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In the very nice paper by GW Stewart:

Stewart, G. W., The efficient generation of random orthogonal matrices with an application to condition estimators. (With mircofiche section), SIAM J. Numer. Anal. 17, 403-409 (1980). ZBL0443.65027.

The author gives the following algorithm for generating a Haar-uniform orthogonal matrix: Let $M$ be an $n\times n$ matrix with i.i.d gaussian entries. Let $M= QR$ be the $QR$ decomposition of $M$ ($Q$ is orthogonal, $R$ is upper-triangular, with positive diagonal entries). Then $Q$ is Haar-uniform.

My question is: What is the distribution of $R?$

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    $\begingroup$ I don't know the answer, but you may be interested in this paper by Francesco Mezzadri: ams.org/notices/200705/fea-mezzadri-web.pdf $\endgroup$
    – Marcel
    Aug 11, 2019 at 23:26
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    $\begingroup$ If your replace QR by Schur decomposition, and orthogonal by unitary with M iid complex Gaussian, the answer is super nice: it is $C_N \prod_{i<j} |R_{ii}-R_{jj}|^2\cdot e^{-\sum |R_{ij}|^2}$. This is known from forever and appears e.g. in section 3 of projecteuclid.org/euclid.ecp/1539309734. (Note the ambiguity with ordering etc). I believe, but have not checked, the same works in the orthonormal case. I have not worked out the QR case but suspect it should be similarly computable. $\endgroup$ Aug 12, 2019 at 1:05
  • $\begingroup$ @oferzeitouni Thanks! That sounds very plausible, I will try to see if I can make head or tail of the computation... $\endgroup$
    – Igor Rivin
    Aug 12, 2019 at 3:42

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