Find the analytical form or the order of the spectral radius (or eigenvalues) of a special sparse matrix.

Let $$\hat{\bf H}$$ be a $$p\hat{N}\times p \hat{N}$$ sparse matrix consisting of $$p\times p$$ blocks, where each block is of size $$\hat{N}\times\hat{N}$$. The values in $$\hat{\bf H}$$ is illustrated below (empty places are zero): Asking for help to find the analytic form of the spectral radius (or eigenvalues) of this matrix. If analytic form is hard to calculate, would it be possible to at least determine the order of the spectral radius approaching 1? I did a simple numerical experiment as the following,

If we fix $$p = 5$$ and let $$\hat{N}$$ go from 5 to 100, we have If we fix $$\hat{N} = 10$$ and let $$p$$ go from 5 to 100, we have • Are non-entered values zero? – coffeemath Dec 8 '18 at 3:35
• @coffeemath Yes. Thanks! – Tony Dec 8 '18 at 3:37
• Just to be clear, there are only $2p-2$ non-zero columns, right? – JimmyK4542 Dec 18 '18 at 7:40
• So it is fairly easy to determine that the singular values of $\widehat{H}$ are $\sqrt{\dfrac{N}{2}}$ with multiplicity $p-2$, $\sqrt{\dfrac{N(2N+1)}{6(N+1)}}$ with multiplicity $2$, $\sqrt{\dfrac{(N-1)N}{6(N+1)}}$ with multiplicity $p-2$, and $0$ with multiplicity $N-2p+2$. Computing the eigenvalues seems a lot harder. Although since all the entries of $\widehat{H}$ are non-negative and the maximum row sum is $1$, we have $\|\widehat{H}\|_1 = 1$, and thus, $\rho(\widehat{H}) \le \|\widehat{H}\|_1 =1$. – JimmyK4542 Dec 18 '18 at 8:30
• Some (probably useless) simplifications. Let us write $n$ for $\widehat{N}$. After some tedious transformations, $\widehat{H}$ can be symmetried to $S\oplus0_{(n-2)p\times(n-2)p}$, where $T$ is the block-tridiagonal matrix of the form $$S=\frac1{n+1}\pmatrix{0&A\\ A^T&0&\ddots\\ &\ddots&\ddots&A\\ &&A^T&0} \text{ where } A=\frac12\pmatrix{n+1&\sqrt{n^2-1}\\ -\sqrt{n^2-1}&-(n-1)}.$$ ($S$ has $p$ block rows; hence its size is $2p\times2p$, which, unlike the size of $\widehat{H}$, does not depend on $n$.) Thus $\widehat{H}$ always has a real spectrum. – user1551 Dec 21 '18 at 10:46