VECTOR SPACE

VECTOR SPACE, a technical essay by Frederick Glenn

A vector space has the following properties:

Really important properties you need to check for:

if u and v are in V, then so is u +v.

if u is in V, so is ku for any number k.

This implies that the space is linear, or flat (but multi-dimensional). We live in a flat world of 3 dimensions. Our world meets the definition of a vector space. In our world, vectors add and subtract and the result still falls in our world. We can multiply vectors in our world by scalars and still come up with vectors that land in our world.

Our world has 3 dimensions usually defined by the cartesian coordinates ( x,y,z) or some times by other coordinate systems such as polar ( r, q , f ). No matter what coordinate system we choose, we still live in a vector space of dimension 3.

It is interesting to note that the mathematics we are used to works in our world (a coincidence?). That is to say that vector addition and scalar multiplication are both associative and distributive:

Pretty important properites of our vector space:

the zero vector is in V. if you add zero to anything, it doesn't change them.

if u is in V, then so is -u.

Other properties (These are usually so obvious you might forget about them)

u+v=v+u

(u+v)+w=u+(v+w)

c(v+u)=cv+cu (c is a scalar, u and v are vectors)

(c+d)v=cv+dv

1u=u

c(d(u)=(cd)u

Let’s imagine a flat world of 2 dimensions. In vector space notation this is usually called R² where the 2 represents a 2 dimensional space. In this world, all vectors that are in our flatland also end up in flatland (R² ) if we add (or subtract them). Likewise scalar multiplication of vectors in R²end up in flatland. Our flatland world is a vector space by definition if it is infinite in extent.

But what if flatland had a black hole in it. That is, if we cut a hole in the sheet at some point (Hole-y land). We can imagine 2 vectors that would add up and land in the hole which is not part of hole-y land. Therefore, again by definition, hole-y land is not a vecotr space.

You can imagine all kinds of defined spaces that are not Vector spaces. The set of positive real numbers is not, the plane of points extending to the right of the origin is not. Hole-y land is not.

Our local space is flat. A good test of this is to measure the interior angles of a triangle. The sum of the interior angles is 180^o in our world’s Euclidean space (named after the Greek geometer Euclid). Try drawing a triangle on a curved space such as a globe. The interior angles add up to something less than 180^o. If you draw a triangle on a saddle-shaped space, the interior angles add up to something more than 180^o.

But I digress..

Mathematicians talk about sub-spaces:

A subspace of a vector space is a vector space that is inside another vector space.

Some Examples to look at:

R² - subspaces are lines through the origin, and also the point (0,0).

R² - subspaces are lines and planes through the origin

Vector space of quadratic polynomials, and some subspaces.

Magic squares

Exercise

Is the set of points (x,y) with x=3 a subspace of R²?

If it is, prove it. If it's not, give an example to show why not.

Answer

It is not a subspace, since (3,0) and (3,9) are both points in this set, but (3,0)+(3,9)=(6,9) is not a point in the set.

Note, this set of points also fails several other properties that a vector space should satisfy. Eg, it fails the second property, for example, 5*(3,1)=(15,5), so (3,1) is

in this set, but not all it's multiples are. It also fails the third property, as (0,0) is not in this set. And it fails the fourth, since (3,1) is in the set, but (-3,-1) is not. But we

don't need to check all these; even if only one fails, it still is not a subspace.

BASIS and DIMENSION

The basis of a vector space is simply the set of vectors you use to define points in that space. This is simply the standard coordinate system you are used to in our world. In our world’s vector space the basis vectors are the unit vectors pointing in the x,y and z directions. In a four dimensional world, there would be 4 basis vectors (don’t try to picture this).

The basis vectors of a vector space must be independent. That is they must be in some sense orthogonal to each other. The projection of one onto the other must be zero ( dot product is zero ). It wouldn’t work to use the vectors x, 3x, and y to define our world. The vectors x and 3x are not independent and could not be used to define the z plane.

