Vector space and number of subspaces

In summary, the conversation discusses the question of how many two-dimensional subspaces (F_3)^4 has, with the initial solution being 6 subspaces with dimension 2. However, the person is unsure if a different basis could generate additional subspaces, leading to a discussion of the possibility of a vector not being able to be written as a linear combination of the basis vectors. The person then presents another attempt at a solution, but is unable to make progress and asks for help. They clarify that F_3 refers to the finite field with elements {0,1,2}.
  • #1
Buri
273
0
Homework Statement

How many two dimensional subspaces does (F_3)^4 have?

The attempt at a solution

I chose an arbitrary basis so B = (v1,v2,v3,v4) for (F_3)^4 and then basically did 4C2 = 6 so it has 6 subspaces with dimension 2. However, thinking over this problem I've realized that I'm not exactly sure whether a different basis C can possibly generate another subspace that B generated. See if C were to generate another subspace that B didn't, wouldn't this mean that there is a vector in this subspace that can't be written as a linear combination of the vectors in B? And therefore, contradicting the fact that B is actually a basis for (F_3)^4?

This was a bonus question on my term test I just finished writing, so a general idea of how to arrive to a correct solution would be great.

Thanks!

EDIT: Another attempt at a solution

I've tried something else. Let U be subspace generated by two basis elements of B. Now I'd like to show that U is also generated by exactly two basis elements of C. So if I let x be an arbitrary vector in U we could write it as a linear combination of two basis elements of B:

x = a1v1 + a2v2 where a1 and a2 cannot both be zero.

So I would like to show that x = b1u1 + b2u2 where again b1 and b2 cannot both be zero.

Now u1 and u2 can be written as a linear linear combination of the vectors in B since they are in the vector space, but then I'd have to show that the some of the coefficients in this linear combination have to be zero (otherwise x would be a linear combination of 3 or 4 vectors of B contradicting that it was generated by only two vectors of B). But I can't get any where with this. Maybe the assertion isn't even true?
 
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  • #2
apologies if its obvious, but what is [itex] F_3 [/itex]?
 
  • #3
Its the finite field with elements {0,1,2} and arithmetic is performed modulo 3.
 
  • #4
Any ideas?
 
  • #5
Try repeating your argument for a simpler problem, e.g. how many one-dimensional subspaces does F34 have? Or better, how about F32? Or maybe what one-dimensional subspaces does R2 have?
 

FAQ: Vector space and number of subspaces

1. What is a vector space?

A vector space is a set of vectors that can be added and multiplied by scalars (numbers), satisfying certain properties such as closure, commutativity, and associativity.

2. What is the dimension of a vector space?

The dimension of a vector space is the number of linearly independent vectors needed to span the space. It is usually denoted by the symbol "n".

3. How do you find the number of subspaces in a vector space?

The number of subspaces in a vector space is equal to 2^n, where n is the dimension of the vector space.

4. What is a subspace?

A subspace is a subset of a vector space that also satisfies the vector space properties of closure, commutativity, and associativity. It can be thought of as a smaller vector space within a larger one.

5. Can a subspace have a dimension greater than the original vector space?

No, the dimension of a subspace must always be less than or equal to the dimension of the original vector space. This is because a subspace is a subset of the original vector space, and therefore cannot have more linearly independent vectors than the original space.

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