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?

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    $\begingroup$ I don't see why this was downvoted, it really is a very good question. I think that if students asked questions like this more often, they would get through their studies quicker. $\endgroup$ – goblin GONE Jul 12 '15 at 1:35
  • $\begingroup$ I think you're conflating two different definitions of "vector": the physicist's definition, and the mathematician's definition. The vector space axioms constitute the mathematician's definition. In order to get a "physical vector", you need to add additional structure on top of that: the ability to calculate a norm, and the behavior under rotation and translation. $\endgroup$ – David Z Jul 12 '15 at 6:05
  • $\begingroup$ You are talking about a maths.kisogo.com/index.php?title=Normed_space and not all vectors have a "direction" to speak of. See maths.kisogo.com/index.php?title=Basis_and_coordinates for a discussion on that (well worth reading) $\endgroup$ – Alec Teal Jul 12 '15 at 12:06

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.

Definition 0. Given a vectors $x,y \in V$, we say that $x$ and $y$ have the same direction iff:

  • there exists $r \in \mathbb{R}_{\geq 0}$ such that $x = ry,$ and
  • there exists $r \in \mathbb{R}_{\geq 0}$ such that $y = rx$.

(The $r$'s don't have to be the same.)

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:

Definition 1. The direction of $x \in V$ is the unique open ray $R \subseteq V$ such that $x \in R$.

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.

Definition 2. Suppose $V$ is a real inner product space. Then the norm (or "magnitude") of $x \in V$, denoted $\|x\|$, is defined a follows:

$$\|x\| = \langle x,x\rangle^{1/2}$$

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!

Definition 3. Suppose $V$ is a real inner product space. Then the angle between of $x,y \in V$, denoted $\mathrm{ang}(x,y)$, is defined a follows:

$$\mathrm{ang}(x,y) = \cos^{-1}\left(\frac{\langle x,y\rangle}{\|x\|\|y\|}\right)$$

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 $0$.

  • $\begingroup$ Very nice and informative answer. $\endgroup$ – Jonathan Hebert Jul 12 '15 at 3:22
  • $\begingroup$ @JonathanHebert, thanks! I'm rather proud of it :) $\endgroup$ – goblin GONE Jul 12 '15 at 3:23
  • $\begingroup$ Thank you for your awesome answer! $\endgroup$ – Omar Nagib Jul 12 '15 at 6:38
  • $\begingroup$ @goblin Can you elaborate on the fact that, such element of finite dimensional inner-product space (that is, the vector in the sense used in physics) is invariant under coordinate rotation and translation? $\endgroup$ – Omar Nagib Jul 12 '15 at 6:48
  • $\begingroup$ @OmarNagib, thanks for the kind words. Unfortunately I don't know much physics, so I can't really help you there, but I'll let you know if anything shows up. Also, I've had a bit of a snoop around wikipedia, and the whole issue of invariance in physics really isn't dealt with in a mathematically precise way anywhere that I can see. I think its probably quite an advanced concept. $\endgroup$ – goblin GONE Jul 12 '15 at 6:59

The word "vector" is used to mean different things in math. Sometimes "vector" means "an ordered $n$-tuple of real numbers". Sometimes it means an equivalence class of directed line segments (with respect to a certain equivalence relation). Sometimes it means an element of a particular vector space. Which definition is being used depends on context. If a vector space does not have a norm defined on it, then indeed the elements of that vector space do not have magnitudes. In that context the "magnitude" of a vector has not been defined.

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    $\begingroup$ Although my knowledge of linear algebra is pretty archaic, it seems to me, that the "different" definitions you provided for vectors are not so different. you said a vector is maybe 1) a $n$-tuple of real numbers 2) an element of vector space and 3) a directed line segment. How these definitions are different? they seem to me to describe exactly the same object. $\endgroup$ – Omar Nagib Jul 12 '15 at 1:24
  • $\begingroup$ Btw I said an equivalence class of directed line segments, not a directed line segment. That's an important distinction. You commonly encounter vector spaces whose elements are not ordered $n$-tuples of real numbers, and yet the elements of such a vector space are still called "vectors" in that context. $\endgroup$ – littleO Jul 12 '15 at 1:29

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