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If $A$ is chosen uniformly at random over all possible $m \times n$ (0,1)-matrices, what is the expected size of the absolute value of the determinant of $AA^T$. We can assume $m < n$ and all arithmetic is over the reals.

In Expected determinant of a random NxN matrix the expected size of the determinant for random (0,1) matrices was given as $\sqrt{(n+1)!}/ 2^n$ by David Speyer.


As pointed out by Richard Stanley, my question is incorrect in its reference to the related problem solved by David Speyer (and others). I should have said that when $A$ is a random $n$ by $n$ (0,1)-matrix then $\sqrt{E((\det A)^2)} = \sqrt{(n+1)!}/ 2^n$

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By the Cauchy-Binet theorem, $\det AA^T=\sum (\det B)^2$, where $B$ ranges over all $m\times m$ submatrices of $A$. The expected value of $(\det B)^2$ is $(m+1)!/4^m$, so the expected value of $\det AA^T$ is ${n\choose m}(m+1)!/4^m$. This is also the expected value of $|\det AA^T|$ since $\det AA^T\geq 0$.

Incidentally, it is misleading to say that the expected size of the determinant of a random $n\times n$ $(0,1)$-matrix $A$ is $\sqrt{(n+1)!}/2^n$. It is true that $(n+1)!/4^n$ is the expected value of $(\det A)^2$, but this doesn't mean that $\sqrt{(n+1)!}/2^n$ is the expected value of $|\det A|$.

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  • $\begingroup$ Thank you very much and thank you for the correction! I am sorry my question was a little sloppy. $\endgroup$ – Lembik Jul 2 '15 at 13:49

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