8
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ 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 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
12
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ 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 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 Jul 12 '15 at 6:59
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.