Eigenvalue of block matrix of order $2n$ How to find eigenvalues of following block matrix?
$$P=\begin{bmatrix} A & B \\ B & A \end{bmatrix}$$
Where,
$A=\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \ddots & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix}_n$
$B=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix}_n$
I had read one result for block matrix which says that eigenvalues of matrix $P$ is union of eigenvalues of $A+B$ and $A-B$
Here,
$A+B=\begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \ddots & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix}_n$
$A-B=\begin{bmatrix} -1 & 1 & 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \ddots & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix}_n$
How to find eigenvalues of $A+B$ and $A-B$?
I know that eigenvalues of $A$ are $2\cos\frac{2\pi j}{n},j=1,2,\cdots,n$ 

Note that the eigenvalues of $P$ are real and in $[-3,3]$.
Let $U_n$ be the matrix that is derived from $A_n$ by putting the entries $[1,n],[n,1]$ equal to $0$. 
Then $\det(A_n\pm B_n-\lambda I_n)=\det(A_n-\lambda I_n)\pm \det(U_{n-1}-\lambda I_{n-1})=p_n(\lambda)\pm q_{n-1}(\lambda)$.
The roots of $p_n$ are $2\cos(\frac{2\pi j}{n}),j=1,\cdots,n$.
The roots of $q_{n-1}$ are $2\cos(\frac{\pi j}{n}),j=1,\cdots,n-1$.
Remark 1. When $n$ is even, $p_n$ is even and $q_{n-1}$ is odd. Thus $\det(P-\lambda I_{2n})=p_n^2(\lambda)-q_{n-1}^2(\lambda)$ is even.
Remark 2. The roots of $p_n,q_{n-1}$, for $1\leq j\leq n-1$ and even, are the same. Then we explicitly know half of the eigenvalues of $A\pm B$. Finally, we know half of the eigenvalues of $P$ (they have multiplicity $2$).
