What is the expected value of the determinant over the uniform distribution of all possible 10 NxN matrices? What does this expected value tend to as the matrix size N approaches infinity?

$\begingroup$ What ring or field do the entries come from? What distribution on the entries are you assuming? $\endgroup$– Ben WeissCommented Jan 26, 2010 at 4:22

2$\begingroup$ @Ben: The field is probably the reals. The OP specified the distribution: i.i.d. uniform on 0 and 1. $\endgroup$– aorqCommented Jan 26, 2010 at 4:27

$\begingroup$ does googling for the words in the title not yield anything you can work with? $\endgroup$– Yemon ChoiCommented Jan 26, 2010 at 4:38

2$\begingroup$ OP stands for Original Poster. $\endgroup$– Qiaochu YuanCommented Jan 26, 2010 at 4:48

1$\begingroup$ I assume that is what Charles means by "on 0 and 1." Anyway, as for the original question it might be more interesting to compute the variance. $\endgroup$– Qiaochu YuanCommented Jan 26, 2010 at 5:26
7 Answers
As everyone above has pointed out, the expected value is $0$.
I expect that the original poster might have wanted to know about how big the determinant is. A good way to approach this is to compute $\sqrt{E((\det A)^2)}$, so there will be no cancellation.
Now, $(\det A)^2$ is the sum over all pairs $v$ and $w$ of permutations in $S_n$ of $$(1)^{\ell(v) + \ell(w)} (1/2)^{2n\# \{ i : v(i) = w(i) \}}$$
Group together pairs $(v,w)$ according to $u := w^{1} v$. We want to compute $$(n!) \sum_{u \in S_n} (1)^{\ell(u)} (1/2)^{2n\# (\mbox{Fixed points of }u)}$$
This is $(n!)^2/2^{2n}$ times the coefficient of $x^n$ in $$e^{2xx^2/2+x^3/3  x^4/4 + \cdots} = e^x (1+x).$$
So $\sqrt{E((\det A)^2)}$ is $$\sqrt{(n!)^2/2^{2n} \left(1/n! + 1/(n1)! \right)} = \sqrt{(n+1)!}/ 2^n$$

$\begingroup$ Previous version had many sign errors, I think they are fixed now. $\endgroup$ Commented Jan 26, 2010 at 15:40

5$\begingroup$ There is another simple proof based on Cauchy Binet theorem. Turan published (in Chinese) a formula for the sum of the 4th power. (I think there are simpler proof but I dont know a reference). I am not aware of a formula for higher powers. $\endgroup$ Commented Jan 26, 2010 at 16:15


$\begingroup$ I don't know what $\ell(v)$ stands for. I don't know what $i$ stands for in the sum over $u$. $\endgroup$ Commented Feb 19, 2018 at 1:41

$\begingroup$ $\ell(v)$ is the length of the permutation $v$. In this case, it only appears in exponents: $(1)^{\ell(v)}$ is the sign of $v$. $\endgroup$ Commented Feb 19, 2018 at 2:58
If $N \ge 2$, then the expected value is $0$ since interchanging two rows preserves the distribution but negates the determinant.
It is a little more convenient to work with random (1,+1) matrices. A little bit of Gaussian elimination shows that the determinant of a random n x n (1,+1) matrix is $2^{n1}$ times the determinant of a random n1 x n1 (0,1) matrix. (Note, for instance, that Turan's calculation of the second moment ${\bf E} \det(A_n)^2$ is simpler for (1,+1) matrices than for (0,1) matrices, it's just n!. It is also clearer why the determinant is distributed symmetrically around the origin.)
The log $\log \det(A_n)$ of a (1,+1) matrix is known to asymptotically be $\log \sqrt{n!} + O( \sqrt{n \log n} )$ with probability $1o(1)$; see this paper of Vu and myself. A more precise result should be that the logarithm is asymptotically normally distributed with mean $\log \sqrt{(n1)!}$ and variance $2 \log n$. This result was claimed by Girko; the proof is unfortunately not quite complete, but the result is still likely to be true.

