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Let $\textsf V = (\mathbb{Z}_2)^3$ and $F=\mathbb{Z}_2=\{0,1\}$.

How many different ordered basis does $\textsf V$ (over $F$) have? First of all, we know that the cardinality of $\textsf V$ is $2^3 = 8$ with

$$\textsf V = (\Bbb Z_2)^3 = \left\{ \begin{pmatrix} 0\\0\\0 \end{pmatrix}, \begin{pmatrix} 1\\0\\0 \end{pmatrix}, \begin{pmatrix} 0\\1\\0 \end{pmatrix}, \begin{pmatrix} 0\\0\\1 \end{pmatrix}, \begin{pmatrix} 1\\1\\0 \end{pmatrix}, \begin{pmatrix} 1\\0\\1 \end{pmatrix}, \begin{pmatrix} 0\\1\\1 \end{pmatrix}, \begin{pmatrix} 1\\1\\1 \end{pmatrix} \right\}$$

Since $\dim (\textsf V) = 3$, basis come in $3$ vectors. I have manually listed all possible permutation, eliminating those that are linearly dependent, and the zero vector.

I got $210 - 18 = 192$ different ordered basis (unfortunately, I don't know how to show it more systematically).

Kindly help me with this. Next : how many distinct two-dimensional subspaces (over $F$) does $\textsf V$ have?

I think there are three two-dimensional subspaces, but I don't know how to show it. I get stuck here. Please help!

azif00
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Yu-Chen
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  • First, you can choose an unordered basis for $V$ and then permute it. It will give you the total number of ordered basis. – Kumar Sep 09 '19 at 09:38
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    Your question is answered (in quite a bit more generality) here: https://math.stackexchange.com/q/70801/101420 – Vincent Sep 09 '19 at 10:07
  • The 2-dimensional subspaces are the kernels of non-zero linear transformations $T:V\to\Bbb{Z}_2$. There are $2^3=8$ linear transformations. Leaving out the zero map gives $7$ as the answer. – Jyrki Lahtonen Sep 10 '19 at 02:41
  • By the way, the number of ordered bases is $7\cdot6\cdot4=168$, but that is not relevant here. The number of ordered pairs of linearly independent vectors is $7\cdot6=42$: the first cannot be zero, 7 choices, the second cannot be neither zero nor the one already chosen, 6 choices. Similarly we see that $3\cdot2$ pairs give rise to the same 2-d subspace. So $42/6=7$. – Jyrki Lahtonen Sep 10 '19 at 02:46
  • If you need to list them pick $a_1,a_2,a_3\in\Bbb{Z}_2$, at least one non-zero. The space consists of the zeros $(x_1,x_2,x_3)$ of $a_1x_1+a_2x_2+a_3x_3=0$. – Jyrki Lahtonen Sep 10 '19 at 02:48
  • Can you elaborate on this one? "the number of ordered bases is 7⋅6⋅4=168" I am still confused – Yu-Chen Sep 11 '19 at 13:51

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