in \mathbb V^3_p$ then $r \vec v \in \mathbb V^3_p$. In other words, the scalar multiplication operation preserves the initial point of the free vector. Under this condition, the formula I gave above is the unique example.