Geometric meaning of vectors

So far, we have been treating vectors purely algebraically. We can also give a geometric interpretation of vectors. Geometrically, vectors can be visualized as ``arrows'' in space, and two arrows are considered ``equivalent'' as long as their lengths and directions are the same, irrespective of where they are located in space.

Think of an ``arrow'' in the 2-dimensional space (Figure fig:arrow). An arrow can be defined by its source (tail) and target (head) points. The source of an arrow is the point where the arrow starts; the target is the point where the arrow ends. So, an arrow $\mathbf{a}$ can be considered as a pair of 2-dimensional points: $\mathbf{a}=(s,t)$ where $s=(x_s,y_s)\in {\mathbb{R}}^2$ and $t=(x_t,y_t)\in {\mathbb{R}}^2$ represent the source and target, respectively. That is, an arrow is a pair of pairs of real numbers: $\mathbf{a}=((x_s,y_s),(x_t,y_t))\in {\mathbb{R}}^2\times {\mathbb{R}}^2(\simeq {\mathbb{R}}^4)$.

Figure fig:arrow. An arrow in the 2D space.

We introduce a relation on the set ${\mathbb{R}}^2\times {\mathbb{R}}^2$. Let $\mathbf{a}=(s,t)=((x_s,y_s),(x_t,y_t))$ and $\mathbf{b}=(u,v)=((x_u,y_u),(x_v,y_v))$ be two arrows. Then we write $\mathbf{a}\sim \mathbf{b}$ if $x_t-x_s=x_v-x_u$ and $y_t-y_s=y_v-y_u$. This relation $\sim$ is an equivalence relation. In fact,

1. (Reflexivity) For each arrow $\mathbf{a}=(s,t)=((x_s,y_s),(x_t,y_t))$, clearly we have $x_t-x_s=x_t-x_s$ and $y_t-y_s=y_t-y_s$. Thus $\mathbf{a}\sim \mathbf{a}$.
2. (Symmetry) Suppose $\mathbf{a}\sim \mathbf{b}$ then $x_t-x_s=x_v-x_u$ and $y_t-y_s=y_v-y_u$, so trivially $\mathbf{b}\sim \mathbf{a}$.
3. (Transitivity) Suppose $\mathbf{a}\sim \mathbf{b}$ and $\mathbf{b}\sim \mathbf{c}$ where $\mathbf{c}=(p,q)=((x_p,y_p),(x_q,y_q))$. Then $x_t-x_s=x_v-x_u$ and $y_t-y_s=y_v-y_u$ as well as $x_q-x_p=x_v-x_u$ and $y_q-y_p=y_v-y_u$ so that $x_q-x_p=x_t-x_s$ and $y_q-y_p=y_t-y_s$. Hence $\mathbf{a}\sim \mathbf{c}$.

Geometrically, two arrows are equivalent if and only if they overlap exactly when translated (moving without rotation). Another way of saying this is that two arrows are equivalent if and only if they have the same ``direction'' and the same length.

We define each equivalence class of ${\mathbb{R}}^2\times {\mathbb{R}}^2$ by $\sim$ to be a (2-dimensional) vector. Note that each arrow $\mathbf{a}=((x_s,y_s),(x_t,y_t))$ is equivalent to the one $((0,0),(x^{\prime },y^{\prime }))$ with $x^{\prime }=x_t-x_s,y^{\prime }=y_t-y_s$. So we can represent each arrow by the one whose source is the origin. This means that each vector can be effectively represented by its target. That is, writing $[\mathbf{a}]=(x^{\prime },y^{\prime })$ is just fine as a representation of a vector (an equivalence class) $[\mathbf{a}]\in ({\mathbb{R}}^2\times {\mathbb{R}}^2)/\sim$. In other words, when dealing with vectors, we don't care where it starts or ends as long as their direction and length are unchanged. After all, we can express any 2-dimensional vectors in ${\mathbb{R}}^2$. This means that the set of equivalence classes of arrows $({\mathbb{R}}^2\times {\mathbb{R}}^2)/\sim$ is more or less the ``same'' as ${\mathbb{R}}^2$.

Vector addition

Let $\mathbf{a}=(a_1,a_2)$ and $\mathbf{b}=(b_1,b_2)$ be elements of $Arrow/\sim$ (From now on, we omit brackets to represent equivalence classes for simplicity). We define the addition between these two vectors by

$$\mathbf{a}+\mathbf{b}=(a_1+b_1,a_2+b_2).$$

Geometrically, this means the following.

1. Let representatives of $\mathbf{a}$ and $\mathbf{b}$ be $((0,0),(a_1,a_2))$ and $((0,0),(b_1,b_2))$, respectively.
2. Translate $\mathbf{b}$ so its tail is at the head of $\mathbf{a}$. That is, the representative of $\mathbf{b}$ is now $((a_1,a_2),(a_1+b_1,a_2+b_2))$. Note that translation does not change the equivalence class of the arrow: $[((a_1,a_2),(a_1+b_1,a_2+b_2))]=[((0,0),(b_1,b_2))]=(b_1,b_2).$
3. Make an arrow from the tail of $\mathbf{a}$ to the head of $\mathbf{b}$. This arrow is $((0,0),(a_1+b_1,a_2+b_2))$.

The resulting arrow is (a representative of) $\mathbf{a}+\mathbf{b}$.

Scalar multiplication

Let $\mathbf{a}=(a_1,a_2)\in Arrow/\sim$ and $\lambda \in \mathbb{R}$. We define the scalar multiplication by

$$\lambda \mathbf{a}=(\lambda a_1,\lambda a_2).$$

The resulting vector is parallel to $\mathbf{a}$ but scaled by $s$.

Higher dimensional vectors

We can extend the same argument to arrows of any dimension. An arrow in the $n$-dimensional space can be represented as a pair of points in ${\mathbb{R}}^n$: $((s_1,s_2,\cdots ,s_n),(t_1,t_2,\cdots ,t_n))\in {\mathbb{R}}^n\times {\mathbb{R}}^n\simeq {\mathbb{R}}^{2n}$. An equivalence relation is introduced between two arrows, $\mathbf{a}=((s_1,s_2,\cdots ,s_n),(t_1,t_2,\cdots ,t_n))$ and $\mathbf{b}=((u_1,u_2,\cdots ,u_n),(v_1,v_2,\cdots ,v_n))$: 

$$\mathbf{a}\sim \mathbf{b}⟺t_i-s_i=v_i-u_i,i=1,2,\cdots ,n.$$

Then, the equivalence classes of ${\mathbb{R}}^n\times {\mathbb{R}}^n$ by $\sim$ are defined to be vectors. That is, vectors are elements of $({\mathbb{R}}^n\times {\mathbb{R}}^n)/\sim \simeq {\mathbb{R}}^n$. Addition and scalar products are defined in the same manner as in the 2-dimensional case.