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Here is the question and it is solution:

Consider the subset $S \subset \mathbb C^3$ consisting of vectors whose only entries are either $0$ or $1.$ How many different bases for $\mathbb C^3$ can we form out of this subset?

$\textbf{Solution.}$

Since the entries of the vectors of $S \subset \mathbb C^3$ are either $0$ or $1,$ then $|F| = 2.$ Since $V = \mathbb C^3$ is of dimension $3,$ So each vector in $v(3,2)$ can be described as a $3$-tuple $(v_1, v_2, v_3)$ where each component $v_j$ is either $0$ or $2,$ so the maximum number of vectors in this vector space are $2^3.$\

Since a basis for $V$ is a linearly independent set, so it can not contain the zero vector, i.e., the first element of the basis can be chosen by $2^3 - 1$ ways. Now, for the second element in this same basis, we should exclude all linear combinations of the first element to pertain linear independence. i.e., the second element can be chosen by $(2^3 - 2)$ ways. For the third element of the same basis ( this is why I multiply the numbers together) we should exclude all linear combinations of the first and the second i.e., $(2^3 - 2^0)(2^3 - 2)(2^3 - 2^2)$ is the number of ordered basis. And the number of unordered basis is $28$ (by dividing by 3!).

I was told that the number I got close to the correct one but not the correct one. Can someone check my solution and correct it please?

Thanks!

Idonotknow
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  • What exactly do you mean when you say "then $|F| = 2$"? What is $F$, and what relevance does it have to this problem? – Ben Grossmann Apr 28 '22 at 12:29
  • Also, bases are usually ordered, so it's strange that you divide by $3!$ – Ben Grossmann Apr 28 '22 at 12:34
  • The relationship between your answer and the correct one is the relationship between this sequence and this one. – Ben Grossmann Apr 28 '22 at 12:37
  • I mean the number of the field elements and I am solving the question very generally and counting both ordered and unordered basis which is in my opinion (and I could be wrong) good. @BenGrossmann – Idonotknow Apr 28 '22 at 12:37
  • @BenGrossmann I am sorry, still I do not understand what is the correct solution. – Idonotknow Apr 28 '22 at 12:38
  • The correct solution is 29. There is no known "closed form" formula – Ben Grossmann Apr 28 '22 at 12:39
  • why there is no known closed formula? also how can I conclude this 29? – Idonotknow Apr 28 '22 at 12:42
  • Your answer is the correct number of bases for $S \subset \Bbb F_2^3$, where $\Bbb F_2$ is the field of two elements. However, a set of $(0,1)$ vectors over $\Bbb F_2$ might be linearly dependent even though the corresponding set over $\Bbb C$ is linearly independent. – Ben Grossmann Apr 28 '22 at 12:42
  • @BenGrossmann ok, so could you please provide me with a correct detailed logical conclusion for the number 29? – Idonotknow Apr 28 '22 at 12:44
  • I don't have the time to do this, but here is an outline of one approach. It is equivalent to count the number of invertible $3 \times 3$ matrices with $(0,1)$ entries (and then divide by $3!$ if desired). Consider the possible values for the determinant of such a matrix; argue that the determinant is an integer in ${-3,-2,\dots,3}$. Your argument counts the number of such matrices with odd determinant. Now, show that there are $3$ such matrices with determinant $2$ and $3$ with determinant $-2$ (corresponding to one unordered basis). – Ben Grossmann Apr 28 '22 at 12:52
  • Specifically, the "extra" unordered basis is ${(1,0,1),(0,1,1),(1,1,0)}$. – Ben Grossmann Apr 28 '22 at 12:53
  • @BenGrossmann Why this problem is the same as the problem of the matrices you mentioned in the previous comment? – Idonotknow Apr 28 '22 at 17:01
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    A set of vectors (in $\Bbb F^n$ for some field $\Bbb F$) is a basis if and only if the matrix that has these vectors as columns is invertible, i.e. has non-zero determinant. – Ben Grossmann Apr 28 '22 at 18:04

1 Answers1

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In the last step, when you subtract out $2^2$, you're using that for most choices of the first two basis vectors, either $u + v$ or $u - v$ (but not both) is in $S$. This is not always true (eg if $u, v \in \{(1, 1, 0), (1, 0, 1), (0, 1, 1)\}$), and so sometimes there are $5$ choices for the third basis vector instead of $4$.

ronno
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