Showing an infinite dimensional vector space has an arbitrary dimensional quot. space

  • Thread starter jdinatale
  • Start date
  • #1
155
0
aaaa.png


I'm trying to understand this problem. Let's take an infinite dimensional vector space, say [itex]\mathbb{R}^2[/itex] and let [itex]n = 4[/itex]. This problem states we can find a subspace $U$ such that dim(\mathbb{R}^2/U) = 4$. Well, one subspace of $\mathbb{R}^2$ is [itex]U = \{(x, y) : x*y \geq 0\}[/itex] (i.e. the first and third quadrant). So [itex]\mathbb{R}^2 / U = \{v + U : v \in \mathbb{R}^2\}[/itex]. But that means [itex]\mathbb{R}^2 / U = \mathbb{R}^2[/itex], right?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619


aaaa.png


I'm trying to understand this problem. Let's take an infinite dimensional vector space, say [itex]\mathbb{R}^2[/itex] and let [itex]n = 4[/itex]. This problem states we can find a subspace $U$ such that dim(\mathbb{R}^2/U) = 4$. Well, one subspace of $\mathbb{R}^2$ is [itex]U = \{(x, y) : x*y \geq 0\}[/itex] (i.e. the first and third quadrant). So [itex]\mathbb{R}^2 / U = \{v + U : v \in \mathbb{R}^2\}[/itex]. But that means [itex]\mathbb{R}^2 / U = \mathbb{R}^2[/itex], right?

That's pretty much all wrong so far. Just because a vector space has an infinite number of vectors doesn't make it infinite dimensional. If the scalar field is R, R^2 isn't infinite dimensional, what is the dimension? U isn't a subspace, why not? Try and review some definitions.
 
  • #3
155
0


That's pretty much all wrong so far. Just because a vector space has an infinite number of vectors doesn't make it infinite dimensional. If the scalar field is R, R^2 isn't infinite dimensional, what is the dimension? U isn't a subspace, why not? Try and review some definitions.

I feel really stupid for claiming that [itex]R^2[/itex] is infinite dimensional. Let me try again. Let [itex]V[/itex] be the set of all polynomials with coefficients in [itex]R[/itex]. Let [itex]n = 4[/itex].

We have to find a subspace of V, U such that V / U has degree 4.

But V / U is the set of all cosets of U in V. Let's say U is the set of all cubic polynomials. Wouldn't V / U = v + U = V?
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619


I feel really stupid for claiming that [itex]R^2[/itex] is infinite dimensional. Let me try again. Let [itex]V[/itex] be the set of all polynomials with coefficients in [itex]R[/itex]. Let [itex]n = 4[/itex].

We have to find a subspace of V, U such that V / U has degree 4.

But V / U is the set of all cosets of U in V. Let's say U is the set of all cubic polynomials. Wouldn't V / U = v + U = V?

Well, no it wouldn't exactly be V. But it would be infinite dimensional, if that's what you mean. Let's skip the 'infinite dimensional' part for a bit. Suppose you are given the ten dimensional space V=R^10 and you want to show me that there is a subspace U of V such that V/U has dimension 4. How would you do that?
 
  • #5
155
0


I would let [itex]U = \{(a, b, c, d, e, f, 0, 0, 0, 0) : a, b, c, d, e, f \in \mathbf{R}\}[/itex] Because I think there's a relationship where dim(V) = dim(U) + dim(V/U). Based on that equation, dim(V/U) = 4.

But that doesn't make sense to me because the set of all cosets of U in V
would be all (a, b, c, d, e, f, 0, 0, 0, 0) + V which would be V itself, right?
 
  • #6
Dick
Science Advisor
Homework Helper
26,260
619


I would let [itex]U = \{(a, b, c, d, e, f, 0, 0, 0, 0) : a, b, c, d, e, f \in \mathbf{R}\}[/itex] Because I think there's a relationship where dim(V) = dim(U) + dim(V/U). Based on that equation, dim(V/U) = 4.

But that doesn't make sense to me because the set of all cosets of U in V
would be all (a, b, c, d, e, f, 0, 0, 0, 0) + V which would be V itself, right?

The union of all of the vectors in all of the cosets is V, yes. But that's not the point. V/U is a set of cosets, not a set of vectors. (a, b, c, d, e, f, 0, 0, 0, 0) is one coset, (a, b, c, d, e, f, 1, 0, 0, 0) is another coset. (a, b, c, d, e, f, 2, 0, 0, 0) is another coset. (a, b, c, d, e, f, 0, 1, 0, 0) is yet another coset. Each coset is itself a six dimensional space. The SET OF COSETS is 4 dimensional. Does that make sense?
 

Related Threads on Showing an infinite dimensional vector space has an arbitrary dimensional quot. space

Replies
8
Views
3K
  • Last Post
Replies
3
Views
13K
Replies
3
Views
3K
  • Last Post
Replies
2
Views
3K
Replies
4
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
869
  • Last Post
Replies
18
Views
3K
Top