# Average size of determinants of integer matrices?

I am interested in estimating how large determinants of matrices tend to be 'on average' given the following model: suppose we form $n \times n$ matrices $M$ such that all of the entries of $M$ are integers, and the entries of the $i$th row of $M$ are bounded by some positive parameter $k_i$. Then by expressing the determinant as a polynomial in the entries, we see that $$\displaystyle |\det(M)| \leq n! (k_1 \cdots k_n).$$ However, this estimate seems to be too large for an estimate for the mean, as on average one would expect significant cancellation to make the determinant small. Thus to pose my question formally:

Consider the set of $n \times n$ matrices with integer entries such that the absolute value of each entry in the $i$th row is bounded by the parameter $k_i > 0$. Let $M(k_1, \cdots, k_n)$ denote this set of matrices. Then what is the average value of the absolute value of determinant of the elements in $M(k_1, \cdots, k_n)$? Let $\mu(k_1, \cdots, k_n)$ denote this average. Given $\epsilon > 0$, can one give an estimate to how many matrices $M$ in $M(k_1, \cdots, k_n)$ satisfy $$\displaystyle (1 - \epsilon)(\mu(k_1, \cdots, k_n)) \leq |\det(M)| \leq (1 + \epsilon)(\mu(k_1, \cdots, k_n))?$$ Thanks for any insight on the matter.

• Hadamard bound gives one estimate. Elsewhere on MathOverflow are related questions for matrices with all elements bounded by 1. Searching for 'expected determinant' or even 'Hadamard determinant' on MathOverflow is likely to be fruitful. Gerhard "Ask Me About Determinant Spectra" Paseman, 2013.04.12 – Gerhard Paseman Apr 12 '13 at 23:27
• Hadamard's inequality seems to give a slightly better better bound for the determinant, but I don't see how it can resolve the issue of how large these determinants tend to be on average. – Stanley Yao Xiao Apr 12 '13 at 23:34
• Also, I suspect the answer will be about the square root of the maximum value, which if memory serves corresponds to the literature for matrices with entries bounded in absolute value by one. Gerhard "Ask Me About System Design" Paseman, 2013.04.12 – Gerhard Paseman Apr 12 '13 at 23:35
• As is typical, it's much easier to estimate the average square of the determinant. If you expand out the formula the only nonzero contributions come from when one term in the sum for the determinant appears twice, which gives a formula of approximately $n! (k_1\dots k_n)^2/3^n$. – Will Sawin Apr 12 '13 at 23:36

As noted in Will's comment above, it's easy to compute the expected square of the determinant. More precisely, we have $$E(\det M^2)=n! \prod \frac{k_i (k_i+1)}{3}.$$
Let $M'$ be formed from $M$ by dividing each row by $\left(\frac{k_i(k_i+1)}{3}\right)^{1/2}$. Now each entry has mean $0$ and variance $1$, and furthermore the entries are bounded. The determinants of such matrices as the size of the matrix tends to infinity have been well studied, by Girko, Tao and Vu, and Nguyen and Vu, among others.
For example, it follows from Theorem 1.1 in the Nguyen and Vu paper linked above that $\log |det(M')|$ is asymptotically normal with mean $\frac{1}{2} \log((n-1)!)$ and variance $\log n$. Taking this and rescaling back to $M$, we have that with probability tending to $1$ as $n$ tends to infinity that $$det M^2 = \ n^{-1+o(1)} E(det M^2).$$
A probably very naive idea is to approximate the average by the absolute value of the determinant by a typical random element in your set. The Euclidean length of the $i-$th row in an average element is probably computable (and given perhaps by something like $k_i\sqrt{n}/2$, more precisely, it should be close to $k_i\int_0^1\cdots\int_0^1\sqrt{t_1^2+t_2^2+\dots+t_n^2}dt_1dt_2\cdots dt_n$ for large $k_i$). Making the (also very naive) assumption that the $n$ rows of a typical matrix, rescaled to be of unit length (with respect to the Euclidean norm) are $n$ independent equidistributed random variables of the unit sphere, one can probably compute the determinant by geometric arguments (I guess someone has thought of this and done the computations since it is a rather natural question).