# Finding the dimension (Linear Algebra)

1. Feb 16, 2010

### EV33

1. The problem statement, all variables and given/known data

My question is how to find the dimension of something.

Let's say: x1-2x2+x3-x4=0

and let's do one with more than 1 equation.

x1+x3-2x4=0
x2+2x3-3x4=0

2. Relevant equations

3. The attempt at a solution

For the second one I figured that if I took the basis of it, then however many vectors were in the basis would be equivalant to the dimensions.

So the transpose is

1 0
0 1
1 2
-2-3

then I row reduced, and took the transpose and got

1 0 0 0
0 1 0 0

This is the basis, so I would say the dimension is 2, because that is the number of column vectors.

For the first one if you go over the same process you will get one vector, and I would assume that means the dimension is 1, but dimension is actually three and I have no idea how.

What should I do different for the first one?

2. Feb 16, 2010

### Mandark

The 2 that you worked out by counting the number of pivots is the rank. Using the rank-nullity theorem you find that the nullity (the dimension you're after) is 4 - 2 = 2 in the first case, which just happens to coincide with the 2 you worked out (by chance). Do you see how this applies in your second case?

3. Feb 17, 2010

### EV33

Well when I do that for the first one I get the nullity to be 3, and the rank to be 1. They add up to 4 which is equal to n. So when I am asked to find dim(W) that is the same thing as asking me to find n? or does dim(W) mean something else?

4. Feb 17, 2010

### Staff: Mentor

It depends on what W is. Typically a question like this would be asking about the dimension of a subspace W of some vector space V.

5. Feb 17, 2010

### EV33

oh ok. For those two problems W is a subspace of R^4 consisting of vectors of the form

x=
x1
x2
x3
x4

determine dim(W) when the components of x satisfy the given conditions. which I posted in the first post.

If it is a subspace it can't be anything greater than R^4, but I don't see how that would change the way I do the problem.

I am not 100% sure on how to do it in the first place.

6. Feb 17, 2010

### EV33

I guess my new problem is that I don't know what dim(W) means?

7. Feb 17, 2010

### vela

Staff Emeritus
The problem is asking you to solve Ax=0, so you're solving for the null space of A and saying what its dimension is.

8. Feb 17, 2010

### EV33

Oh. Thank you. It makes so much more sense now.