# Vectors having many dimensions

1. Oct 7, 2004

### bezgin

While it was hard for me to conceptualize a 3-D world in my head, they told me vectors might have more than 3 dimensions! It was my first linear algebra lesson, I was kinda shocked. How is it possible to define a vector named u, for instance, that have four dimensions? u(p,q,r,s) ????????

2. Oct 7, 2004

### jcsd

It's a fairly natural extension of the concept of a vector to allow it to have four lineraly indpendent componets, of course such a vector can only exist in a four dimensional vector space.

Infact vectors can have an infinite number of compoenets (i.e. vector spaces can be infinite dimensional)!

3. Oct 7, 2004

### StatusX

would an infinite dimensional vector space have more vectors than there are real numbers?

4. Oct 7, 2004

### arildno

A word of caution here, StatusX:
When dealing with sets with infinitely many elements, the ordinary intuition of "largeness" (or more-ness) becomes very shaky indeed.
What you need to develop in order to handle such questions in a mathematically rigourous way, is the concept of "cardinality".
I won't go into this here; others here are far more competent than me to enlighten you on these issues.

5. Oct 7, 2004

### NateTG

Not always. For example, any countably dimensional space of real-valued vectors has the cardinality of the continuum.

The cardinality of a vector space is always larger than the cardinality of a basis of that vector space. This follows readily from a diagonal argument.

6. Oct 7, 2004

### StatusX

is a vector of uncountable dimensions just a function? does that mean there are more functions than real numbers?

sorry for taking over this thread. to answer your original question, it is possible to have vectors of more than 3 dimensions. einstein used 4-dimensional vectors in his theory of relativity. they are pretty much impossible to visualize, but their math is just a straightforward extension of 1, 2, and 3 dimensional vectors. For example, the length of the 4-vector (1,3,2,-1) is sqrt(1+9+4+1) = sqrt(15).

Last edited: Oct 7, 2004
7. Oct 7, 2004

### NateTG

A vector is essentially a function from an index set to a ring. So, every vector is a function, but not all functions are vectors.

For example, let's say we have your familiar 3-dimensional real-valued vector.
$$\vec{v}=<v_x,v_y,v_z>$$

and let's compare it to the function:
$$v:\{x,y,z\} \rightarrow \Re$$
with
$$v(x)=v_x$$
$$v(y)=v_y$$
$$v(z)=v_z$$

It's pretty easy to see that they're rather similar. For a countable-dimesional vector:
$$\vec{v}=<v_1,v_2,v_3...>$$
and a function
$$v:\mathbb{N} \rightarrow \Re$$
with
$$v_n=v(n)$$

I'm not sure what you mean by that question. There are 'more' functions from the reals to the reals than there are real numbers.

8. Oct 7, 2004

### StatusX

thats what i meant by function in both questions, a function from the reals to the reals.

Last edited: Oct 7, 2004
9. Oct 7, 2004

### NateTG

The answer to the question you probably wanted to ask is yes, functions from the reals to the reals can be treated as a kind of uncountable dimensional vector. However, there are many other kinds of uncountable dimensional vectors than functions from the reals to the reals.

Also, the cardinality of the set of functions from the reals to the reals is larger than the cardinality of the reals, so it's typically said that there are more functions from the reals to the reals than there are reals.

10. Oct 8, 2004

### matt grime

But to get back to dimensions and visualization.

The dimension of a vector space can perhaps be thought of as the number of bits of data that are needed to specify things.
For instance I may wish to keep track of the position (3 components), velocity, acceleration, weight, temperature etc of some projectile. I can arrange these in an array with 7 entries: (x,y,z,v,a,w,t)

I can now treat these quantities *abstractly* as if they were vectors just like ordinary position vectors in three dimensions.

This might not actually mean anything realistically in our model in this case: what does it mean to add together the temperatures of two projectiles? But I can do it all the same.

After all, position vectors are just an abstraction of space (is space infinitely divisible?). It's a model for something, that's all.

Sometimes we need to have models of things with 3,4,5 or many more dimensions. You just can't draw them or make a physical model of them in space, but there's no need to do that: we can study them abstractly and learn far more about them that way, and it turns out lots of things are vector spaces, often without a finite number of dimensions.