3$\begingroup$ Update: the central limit theorem for the logdeterminant was worked out carefully by Nguyen and Vu, arxiv.org/abs/1112.0752 $\endgroup$ Commented Jul 2, 2015 at 15:46

$\begingroup$ How exactly was the Gaussian elimination carried over? I mean how can one get from (1,+1) ran. matrix to another (0,1) ran. matrix $\endgroup$ Commented Jun 12, 2022 at 23:36

1$\begingroup$ By multiplying rows and columns by signs, one can normalise the first row and column to be all 1s. If one then subtracts the first row from all the others to make the first column (1,0,...,0), the bottom right n1 x n1 minor is now a random matrix of 0s and 2s, whose determinant is $(2)^{n1}$ times that of a random 01 matrix. The various sign factors one incurs in this process can then be removed by interchanging some rows (assuming $n>2$). $\endgroup$ Commented Jun 13, 2022 at 15:56
For some further results of this nature, see Exercise 5.64 of Enumerative Combinatorics, vol. 2. This exercise deals with the uniform distribution on (0,1)matrices or $(1,1)$matrices, but the arguments can be carried over to other distributions where the matrix entries are i.i.d. The proofs are similar to the argument in David Speyer's comment.
5.64. a. [2+] Let $\mathcal D_n$ be the set of all $n\times n$ matrices of $+1$'s and $1$'s. For $k\in\mathbb P$ let \begin{align*} f_k(n)&= 2^{n^2} \sum_{M\in\mathcal D_n} (\det M)^k \\ g_k(n)&= 2^{n^2} \sum_{M\in\mathcal D_n} (\operatorname{per} M)^k, \end{align*} where $\operatorname{per}$ denotes the permanent function defined by $$\operatorname{per}(m_{ij})= \sum_{n\in\mathfrak{S}_n} m_{1,\pi(1)} m_{2,\pi(2)} \dots m_{n,\pi(n)}.$$ Find $f_k(n)$ and $g_k(n)$ explicitly when $k$ is odd or $k=2$.
b. [3] Show that $f_4(n)=g_4(n)$, and show that $$\sum_{n\ge 0} f_4(n) \frac{x^n}{n!} = (1x)^{3} e^{2x}. \tag{5.120}$$ HINT. We have $$\sum_m (\det M)^4 = \sum_M \left(\sum_{\pi\in\mathfrak S_n} \pm m_{1,\pi(1)}\dots m_{n,\pi(n)}\right)^4.$$ Interchange the order of summation and use Exercise 5.63.
c. [2+] Show that $f_{2k}(n)<g_{2k}(n)$ if $k\ge 3$ and $n\ge 3$.
d. [3] Let $\mathcal D'_n$ be the set of all $n\times n$ 01 matrices. Let $f'_k(n)$ and $g'_k(n)$ be defined analogously to $f_k(n)$ and $g_k(n)$. Show that $f'_k(n)=2^{kn} f_k(n+1)$. Show also that \begin{align*} g'_1(n) &= 2^{n} n!\\ g'_2(n) &= 4^n n!^2 \left(1+\frac1{1!}+\frac1{2!}+\dots+\frac1{n!}\right) \end{align*}
Is it not zero whenever $n \geq 2$? Let $A$ be a $n \times n$ permutation matrix with determinant $1$ (which requires $n \geq 2$). Then the uniform distribution of a random $n \times n$ $(0,1)$matrix $X$ is the same as the distribution of $AX$. The determinant is multiplicative, hence Det$(AX)=$Det$(A)$Det$(X)=$Det$(X)$. Hence the probability of Det$(X)=x$ is the same as the probability of Det$(X)=x$.

$\begingroup$ The permanent isn't multiplicative. $\endgroup$ Commented Jan 26, 2010 at 4:41

1$\begingroup$ That's somewhat of a Freudian slip (since I typically use permanents rather than determinants). Thanks, I fixed it. $\endgroup$ Commented Jan 26, 2010 at 4:54
Unless I'm missing something, this also follows immediately from linearity and multiplicativity of expectation, treating each entry as independently $01$ with probability $1/2$. Every permutation yields the same expected value in the sum, $\pm (1/2)^n$ depending on sign, and the number of even and odd permutations is identical (for $n \ge 2$, as noted above).
It's probably worth mentioning that an old result of Komlos shows that despite this, the probability the determinant is actually 0 is $o(1)$.
Miodrag Zivkovic has actually done a classification on small orders of 01 matrices by rank and absolute determinant value. You may be interested in the tables in his Arxiv paper http://arxiv.org/abs/math.CO/0511636 .
Gerhard "Ask Me About System Design" Paseman, 2010.01.26