eigenvalues of certain block matrices

This question inquired about the determinant of this matrix: $$ \begin{bmatrix} -\lambda &1 &0 &1 &0 &1 \\ 1& -\lambda &1 &0 &1 &0 \\ 0& 1& -\lambda &1 &0 &1 \\ 1& 0& 1& -\lambda &1 &0 \\ 0& 1& 0& 1& -\lambda &1 \\ 1& 0& 1& 0&1 & -\lambda \end{bmatrix} $$ and of other matrices in a sequence to which it belongs. In a comment I mentioned that if we permute the indices 1, 2, 3, 4, 5, 6 to put the odd ones first and then the even ones, thus 1, 3, 5, 2, 4, 6, then we get this: $$ \begin{bmatrix} -\lambda & 0 & 0 & 1 & 1 & 1 \\ 0 & -\lambda & 0 & 1 & 1 & 1 \\ 0 & 0 & -\lambda & 1 & 1 & 1 \\ 1 & 1 & 1 & -\lambda & 0 & 0 \\ 1 & 1 & 1 & 0 & -\lambda & 0 \\ 1 & 1 & 1 & 0 & 0 & -\lambda \end{bmatrix} $$ So this is of the form $$ \begin{bmatrix} A & B \\ B & A \end{bmatrix} $$ where $A$ and $B$ are symmetric matrices whose characteristic polynomials and eigenvalues are easily found, even if we consider not this one case of $6\times 6$ matrices, but arbitrarily large matrices following the same pattern.

Are there simple formulas for determinants, characteristic polynomials, and eigenvalues for matrices of this latter kind?

I thought of the Haynesworth inertia additivity formula because I only vaguely remembered what it said. But apparently it only counts positive, negative, and zero eigenvalues.

Solution 1:

We have $$ \det \left( \begin{array}{cc} A & B\\ C & D \end{array} \right) = \det(A-BD^{-1}C) \det(D), $$ where the matrix $A-BD^{-1}C$ is called a Schur complement. In your case, $A=D=-\lambda I_n$ and $B=C=J_n$ = the order $n$ matrix with all entries equal to 1. So, the RHS is equal to $\det(-\lambda I_n + \frac{n}{\lambda} J_n) \det(-\lambda I_n) = (-n)^n \det(-\frac{\lambda^2}{n}I_n + J_n)$. If I remember correctly, $\det(xI_n + J_n) \equiv x^{n-1}(x+n)$, but you should check whether this is true or not.

Solution 2:

I am not sure whether I understand what you want to ask.. but the following are some facts on the matrix of this type

$\det\begin{bmatrix} A & B \\\\ B & A \end{bmatrix}=\det(A+B)\det(A-B)$. The eigenvalues of $\begin{bmatrix} A & B \\\\ B & A \end{bmatrix}$ are the union of eigenvalues of $A+B$ and the eigenvalues of $A-B$.