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Enumerate all two-dimensional subspaces of the space $(\mathbb Z/2\mathbb{Z})^3$.

Obviously we have $|(\mathbb Z/2\mathbb{Z})^3|=2^3=8$ with

$$(\mathbb Z/2\mathbb{Z})^3=\{v_0,\ldots,v_7\}=\left\{ \begin{pmatrix}0\\0\\0\end{pmatrix}, \begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}, \begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\end{pmatrix}, \begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}0\\1\\1\end{pmatrix}, \begin{pmatrix}1\\1\\1\end{pmatrix}\right\}.$$

However in the solution to the question is the assumption that there are seven different subspaces $\mathcal{C}_k\subset(\mathbb{Z}/2\mathbb{Z})^3,k\in\mathbb N,$ with $\dim(\mathcal{C_k})=2$:

$$ \mathcal C_1:=\operatorname{span}\{v_1,v_2\}=\operatorname{span}\{v_1,v_3\}=\operatorname{span}\{v_2,v_3\}\\ \mathcal C_2:=\operatorname{span}\{v_1,v_4\}=\operatorname{span}\{v_1,v_5\}=\operatorname{span}\{v_4,v_5\}\\ \mathcal C_3:=\operatorname{span}\{v_1,v_6\}=\operatorname{span}\{v_1,v_7\}=\operatorname{span}\{v_6,v_7\}\\ \mathcal C_4:=\operatorname{span}\{v_2,v_4\}=\operatorname{span}\{v_2,v_6\}=\operatorname{span}\{v_4,v_6\}\\ \mathcal C_5:=\operatorname{span}\{v_2,v_5\}=\operatorname{span}\{v_2,v_7\}=\operatorname{span}\{v_5,v_7\}\\ \mathcal C_6:=\operatorname{span}\{v_3,v_4\}=\operatorname{span}\{v_3,v_7\}=\operatorname{span}\{v_4,v_7\}\\ \mathcal C_7:=\operatorname{span}\{v_3,v_5\}=\operatorname{span}\{v_3,v_6\}=\operatorname{span}\{v_5,v_6\} $$

I did recognize the pattern that $\operatorname{span}\{v_a,v_b\}=\operatorname{span}\{v_a,v_c\}=\operatorname{span}\{v_b,v_c\}$ and I understand why this is true, but how do I get both the number of all subspaces AND the subspaces, too? My first thought was that the nullspace can be ignored if one wants to span spaces with two vectors - is that the reason why we have exactly seven subspaces?

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up vote 1 down vote accepted

Although you can count subspaces by counting their bases, as you did, and then grouping together those bases that define the same subspace, you can list the subspaces more efficiently as follows (a technique that works well because you are looking at subspaces of dimension one less than the whole space, which are called hyperplanes in general).

Every subspace is determined by one nontrivial linear homogeneous equation. Moreover two nontrivial linear homogeneous equations define the same subspace if and only if one is obtained from the other by multiplication by a nonzero scalar. But here there is only one nonzero scalar, so different equations give different subspaces. So you can just enumerate the nontrivial linear homogeneous equations. There are $8$ possible equations (do you see which?) one of which is trivial.

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