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Let $n\in\mathbb{N}^*$. I wonder if there is any way to count the number of matrices (of size $n\times n$) only filled with 0 and 1 that is invertible?

(a random example of such matrices: $\begin{bmatrix} 1 & 1 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 \end{bmatrix}$ (I don't know if it is invertible)).

MiKiDe
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1 Answers1

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Any square matrix is invertible if all column vectors are linearly independent. For an $n\times n$ matrix filled with $0$'s and $1$'s, consider how many choices there are for the first column: all it has to be is not all $0$, so there are $2^n-1$ choices. For the second column, all it has to be is linearly independent to the first column, so there are

$\displaystyle{2^n-\underbrace{1}_{\text{the zero vector}}-\underbrace{1}_{\text{whatever the first column vector was}}=2^n-2}$

choices (assuming we're working in a field of characteristic 2). Carrying on, we get

$\displaystyle{(2^n-1)(2^n-2)\cdots(2^n-2^{n-1})}$

ways to build this matrix. Just a quick check, this formula gives 6 for $2\times 2$ matrices, which is correct.

The answer only works in fields of characteristic 2.

Teddy38
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    Isn't this the result for fields of characteristic $2$? The characteristic $0$ problem is much harder. – Will Orrick Sep 03 '17 at 11:36
  • @Will Orrick You're absolutely right. I've edited it. – Teddy38 Sep 03 '17 at 11:43
  • Thank you for your answer! So is that working on $\mathbb{R}$ or $\mathbb{Z}$? (I don't know what is a field of characteristic 2, but if it's working on those ones, it's OK) – MiKiDe Sep 03 '17 at 12:43
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    The result does not apply to $\mathbb{R}$ or $\mathbb{Z}$. The integers modulo $2$ are an example of a field of characteristic $2$. In general, in a field of characteristic $2$ the relation $1+1=0$ holds. Fields in which $1+1+\ldots+1\ne0$, no matter how many terms there are in the sum are said to be of "characteristic $0$". So $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ are characteristic $0$. Do have a look at the linked posts in my comment to your question. This is a hard problem, and no formula is known, although some asymptotic results are known. – Will Orrick Sep 03 '17 at 13:20
  • OK, I understand the difficulty. Thank you very much. – MiKiDe Sep 03 '17 at 17:09