So, how do you define a 4 vector u? Well, you just do, and its properties are deducible from the fact that it's a 4 dimensional vector. You can't visualize it, but you shouldn't need to.

11. Oct 9, 2004

### bezgin

Then the principles of linear algebra is built on an assumption that we can have more than 3 dimensions. The thing is, how can we prove the existence of infinite dimensions while we can't visualize more than 3?

12. Oct 9, 2004

### jcsd

The number of dimensions in 'physical space' really doesn't matter in linear algebra, we're not necessaily trying to model 'physical space'. Whether dimensions 'exist' or not is no concern of math's.

13. Oct 9, 2004

### Hurkyl

Staff Emeritus
Not quite; the principles of linear algebra are built without the assumption that there are no more than 3 dimensions. Linear algebra applies to anything that satisfies the abstract definition of a vector space.

Visualization has nothing to do with it. We (constructively) prove the existance of infinite dimensional vector spaces by defining what it means to be infinite dimensional, and exhibiting a construct that satisfies the definition.

By the way, here's an example of a (countably) infinite dimensional vector space that you might be able to visualize somewhat: the space of all continuous real valued functions. (that is, each vector is a continuous function on the real numbers)

14. Oct 10, 2004

### mathwonk

There is a distinction between a geometric vector space, in which the vectors are arrows emanating from an origin point, and in which addition is done by the parallelogram law, as opposed to an algebraic vector space in which the "vectors" are merely elements of some algebraic set in which it is possible to add two vectors, and also to multiply a vector by a number. In fact the algebraic elements can always be taken to be real valued functions.

Given a geometric vector space, (which most people can only imagine visually in one or two, or three dimensions), one can make it into an algebraic vector space by choosing an axis system, or "basis" as follows: just choose, in three space say, an ordered set of three non zero vectors, all in different directions, i.e. no one is in the plane spanned by the other two.

Use these three vector to set up a three dimensional axis system, so that every vector has its foot at the origina, and its head at some point having three coordinates (x,y,z).

Then we replace the arrow from the origin to the point with coordinates (x,y,z), simply the three coordinates themjselves. I.e. we say that, algebraically speaking, the triple of numbers (x,y,z), is the vector. In this way, the geometry is gone and we are calling as "vectors" simply all ordered triples of real numbers (x,y,z).

Then we just add these component by component and multiply by numbers in a similar way.

Now once we have done this there is nothing to stop us from saying that an ordered 4 tuple of real numbers is a 4 dimensional vector, added and multiplied in the same way. How to think of it geometrically is another problem, which we actually never need to deal with, although it is fun to try.

Then why not make a jump and call all sequences of real numbers vectors too. We can add then term by term, and multiply each term by the same scalar. Remember too that a sequence is also interpreted as a function from the positive integers to the reals. Just as the previous example was a 4 diemensional vector space, this is an infinite dimensional vector space. I.e. the dimension of the space is the number of coordinates of each vector.

Then one could just say that any real valued fucntion on any domain space is a "vector" since one can add functions and also multiply a function by a scalar. So algebraically a vector space is just the set of all real valued functions on any domain set. The dimension is then the number of points in the domain set, not the number of functions on that set.

Abstractly, one can also call a vector space any set of objects that can be added and multiplied by scalars, subject to all the usual laws, but it can be proved that by choosing a basis, all such vector spaces are equivalent to spaces of functions.

The interesting invariant associated to a vector space is not the number of points in it, i.e. not the number of vectors, but the number that are "independent".

I.e. a set of vector is independent if none of them equals a sum of multiples of the others. Then the maximal number of independent ones is called the "dimension", which you see has no geometric dependency at all, and is defined purely algebraically. Wonderful to say however, it exactly coincides with the usual geometrical concept of dimension for sopaces of dieension 1, 2 and 3. It also coincides with the number of points of the domain set in the representation as a function space on a set.

After some practice, one can begin to imagine a family of three dimensional subspaces sliding along a 4th axis in 4 space, or even higher, say a 5 dimensional subspace of a 7 dimensional space. Maybe even an infinite number of mutually perpendicular vectors sticking out of the same point, all in very different directions. Why not?

Last edited: Oct 10, 2004