# $16$ subspaces of $F_2^3$

Given the finite field, $$F_2$$, consider the 3-dimensional vector space $$V$$ which contains all of the $$3$$-tuples consisting of elements in $$F_2$$.

In my textbook, it says that $$V$$ has $$16$$ subspaces, but I can't figure out what they are. Can someone please help me? here is what I have so far:

Subsets of dimension $$1$$: $$(0,0,0)$$

Subsets of dimension $$2$$: $$(0,0,0)$$ mixed with any other element $$\rightarrow 7$$ possibilities

Subsets of dimension $$3$$: $$\{ (0,0,0), (0,0,1), (1,1,0) \}, \{(0,0,0), (0,1,0), (1,0,1) \}, \{ (0,0,0), (1,0,0), (0,1,1) \} \rightarrow 3$$ possibilities

There is the entire $$V$$.

So far, I have these $$12$$ subspaces. What are the last $$4$$?

• You're mistaking cardinality and dimension. For instance, $\{(0,0,0)\}$ is zero dimensional, not one dimensional. The dimension is the cardinality of a basis, not of the space itself. Feb 24 at 0:38

There is one $$0$$-dimensional subspace: $$\{0 \}$$.
There are $$7$$ $$1$$-dimensional subspaces: the span of any nonzero element, consisting of $$0$$ and that element.
A $$2$$-dimensional subspace with basis $$\{b_1, b_2\}$$ consists of $$0, b_1, b_2, b_1+b_2$$. Note that any two distinct nonzero elements are linearly independent. Each $$2$$-dimensional subspace thus has $$3$$ (unordered) bases, and there are $${7 \choose 2} = 21$$ possible bases in all, so $$21/3=7$$ $$2$$-dimensional subspaces.
There is one $$3$$-dimensional subspace, namely the whole space. That's a total of $$16$$ subspaces.