# Expected determinant of a random NxN matrix.

What is the expected value of the determinant over the uniform distribution of all possible 1-0 NxN matrices? What does this expected value tend to as the matrix size N approaches infinity?

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What ring or field do the entries come from? What distribution on the entries are you assuming? – Ben Weiss Jan 26 '10 at 4:22
@Ben: The field is probably the reals. The OP specified the distribution: i.i.d. uniform on 0 and 1. – aorq Jan 26 '10 at 4:27
Thanks Charles, what's OP? – Ben Weiss Jan 26 '10 at 4:29
does googling for the words in the title not yield anything you can work with? – Yemon Choi Jan 26 '10 at 4:38
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. – 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 $$(n!) \sum_{u \in S_n} (-1)^{\ell(u)} (1/2)^{2n-\# (\mbox{Fixed points of }i)}$$

This is $(n!)^2/2^{2n}$ times the coefficient of $x^n$ in $$e^{2x-x^2/2+x^3/3 - x^4/4 + \cdots} = e^x (1+x).$$

So $\sqrt{E((\det A)^2)}$ is $$\sqrt{(n!)^2/2^{2n} \left(1/n! + 1/(n-1)! \right)} = \sqrt{(n+1)!}/ 2^n$$

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Previous version had many sign errors, I think they are fixed now. – David Speyer Jan 26 '10 at 15:40
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. – Gil Kalai Jan 26 '10 at 16:15

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 $2^{n-1}$ times the determinant of a random n-1 x n-1 (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 $1-o(1)$; see this paper of Vu and myself. A more precise result should be that the logarithm is asymptotically normally distributed with mean $\log \sqrt{(n-1)!}$ 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.

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Update: the central limit theorem for the log-determinant was worked out carefully by Nguyen and Vu, arxiv.org/abs/1112.0752 – Terry Tao Jul 2 '15 at 15:46

Miodrag Zivkovic has actually done a classification on small orders of 0-1 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 .

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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.

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Unless I'm missing something, this also follows immediately from linearity and multiplicativity of expectation, treating each entry as independently $0-1$ 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 $o(1)$.

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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$(X)=-x$.

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The permanent isn't multiplicative. – Qiaochu Yuan Jan 26 '10 at 4:41
That's somewhat of a Freudian slip (since I typically use permanents rather than determinants). Thanks, I fixed it. – Douglas S. Stones Jan 26 '10 at 4:54

If $N \ge 2$, then the expected value is $0$ since interchanging two rows preserves the distribution but negates the determinant.

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