Finding the dimension (Linear Algebra)

In summary, the conversation discusses how to find the dimension of a subspace W in R^4, given a system of equations. The process involves solving for the null space of the system, and the number of vectors in the basis of the null space is equivalent to the dimension of W. The concept of rank and nullity are also mentioned in finding the dimension.
  • #1
EV33
196
0

Homework Statement



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


Homework Equations





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 equivalent 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?
 
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  • #2
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
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
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
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
I guess my new problem is that I don't know what dim(W) means?
 
  • #7
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
Oh. Thank you. It makes so much more sense now.
 

1. What is a dimension in linear algebra?

A dimension in linear algebra refers to the number of independent variables needed to describe a vector space. It represents the maximum number of vectors that can be linearly independent in a vector space.

2. How do you find the dimension of a vector space?

The dimension of a vector space can be found by counting the number of linearly independent vectors or by using the rank-nullity theorem, which states that the dimension of a vector space is equal to the sum of its rank and nullity.

3. Can a vector space have an infinite dimension?

Yes, a vector space can have an infinite dimension. This is the case for spaces such as the set of all real numbers or the set of all polynomials with coefficients from a field.

4. What is the relationship between dimension and basis in linear algebra?

A basis is a set of linearly independent vectors that span a vector space. The number of vectors in a basis is equal to the dimension of the vector space. A basis is used to represent all other vectors in the vector space.

5. How does the dimension of a vector space affect its properties?

The dimension of a vector space affects its properties in many ways. For example, the dimension determines the maximum number of linearly independent vectors, the size of the basis, and the number of solutions to a system of linear equations. It also affects the dimension of the matrix representation of linear transformations.

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