Finding number of linearly independent subsets with $3$ elements of a vector space over $\Bbb Z_2$ of dimension $4$.

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Suppose $V$ is a vector space over $\mathbb{Z}$$_2$ of dimension $4$. Then we have to find the number of linearly independent subsets of $V$ containing $3$ elements.

My try is that first I had calculated the number of vectors in $V$. Since $\dim(V)= 4$ given then let $\beta = \{ v_1, v_2, v_3, v_4\}$ be a basis so we can write any vector of V as v= $a_1$$v_1$ + $a_2$$v_2$ + $a_3$$v_3$ + $a_4$$v_4$ uniquely for each $a_i$ in $\mathbb{Z}$$_2$.

$\mathbb{Z}$$_2$ is the field with $2$ elements. So by permutation-combination we get the number of elements of $V$ is $2^4$.

But after that I am not able to think how to proceed. So please help me to solve this.

Answer 1

You calculated $|V| = 2^4$ correctly. Let us now start by choosing one linear independent vector $v_1$:

• There are $2^4 - 1$ choices for $v_1$, all vectors but zero.

To choose the second vector in our set, we have to avoid all vectors dependent on $v_1$, of which there are $2^1$ (all multiples of $v_1):

• There are $2^4 - 2^1$ choices for $v_2$, all vectors not in the subspace generated by $v_1$.

To choose the third vector, we have to avoid all vectors dependent on $v_1$ and $v_2$, which are the vectors $\alpha_1v_1 + \alpha_2 v_2$, $\alpha_i \in \mathbf Z/(2)$, hence there are $2^2$ of them:

• There are $2^4 - 2^2$ choices for $v_3$, all vectors not in the subspace generated by $v_1, v_2$.

Hence there are $15 \cdot 14 \cdot 12 = 2520$ ways to choose a triple $(v_1,v_2,v_3)$ of independent vectors. We are asked to count the number of subsets, each of which corresponds to $3! = 6$ triples (all possible orderings of the three elements). Hence, there are $\frac{2520}6 = 420$ linear independent sets with three elements.