How many subspaces of each dimension does Z_2^3 have?

[Benny]

Hi, I'm wondering how I would decide how many "subspaces of each dimension $$Z_2^3$$ has." The answer is: 1 subspace with dim = 0, 7 with dim = 1, 7 with dim = 2, 1 with dim = 3.

I'm looking for subsets of $$Z_2^3$$ which are closed under addition and scalar multiplication. An arbitrary vector in $$Z_2^3$$ is (a,b,c) where $$a,b,c \in Z_2$$. I can't think of a general way to do this, trial and error is a possibility and quite time consuming. So I think that there is some concept being tested which I'm not seeing.

Any set consisting of a single vector with the zero vector in Z_2 would be a subspace since it would be closed under addition. But what about sets with 3 or more elements. I'm not sure how to approach this question. Can someone please help me out?

 

[Careful]

Obviously, there is only one subspace of dimension one and three:
In general, one has a duality between subspaces of dimension k and n-k. So, you only need to count the number of one dimensional subspaces. There are 2^(3) - 1 = 7 nonzero vectors and they are all linearly independent.

 

[HallsofIvy]

Obviously "Careful" meant to say "there is only one subspace of dimension zero and three".


Be careful, Careful!

 

[Benny]

Hmm, ok thanks for your help.