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I want to calculate how many bases of $\mathbb{C}^3$, as a complex vector space, there are in the subset of vectors whose coordinates are $0$ or $1$. I don't know how to approach this problem systematically, I tried some brute force method but obviously it gets too complex for large $n$, $\mathbb{C}^n$. Any hints?

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  • $\begingroup$ You are considering $\mathbb{C}^3$ as a vector space over $\mathbb{C}$, or over $\mathbb{R}$ ( or some other filed)? $\endgroup$ Jan 6, 2017 at 20:37
  • $\begingroup$ Sorry, I edited the post. $\endgroup$ Jan 6, 2017 at 20:39
  • $\begingroup$ "It gets too complex for large $n$": pun intended? $\endgroup$
    – user384138
    Jan 6, 2017 at 20:55
  • $\begingroup$ According to this MO post, the answer remains unknown for a general $n$. $\endgroup$
    – user1551
    Jan 6, 2017 at 21:03
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    $\begingroup$ Using brute force search, we count 174 nonsingular $\{0,1\}$-matrices of size $3$. Therefore there are 174/6 = 29 unordered bases consisting of $\{0,1\}$-vectors in $\mathbb C^3$. $\endgroup$
    – user1551
    Jan 6, 2017 at 21:19

1 Answer 1

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We can pick the first row in $7$ ways and the second row in $6$ ways so that the first two rows are linearly independant.

Now we just have to seperate into a couple of cases to check how many $1-0$ matrices are contained in the span of the first two rows.

The only way in which a fourth $1-0$ matrix is spanned is if the two rows do not share a $1$ or if one of the rows contains the ones of the other row.

So how many cases is this? First lets analyze the first case: In total there are $3^3$ ways this can happen, but there are $1+7+7$ cases in which the two rows are not linearly independent. So there are only $12$ cases in which the first two rows generate a fourth $1-0$ column.

For the second case first lets look at the cases in which the first row contains the second. In total there are $3^3$ cases, but not all of these satisfy that the two rows are linearly independent. We must subtract $1+7+7$ cases. So there are only $12$ cases to consider. We multiply this by $2$ to consider the case in which the second row contains the first one.

Hence the answer is $36\times 4+ (6\times 7 -36)\times 5=174$ ways.

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