# submatrix of orthogonal matrix

Suppose that $O$ is a uniformly random orthogonal matrix (w.r.t. Haar measure) and $X$ be its top-left $k\times k$ block.

There have been some literature studying the distribution of eigenvalues of $X$. Actually, there are more complete results for submatrix of random unitary matrix, which says that the eigenvalues are distributed like the eigenvalues of Gaussian random matrix (properly normalized). Presumably similar results hold for submatrix of random orthogonal matrix.

But I have not seen much result regarding the singular values of submatrix of orthogonal matrix. Jiang has a paper that studies the singular valued distribution of $X$ when $k=o(\sqrt{n})$, basically based on the fact (his earlier paper) that when $k=o(\sqrt{n})$, $X$ is statistically close to a Gaussian random matrix.

My question is, what happens when $k>\sqrt{n}$? Put the distribution of singular values of $X$ aside, do we know anything about the operator norm of $\|X\|$? Is it concentrated? Are there such results in the literature?

The joint probability distribution of the singular values $\sigma_i$ of $X$ was calculated in Physical Review B 82, 014536 (2010), (arXiv:1004.2438). Denoting $T_i=1-\sigma_i^2$, the set of $N_{\rm min}=\min(k,n-k)$ nonzero $T_i$'s has probability distribution $$P(\{T_i\})\propto\prod_i T_i^{|n-2k|/2}T_i^{-1/2}(1-T_i)^{-1/2}\prod_{j<k}|T_k-T_j|.$$ [Physics context: the $T_n$'s are the transmission probabilities of a chaotic scatterer in the Circular Real Ensemble (CRE), which applies to a superconducting quantum dot without time-reversal or spin-rotation symmetry. The sum $\sum_i T_i$ is the dimensionless thermal conductance $g$ of this system.]
This is the exact joint probability distribution for any $n,k$. Asymptotically for $N_{\rm min}\gg 1$, the density $\rho(T)=\langle \sum_i\delta(T-T_n)\rangle$ approaches the limiting form $$\rho(T)=\frac{n}{2\pi}\left(\frac{T-T_c}{1-T}\right)^{1/2}\frac{1}{T},\;\;T_c<T<1,$$ with $T_c=(1-2k/n)^2$.
Concerning the operator norm of $X$, the distribution of $g=\sum_{i}T_i=k-\sum_{i}\sigma_i^2$ has moments which can be computed using a Selberg integral formula. The first two cumulants are $$\langle g\rangle=\frac{k(n-k)}{n},\;\;{\rm var}\,g=\frac{2k^2(n-k)^2}{(n-1)n^2(n+2)}$$ Higher cumulants vanish asymptotically for $N_{\rm min}\rightarrow\infty$, so the distribution of $g$ tends to a Gaussian in that limit.
• Thanks! Since $X$ is $k$-by-$k$, shouldn't $g = \sum_i T_i$ be $k-\sum_i \sigma_i^2$ instead of $n-\sigma_i^2$? This means that $g = k - \|X\|_F^2$? Yes, the Frobenius norm of $X$ will be statistically close between the case of orthogonal matrix and the Gaussian matrix. – user58955 Mar 14 '17 at 12:55
• So it also looks possible to compute $\sum \sigma_i^4$ or higher orders, right? Expand the sum in terms of $\sum T_i^2$, $\sum T_i^3$, ... and use Selberg integral formula? – user58955 Mar 14 '17 at 13:31
• Oh I see, in the density in your answer above, shouldn't it be $(n-2k)/2$ instead of $(n-k)/2$ for the exponent of $T_i$? It seems that it breaks the $n\times n$ matrix into $(k,n-k)$ and looks at the difference between $k$ and $n-k$. – user58955 Mar 14 '17 at 14:45
• And then we can use Amoto's integral in en.wikipedia.org/wiki/…, I suppose, applied to $k:=1$ and $n:=n-k$? That seems to give the expectation. Then what do we do for $\int t_i^2 P(t_i)$ if we want to look at the variance? – user58955 Mar 14 '17 at 14:48