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Define the set $S_n=\{A_n| A_n \hbox{is invertible 0-1 matrix}\}$. What is the size of $S_n$? When $n=2$, it is easy to see $\sharp S_2=6$. I guess $\sharp S_n=\prod_{k=1}^n(2^n-2^{k-1})$.

Martin Sleziak
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Sunni
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    It would not hurt if you were explicit about whether you want invertibility of the matrix as real matrices or over some other field (possibly the field with two elements) The answer will depend in this. – Mariano Suárez-Álvarez Apr 03 '12 at 02:32
  • the matrices considered are with entries from real field. – Sunni Apr 03 '12 at 02:34
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    Add that information on the body of the question. – Mariano Suárez-Álvarez Apr 03 '12 at 02:35
  • Dear Sunni, your answe to my question in the comment above is contradicted by your acceptance of Daniel's answer below... – Mariano Suárez-Álvarez Apr 03 '12 at 03:29
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    Sunni, please note that if matrices are over $\mathbb{Z}2$ then $S_n$ is essentially the general linear group: $\mathsf{GL}(n, \mathbb{Z}_2),$ which has size $\displaystyle\prod{i=0}^{n-1} (2^n- 2^i).$ But if matrices are over $\mathbb{R},$ then you're looking at $S_n \subset \mathsf{GL}(n, \mathbb{R}).$ The accepted answer by Daniel M. addresses the former case not the latter case. –  Apr 03 '12 at 03:57
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    See http://mathoverflow.net/questions/18636/number-of-invertible-0-1-real-matrices – morgan Apr 03 '12 at 08:17
  • Also: http://math.stackexchange.com/questions/54246/probability-that-a-random-binary-matrix-is-invertible – morgan Apr 03 '12 at 08:25
  • Sorry all... so the conclusion is that my guess in the question is wrong and there is no known formula. – Sunni Apr 03 '12 at 19:21

1 Answers1

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Your guess is almost right ($k$ should start at $0$). This is a particular case of the order of the general linear group over a field $F$.

In your case here is how the proof goes. You first want a non-zero column for your matrix. You have $$2^n-1$$ options to pick such first column. Then, you want to pick a column that is linearly independent from your first column, so you have a total of $2^n$ possible options for this entry minus the multiples of the first column, in this case $2$. For the third column you again have to substract the linear combinations of the first two columns in this case is $2^2$. Doing so you obtain: $$(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})$$Hope this helps.

Daniel Montealegre
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