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 Weiss Jan 26 '10 at 4:22

1$\begingroup$ @Ben: The field is probably the reals. The OP specified the distribution: i.i.d. uniform on 0 and 1. $\endgroup$ – aorq Jan 26 '10 at 4:27

$\begingroup$ Thanks Charles, what's OP? $\endgroup$ – Ben Weiss Jan 26 '10 at 4:29

$\begingroup$ does googling for the words in the title not yield anything you can work with? $\endgroup$ – Yemon Choi Jan 26 '10 at 4:38

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 Yuan Jan 26 '10 at 5:26
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. $\endgroup$ – David E Speyer Jan 26 '10 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. $\endgroup$ – Gil Kalai Jan 26 '10 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 $u$. $\endgroup$ – Gerry Myerson Feb 19 '18 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 $v$. $\endgroup$ – David E Speyer Feb 19 '18 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.

1$\begingroup$ Update: the central limit theorem for the logdeterminant was worked out carefully by Nguyen and Vu, arxiv.org/abs/1112.0752 $\endgroup$ – Terry Tao Jul 2 '15 at 15:46
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*}
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 $o(1)$.
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$(X)=x$.


1$\begingroup$ That's somewhat of a Freudian slip (since I typically use permanents rather than determinants). Thanks, I fixed it. $\endgroup$ – Douglas S. Stones Jan 26 '10 at 4:54
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