The dimension of a vector space is simply the number of basis vectors needed to define it. An example may help:

There can be many different choices of basis for a vector space,

eg, R² has a basis {(1,0), (0,1)}, but it also has a basis {(1,1), (1,-1)}. Whatever basis we choose will always have the same number of elements. Otherwise, the basis vectors would not be independent.

The basis is about the most important thing about a vector space; it means that even though there are infinitely many things in the vector space, you can epxress them

in terms of a few "basic" vectors, that is, in terms of the basis.

SPANNING A SPACE

If we choose our basis vectors correctly, we can reach every point in the vector space. If our vectors are the typical cartesian coordinate vectors x,y, and z we can reach every point in our 3 dimensional space by using them. This is to say that the R³ vector space is SPANNED by the 3 basis vectors x, y, z. Or we could also say that the SPAN of the vectors x,y,z is what we think of as our as our world in 3 dimensions. The span of a set of vectors and the independence of those vectors are closely related. Let’s look at the set of vectors x, y, z, 3x. You might think this spans a 4 dimensional space, but it does not because 3x is not independent of x. So the SPAN of x, y, z, 3x is 3, not 4. We would need a 4^th independent vector to span a space of 4 dimensions.

If we represent vectors by matrices, then the RANK of the matrix is the dimension of the space SPANNED by its column vectors. If the vector x is represented by (1,0,0) , the vector y as (0,1,0) the vector z as (0,0,1) , and the vector 3x as (3,0,0) the resulting matrix would have the RANK of 3 because the 4 vectors only span a 3 dimensional space.

Free Variables and NULLITY

If we have three variables x,y,z, and we add the condition x+y+z=0, then now we have two "free" variables, y and z, because we can write x in terms of them. (x = -y – z)

We also could write y = -x – z. The point is we still have 2 undefined or "free" variables.

Number of variables (x,y,z) - number of conditions (x+y+z-0) = number of free variables (y,z)

In matrix terms, the number of free variables is known as the NULLITY of the matrix.

An example will help:

What is the space SPANNED by the following vectors:

1 1 0

2 , 1 , 1 (there should be braces around each column)

3 1 2

This is another way of writing (1,2,3) (1,1,1) and (0,1,2) vectors.

So the SPAN is the dimension of the space we can reach by adding up these vectors in some combination:

1 1 0

a 2 , b 1 ,c 1 (there should be braces around each column)

3 1 2

where here I’m trying to write a(1,2,3) b(1,1,1) and c(0,1,2)

The set of all vectors you can get to like this is at least a subspace of R³(our world), since if you can get to a point v and a point u, then you can also get to the point v+u using these three vectors. And if you multiply by a constant k, then kv can also be reached with these vectors.

So the question becomes do the three vectors SPAN R³or just a subspace of it?

Examining the vectors we can see that (1,2,3) – (1,1,1) = (0,1,2). The third vector is not independent of the first two. It cannot, therefore, define another dimension.

A basis is a set of vectors that will get you everywhere (ie, they span the subspace), and they have to be linearly independent.

Well, this subspace is where you can get to with (1,2,3), (1,1,1), and (0,1,2), so they do span it.

So that means the dimension is at most 3. But it might be less than three, e.g., if we took a space spanned by (1,1,1), and (2,2,2) and (3,3,3), it could be spanned

with just (1,1,1), so only has dimension 1. To find the dimension, we need to find a linearly independent set of vectors that span. We need to cut down on the number of vectors we give if we can find any relationships between them.

How do we find relationships between the columns of a matrix (some may not be so obvious)? This is done through the process of ROW REDUCTION (a technique I’m not going to get into here).

In general, to find the dimension of the column space of a matrix, we have to row reduce, and then the number of leading ones is equal to the dimension of the

column space. This number is called the rank of the matrix.

So the rank of

1 1 0

2 1 1 (there should be braces around the matrix)

3 1 2

is 2 and its NULLITY is one.