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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 $4$?

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    $\begingroup$ 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. $\endgroup$ Feb 24 at 0:38
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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.

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