Say $Q$ is a random variable which is sampling orthogonal matrices in $m$ dimensions using the Haar measure on $O(m)$. Let $A$ and $B$ be some (fixed) subset of rows and columns of $Q$ such that $\vert A \vert = \vert B \vert = k$.

Now is such an identity true? (If yes then could you kindly give the proof or a reference for it!)

$\mathbb{E}_{Q \sim O(m)}[det^2(Q_{A,B})] = \frac {1}{\binom {m}{k} }$

where by $Q_{A,B}$ we mean the submatrix of $Q$ corresponding to the rows from $A$ and columns from $B$.

If it seems necessary for the proof feel free to assume that $Q \sim SO(m)$

  • $\begingroup$ unexpected result, how did you find it? $\endgroup$ Dec 3, 2017 at 22:01
  • $\begingroup$ I was reading a paper which seemed to claim this. They have a lot of messy notation under which this is hidden. I couldnt derive this on my own and hence wanted to confirm! $\endgroup$ Dec 3, 2017 at 22:36

1 Answer 1


Let's denote $S=\{1,2,\dots m\}$ and $A\subset S$ with $|A|=k$. Then from orthogonality of $Q$ we have $$Q_{A,S}Q_{A,S}^{T}=I_k$$ therefore Cauchy Binet tells us that $$\sum_{B\in \binom{S}{k}}\det(Q_{A,B})^2=1.$$ If you apply expectation on both sides and notice that $E(\det(Q_{A,B})^2)$ doesn't depend on $B$ by invariance of the Haar measure you get your identity.

  • $\begingroup$ Thanks for the reply! I guess I am not understanding everything here! (1) Why is $Q_{A,S}$ is a orthogonal matrix for arbitrary $A$ and $B$? (2) In what sense is this second equation of yours a Cauchy-Binet formula? (proofwiki.org/wiki/Cauchy-Binet_Formula). Are you doing this implicitly as, $det(Q_{A,B})^2 = det(Q_{A,B)det((Q_{A,B})^T) = det(Q_{A,B)det(Q^T_{B,A})$ $\endgroup$ Dec 4, 2017 at 0:57
  • $\begingroup$ (3) How is Invariance of Haar measure implying that, $\mathbb{E}[det^2(Q_{A,B)]$ is independent of $B$? (at a fixed $A$) (Haar invariance is under shifting of the integrand argument by a fixed element of the group.) $\endgroup$ Dec 4, 2017 at 0:57
  • $\begingroup$ 1) $Q_{A,S}$ is a $k\times m$ matrix, it is a submatrix of $Q$. 2) Yes, that's exactly how Cauchy-Binet is applied here. 3) Use Haar invariance with respect to shifting by a permutation matrix (so the columns can be permuted however you wish). $\endgroup$ Dec 4, 2017 at 1:12
  • $\begingroup$ But if you choose a specific $Q_{A,B}$ (...which in your case is really, $(Q_{A,S})_{A,B}$...) now the permutation required by different $Q_{A,B_1}$ and $Q_{A,B_2}$ to get to $Q_{A,B}$ (i.e the permutation that moves the columns such that the indices of $B_i$ match the indices of $B$) is different for each $B_i$. So this is not a translation by a fixed permutation matrix. I dont see how this is Haar invariance. Can you maybe write some explicit equations? $\endgroup$ Dec 4, 2017 at 1:24
  • $\begingroup$ @gradstudent: Your goal is to show that $E(\det(Q_{A,B_1})^2)=E(\det(Q_{A,B_2})^2)$. You just need one permutation $\endgroup$ Dec 4, 2017 at 1:28

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