I'm working on a problem where I want to find the supremum over the orthogonal group $O_n(\mathbb{R})$ of the sum of the upper triangular elements of matrices in this group, specifically we want to compute
$$ m_n=\sup_{M \in O_n(\mathbb{R})} \sum_{1 \leq i \leq j \leq n} m_{ij} $$
I have shown that this is equivalent to finding the supremum of $\text{Tr}(MB)$ where $M$ ranges over orthogonal matrices and $B$ is the "1s" upper triangular matrix :
$$ B = \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 1 & \cdots & 1 \\ 0 & 0 & 1 & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix} $$
I then use the SVD decomposition of $B$, writing it as $B=UA^2V^T$. Then, the trace calculation goes as follows:
\begin{align*} \text{Tr}(MB) &= \text{Tr}(MUA^2 V^T) \\ &= \langle MUA, VA \rangle \\ &\leq ||A||_2^2 \quad (\text{by orthogonality and Cauchy-Schwarz inequality}) \\ &= \text{Tr}(A A^T) \\ &= \text{Tr}(S) \end{align*}
where $S$ is the diagonal matrix with the singular values of $B$. The bound is achieved by $VU^T$, leading to the conclusion that the answer to the problem is $\text{Tr}(S)$, then
$$ m_n = \sum_{\lambda \in \operatorname{Sp}(C)} \sqrt{\lambda} $$
With $C$ is the matrix such that $C_{i,j}=\min(i,j)$. (C is not exactly BB^T, but i changed order of rows/columns because it is easier to work with the matrix like this :) )
That's super cool because now i can simulate the value of $m_n$ ! Though, I have no closed analytic formula for $Sp(C)$... And my simulation seems to show that $m_n \sim\frac{n \log(n)}{\pi} $ which is super exciting.
Do you have an idea of a way to conclude, using my first ideas, to get the asymptotic equivalent ? Or maybe another way of getting it ?
Note that the singular values of $\sigma_i(B)=\sigma_i(B^{-1})^{-1}=\lambda_i^{\uparrow}\left((B^{-1})^TB^{-1}\right)^{-1/2}$ and $$ (B^{-1})^TB^{-1}=\pmatrix{1&-1\\ -1&2&\ddots\\ &\ddots&\ddots&\ddots\\ &&\ddots&\ddots&-1\\ &&&-1&2} $$ is a tridiagonal matrix that is almost Toeplitz. Its eigenvalues can hence be determined by solving a linear recurrence relation. They are known to be $4\sin^2\left(\frac{j-\frac12}{n+\frac12}\frac\pi2\right)$ for $j=1,\ldots,n$. It follows that $\sum_{j=1}^n\sigma_j(B)=\sum_{j=1}^n\frac12\sin\left(\frac{j-\frac12}{n+\frac12}\frac\pi2\right)^{-1}$.