Finding the dimension (Linear Algebra)

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Homework Help Overview

The discussion revolves around finding the dimension of subspaces in linear algebra, specifically through the context of solving systems of linear equations. The original poster presents two equations and seeks to understand the dimension of the solution space.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the rank and nullity of matrices, questioning how these concepts apply to the dimensions of the subspaces defined by the equations. There is an attempt to clarify the meaning of dimension in the context of the null space.

Discussion Status

The discussion is active, with participants providing insights into the rank-nullity theorem and its implications for determining dimensions. Some participants express uncertainty about the definitions and the implications of their calculations, while others offer clarifications on the nature of the subspace and the problem's requirements.

Contextual Notes

There is some confusion regarding the definitions of dimension and null space, as well as the implications of the equations provided. The original poster is working within the constraints of a linear algebra homework assignment, which may impose specific methods or interpretations.

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

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