# Determinants of binary matrices

I was surprised to find that most $$3\times 3$$ matrices with entries in $$\{0,1\}$$ have determinant $$0$$ or $$\pm 1$$. There are only six out of 512 matrices with a different determinant (three with $$2$$ and three with $$-2$$) and these are $$\begin{bmatrix}1 & 1 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1\end{bmatrix}$$ and the different permutation of this.

Hadamard's maximum determinant problem for $$\{0,1\}$$ asks about the largest possible determinant for matrices with entries in $$\{0,1\}$$ and it is known that the sequence of the maximal determinant for $$n\times n$$ matrices for $$n=1,2,\dots$$ starts with 1, 1, 2, 3, 5, 9, 32, 56, 144, 320,1458 (https://oeis.org/A003432). It is even known that the number of different matrices realizing the maximum (not by absolute value) is 1, 3, 3, 60, 3600, 529200, 75600, 195955200, (https://oeis.org/A051752).

My question is: what is the distribution of determinants of all $$n\times n$$ $$\{0,1\}$$-matrices?

Here is the data for (very) small $$n$$: $$\begin{array}{lccccccc} n & -3 & -2 & -1 & 0 & 1 & 2 & 3\\\hline 1 & & & & 1 & 1 & & \\ 2 & & & 3 &10 & 3 & & \\ 3 & & 3 & 84 & 338 & 84 & 3 & \\ 4 & 60 & 1200 & 10020 & 42976 & 10020 & 1200 & 60 \end{array}$$

• Will Orrick probably has the latest data. Miodrag Zivkovic did an analysis in 2005 for n up to 9. While lower determinant values are heavily weighted, it is not known how quickly the counts for off as the determinant value grows. You can find related questions here on MathOverflow. Gerhard "Search For Determinant Spectrum Problem" Paseman, 2019.01.15. – Gerhard Paseman Jan 15 at 20:33
• – Timothy Chow Jan 16 at 3:07
• @WillOrrick Of course - fixed! – Dirk Jan 16 at 13:51
• I think it would be good to clarify the statement of the question, since there are different possible interpretations. The way I interpreted it was, roughly speaking, to understand the limiting shape of the histograms that are implicit in the data shown for $n=1,2,3,4.$ With this interpretation, understanding the precise asymptotics of the number of $n \times n$ matrices with a given determinant is probably irrelevant (unless that value occurs a positive percentage of the time). However, the data suggest that the most common determinant value is $0$, which occurs $o(2^{n^2})$ times (right?) – Matt Young Jan 16 at 14:40
• The limiting shape of the histogram is known: in 2011 Hoi H. Nguyen and Van Vu proved that the log of the absolute value of the determinant approaches a normal distribution with mean $\frac{1}{2}\log (n-1)!$ and variance $\frac{1}{2}\log n$. (Note how large the mean is and how sharp the peak is--data for sizes less than $10$ are misleading.) This result applies to random $\{-1,1\}$ matrices and to many other classes of random matrices. Using the mapping mentioned in my comment to Timothy Chow's answer, you can adapt this result to $\{0,1\}$ matrices. – Will Orrick Jan 16 at 17:35

• Thanks for the pointers! That's somehow what I expected. One thing: As you say, there are some results for $\pm1$-matrices. I fail to see how these results help for $\{0,1\}$-matrices. Or do you mean to say that $\{0,1\}$ matrices are harder than $\pm1$-matrices (and also I here I don't see why this should be so). – Dirk Jan 16 at 9:24
• See the Wikipedia article on Hadamard's maximal determinant problem for the mapping between $\{0,1\}$ matrices and $\{-1,1\}$ matrices. For every $n\times n$ $\{0,1\}$ matrix with determinant $D$, there are $2^{2n+1}$ $\{-1,1\}$ matrices with determinant $2^nD$. The case $n=1$ is slightly different. – Will Orrick Jan 16 at 13:30