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

I'm trying to understand this problem. Let's take an infinite dimensional vector space, say $\mathbb{R}^2$ and let $n = 4$. 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 $U = \{(x, y) : x*y \geq 0\}$ (i.e. the first and third quadrant). So $\mathbb{R}^2 / U = \{v + U : v \in \mathbb{R}^2\}$. But that means $\mathbb{R}^2 / U = \mathbb{R}^2$, right? Answers and Replies Dick Science Advisor Homework Helper I'm trying to understand this problem. Let's take an infinite dimensional vector space, say $\mathbb{R}^2$ and let $n = 4$. 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 $U = \{(x, y) : x*y \geq 0\}$ (i.e. the first and third quadrant). So $\mathbb{R}^2 / U = \{v + U : v \in \mathbb{R}^2\}$. But that means $\mathbb{R}^2 / U = \mathbb{R}^2$, 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.

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 $R^2$ is infinite dimensional. Let me try again. Let $V$ be the set of all polynomials with coefficients in $R$. Let $n = 4$.

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?

Dick
Homework Helper

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

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?

I would let $U = \{(a, b, c, d, e, f, 0, 0, 0, 0) : a, b, c, d, e, f \in \mathbf{R}\}$ 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?

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