According to my understanding, A vector is an element of a set called the vector space which satisfies a list of axioms like : closure under vector addition, closure under scalar multiplication, associativity, commutativity, distributivity, the existence of : a zero vector(additive identity), multiplicative identity , additive inverse and so on.
They're quite useful in physics for at least two reasons:
1)a vector is a quantity that has both magnitude and direction, and such a quantity pops up a lot in physics.
2)a vector is invariant under co-ordinate rotation and translation. Now this is pretty important because: if a physical law can be described by vector equation (e.g. Newton's second law) then this law is invariant under co-ordinate rotation and translation, a property that every physical law should satisfy.
My question is : How does an element in vector space(i.e. a vector) which satisfies the aforementioned list of axioms imply that this element(this vector) is a quantity that has both magnitude and direction?
In addition to that, How does satisfying the aforementioned axioms make this element(this vector) invariant under rotation and translation of coordinates?
Best Answer
Elements of a real vectorspace certainly have direction, but they don't really have a magnitude. Well actually, they... kind-of have a magnitude. But for a proper magnitude, you need further structure, such as a norm or inner product. Let me explain.
Vector Spaces.
Suppose $V$ is a real vectorspace.
This induces an equivalence relation on $V$, so we get a partitioning of $V$ into cells. Each cell is an open ray, so long as we regard $\{0\}$ as an open ray. You may wish to exclude $\{0\}$ from its privileged position as a ray, in which case you should only deal with non-zero vectors; that is, you need to be dealing with $V \setminus \{0\}$ rather than $V$.
Irrespective of which conventions are used, we can make sense of direction using these ideas:
Notice that the equivalence relation of having the same direction is preserved under scalar multiplication; what I mean is that if $v$ and $w$ have the same direction, then $av$ and $aw$ have the same direction, for any $a \in \mathbb{R}$. Geometrically, this means that if we scale a ray, we'll end up with a subset of another ray.
As for magnitude; well, if you choose a ray $R \subseteq V$, then we can partially order $R$ as follows. Given $x,y \in R$, we define that $x \geq y$ iff $x = ry$ for some $r \in \mathbb{R}_{\geq 1}$. So some vectors along this ray are longer than others, hence magnitude.
Inner Product Spaces.
Actually, this isn't the whole story. The problem with vector spaces is that if $x$ and $y$ don't belong to the same ray (nor to the the "negatives" of each others rays), then there's no way of comparing the magnitudes of $x$ and $y$. We can't say which is longer! Now there are mathematical situations where this limitation is desirable, but physically, you probably don't want this. A related issue is that you can't really make sense of angles in a (mere) vector space; at least, not without some further structure.
For this reason, when physicists say "vector", what they usually mean is "element of a finite-dimensional inner-product space." This is a (finite-dimensional) vector space $V$ with further structure; in particular, it comes equipped with a function
$$\langle-,-\rangle : V \times V \rightarrow \mathbb{R}$$
that is required to satisfy certain axioms resembling the dot product. Especially important for us is that these axioms include a "non-negativity" condition:
$$\langle x,x\rangle \geq 0$$
Using this, we can define the magnitude of vectors as follows.
This allows us to compare the magnitudes of vectors that don't live in the same ray; we simply define that $x \geq y$ means $\|x\| \geq \|y\|.$ When confined to a single ray, this agrees with our earlier definition! Be careful though, because the relation $\geq$ we just defined is only a preorder.
In fact, the inner product gives us more than just magnitudes; it also gives angles!
It can be shown that vectors $x$ and $y$ have the same direction (in the sense described at the beginning of my post) iff the angle between them is $0$. In fact, you can modify the above definition so that it defines the angle between any two non-zero open rays. In this case, it turns out that two rays are equal iff the angle between them is $0$.