st.statistics - Statistics of a distribution on unitary matrices - MathOverflow
Statistics of a distribution on unitary matrices
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I recently wrote some code to generate random matrices $A\in U(n)$ with the property that the entries in $A$, the entries in the normalised eigenvectors of $A$, and the eigenvalues, all lie in $\mathbb{Q}(i)$. (This is convenient because it allows me to test certain ideas by exact calculation.) I will describe the process below, but the details are quite ad hoc and are not important for my main question. Although my process is certainly sufficiently random for the purposes I had in mind, I was nonetheless wondering how close it is to a uniform distribution with respect to Haar measure. What is the best way to formulate that question, and to test it by calculating statistics from random samples?
For the record, my process is as follows. I fix an integer $N>0$, then generate random rational numbers $q=a/b$ with $a$ uniformly random in $[-N^2,N^2]$ and $b$ uniformly random in $[1,N]$. This gives random points $z=(1+iq)/(1-iq)$ on $S^1$. These are probably not so well distributed, but I generate three of them, take their product, and multiply by $i^k$ with $k$ uniformly random in $\{0,1,2,3\}$. This gives points on $S^1$ which I guess are quite uniformly distributed. I then take $n$ of these and use them as the diagonal entries of a diagonal matrix $D$. I also make a random Hermitian matrix $H$ where the real and imaginary parts of the entries are random rationals as discussed above, and I put $U=(1+iH)(1-iH)^{-1}$ and $A=UDU^{-1}$. (I would also be interested in suggestions for more elegant or efficient algorithms.)
st.statistics<br>random-matrices
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edited 41 mins ago
asked 1 hour ago
Neil Strickland
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$\begingroup$<br>Why is $A$ unitary? I get $A A^* = (1+i H) D (1-iH)^{-1} (1+iH)^{-1} D^{-1} (1-iH)$ which is conjugate to $[(1-iH) (1+iH), D]$ which is usually not the identity.<br>$\endgroup$
Will Sawin
Will Sawin
2026-06-16 13:09:03 +00:00
Commented<br>51 mins ago
$\begingroup$<br>@WillSawin sorry, that was not what I meant, I have corrected it<br>$\endgroup$
Neil Strickland
Neil Strickland
2026-06-16 13:18:56 +00:00
Commented<br>41 mins ago
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I would suggest starting with quantities whose Haar expectations are known explicitly. A particularly natural family is given by the trace moments. For Haar measure on $U(n)$,<br>$$\mathbb{E}[\operatorname{Tr}(U^k)] = 0 \qquad (k\neq 0),$$<br>and<br>$$\mathbb{E}\left[|\operatorname{Tr}(U^k)|^2\right]=\min(k,n).$$<br>These are easy to compute numerically and already give a fairly sensitive test.<br>Another useful class of statistics comes from matrix entries. For Haar-distributed $U$,<br>$$\mathbb{E}(|U_{ij}|^2)=\frac1n, \qquad \mathbb{E}(|U_{ij}|^4)=\frac{2}{n(n+1)}.$$<br>More generally, one can compare higher mixed moments using Weingarten calculus. Also, maybe running test on the eigenangle spacing statistics will be useful.
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edited 3 mins ago
answered 9 mins ago
Shahrooz
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