Polynomials are typically a good way to think of vector spaces.  You can add or subtract two polynomials and you end up with a polynomial of the same degree.  But if you try to multiply them you will (typically) get a polynomial of greater degree.  If you try to divide, you may not end up with a polynomial.  Also, you can multiply and divide by scalars.

Ricky: Yes, I was going to touch on polynomial space in a bit, I wanted to get established first, however, I think I may have to address this first;

Jane: I don't at all like the way you expressed this, so I guess I'll have to take a bit of a detour.

Consider R, the real numbers. We grew up with them and their familiar properties, and some of us became quite adapt at manipulating them in the usual ways. But we never stopped to think of what we really meant by R, and what the properties really represent. Abstract algebra, which is what we are doing here, tries to tease apart those particular properties that R shares with other mathematical entities.

It turns out that R is a ring, is a field, is a group, is a vector space, is a topological space......., but there are groups that are not fields, vector spaces that are not groups and so on. So it pays to be careful when talking about R, by which I mean one should specify whether one is thinking of R as a field or as a vector space.

So, what you claim is not true, in the sense that the field R is a field, not a vector space. What is true, however, is that R, Q, C etc have the properties of both fields and vector spaces. So for example, if I say that -5 is a vector in R, I mean there is an object in R that has magnitude +5 in the -ve direction. Multiplication should be thought of as applying a directionless scalar, say 2, which is an object in R viewed as a scalar field.

I hope this is clear. Sorry to ramble on, but it's an important lesson - R, C and Q are dangerous beasts to use as examples of, well anything, really, because they are at best atypical, and at worst very confusing.

Sorry for the delay in proceeding, folks, I had a slight formatting problem for which I I have found a partial fix - hope you can live with it. But first this;

When chatting to jane, I mentioned that the real line R is a vector space, and that we can regard some element -5 as a vector with magnitude +5 in the -ve direction. But of course there is an infinity of vector in R with magnitude +5 in the -ve direction. I should have pointed out that, in any vector space, vectors with the same magnitude and direction are regarded as equivalent, regardless of their limit points (they in fact form an equivalence class, which we can talk about, but it's not really relevant here)

We agreed that a vector in the {x,y} plane can be written

, where

are scalar. We can extend this to as many Cartesian dimensions as we choose. Let's write

, i = 1,2,...,n. (note that the superscripts here are not powers, they are simply labels.

Now, you may be thinking I've taken leave of my senses - how can we have more that 3 Cartesian coordinates? Ah, wait and see. But first this. You may find the equation

a bit daunting, but it's not really. Suppose n = 3. All it says is that there is an object called v whose direction and magnitude can be expressed by adding up all the units (the

) that v projects on the coordinates

. Or, if you prefer, v has projection 

along the 

th coordinate.

Now it is a definition of Cartesian coordinates that they are to be perpendicular, right? Then, to return to more familiar territory, the x-axis has zero projection of the y-axis, likewise all the other pairs. This suggests that I can write these axes in vector form, so take the x-axis as an example. x = ax + 0y + 0z, this is the definition of perpendicularity (is this a word?) we will use. So, hands up - what's "a"? Yeah, well this alerts us to a problem, which I'll state briefly before quitting for today.

You all said "I know, a = 1" right? But an axis, by definition, extends to infinity, or at least it does if we so choose. So, think on this; an element of a Cartesian coordinate system can be expressed in vector form, but not with any real sense of meaning. The reason is obvious, of course: Cartesian (or any other) coordinates are not "real", in the sense that they are just artificial constructions, they don't really exist, but we have done enough to see a way way around this.

More later, if you want.

Just a concern; when you say that

you are assuming an orthonormal basis in cartesian coordinates (where the metric tensor is just the identity,

). Perhaps you should make this assumption clearer. This is certainly fine in an introduction, as the inner product is given as

usually when first introduced (at that stage usually you are not worried about vector spaces which use other inner products). Overall I just feel that the inner product section is a bit shaky. For example, you never tell the reader that

even though you substitute it into your formula out of nowhere.

So, lemme see if I can make the inner product easier.

First recall the vector equation I wrote down:

. I hope nobody needs reminding this is just shorthand for

. The scalars

are called the components of v. Note that the

do not enter explicitly into the sum; they merely tell us in what direction we want to evaluate our components. (v is a vector, after all!).

So we can think of our vector projecting

units on the ith axis. But we can ask the same question of two vectors: what is the "shadow" that, say, v casts on w? Obviously this is at a maximum when v and w are parallel, and at a minimum when they are perpendicular (remember this, it will become important soon enough.

OK, so in the equation I wrote for the inner product

what's going on? specifically, why did I switch indices on b, and what happened to the coordinates, the x's?

Well, it makes no sense to ask, in effect, what is v's projection in the x direction on w's projection on the y axis. So we only use like pairs of coordinates and components, in which case we don't need to write down coordinates. In fact it's better if we don't, since we want to emphasize the fact that the inner product is by definition scalar. This what the Kroenecker delta did for us. It is the exact mathematical equivalent of setting all the diagonal entries of a square matrix to 1, and all off-diagonal entries to 0!

On last thing. Zhylliolom wrote the inner product as something a bit like this <v,w>. I prefer (v, w), a symbol reserved in general for an ordered pair. There is a very good reason for this preference of mine, which we will come to in due course. But as always, I've rambled on too long!

Last edited by ben (2007-04-19 08:13:29)