Show that all eigenvalues of this pentadiagonal matrix are double degenerate I am trying to show in general that the following pentadiagonal matrix $\mathbf{M}$ has double degenerate eigenvalues,
\begin{equation}
  \mathbf{M} =
  \left[ \begin{array}{cccccccc}
   4 & 0 & a & 0 & 0 & 0 & 0 & 0 \\
   0 & 3 & 0 & a & 0 & 0 & 0 & 0\\
   a & 0 & 2 & 0 & a & 0 & 0 & 0\\
   0 & a & 0 & 1 & 0 & a & 0 & 0\\
   0 & 0 & a & 0 & 1 & 0 & a & 0\\
   0 & 0 & 0 & a & 0 & 2 & 0 & a\\
   0 & 0 & 0 & 0 & a & 0 & 3 & 0\\
   0 & 0 & 0 & 0 & 0 & a & 0 & 4
  \end{array} \right]\!,
\end{equation}
where $a$ is just some real number. It doesn't particularly matter what the matrix elements are on the diagonal, so long as they are symmetric about the anti-diagonal the eigenvalues will still be double degenerate. Numerically, I find that a pentadiagonal matrix of this form and of arbitrary size always yields double degenerate eigenvalues. 
In block form we can express $\mathbf{M}$ as
\begin{equation}
  \mathbf{M} =
  \left[ \begin{array}{cc}
   \mathbf{A} & \mathbf{B} \\
   \mathbf{J}^{-1}\mathbf{B}\mathbf{J} & \mathbf{J}^{-1}\mathbf{A}\mathbf{J} \\
  \end{array} \right]\!,
\end{equation}
where 
\begin{equation}
\mathbf{A}=\left[\begin{array}{cccc}
4 & 0 & a & 0 \\
0 & 3 & 0 & a \\
a & 0 & 2 & 0 \\
0 & a & 0 & 1  
\end{array} \right],
%
\; \; \mathbf{B}=\left[ \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
a & 0 & 0 & 0 \\
0 & a & 0 & 0  
\end{array} \right]
\end{equation}
and $\mathbf{J}$ is the exchange matrix, hence the top-left and bottom-right blocks are permutation-similar (as are the top-right and bottom-left blocks). However, this has not helped me show that all eigenvalues are double degenerate. 
So far I have that the truncated 4x4 matrix
\begin{equation}
\mathbf{M}_{\text{trunc}}=\left[\begin{array}{cccc}
2 & 0 & a & 0 \\
0 & 1 & 0 & a \\
a & 0 & 1 & 0 \\
0 & a & 0 & 2  
\end{array} \right]
\end{equation}
has a characteristic polynomial
\begin{equation}
p(\lambda)=(a^2+(1-\lambda)\lambda + 2(\lambda-1))^2
\end{equation}
which has roots which are evidently double degenerate. 
Any help at all would be much appreciated!
 A: Shortly speaking, such a matrix can be essentially split into two submatrices formed by odd and even rows and columns, and those two matrices are obtained from one another by conjugating by $\mathbf{J}$. To be more detailed: consider the matrix
\begin{equation}
  \mathbf{S} =
  \left[ \begin{array}{cccccccc}
   1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
   0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
   0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
   0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\
   0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
   0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\
   0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
   0 & 0 & 0 & 0 & 1 & 0 & 0 & 0
  \end{array} \right]\!.
\end{equation}
Then $\mathbf{S^{-1}MS}$ is of the form
\begin{equation}
  \left[ \begin{array}{cc}
   \mathbf{C} & \mathbf{0} \\
   \mathbf{0} & \mathbf{C} \\
  \end{array} \right]\!,
\end{equation}
where\begin{equation}
  \mathbf{C} =
  \left[ \begin{array}{cccccccc}
   4 & a & 0 & 0 \\
   a & 2 & a & 0 \\
   0 & a & 1 & a \\
   0 & 0 & a & 3 
  \end{array} \right]\!,
\end{equation}
so the eigenvalues of $\mathbf{S^{-1}MS}$ (which are the same as the eigenvalues of $\mathbf{M}$) are evenly degenerate. I don't know how to prove that the multiplicities of eigenvalues of $\mathbf{M}$ are exactly 2 (i.e., that the eigenvalues of $\mathbf{C}$ are non-degenerate), though, and whether it's actually true for all values of $a$. 
A: I should have made this only a comment, sorry:
Mathematica easily grinds out the characteristic polynomial of $\bf M$,    $$a^8-6 a^6 \lambda ^2+34 a^6 \lambda -46 a^6+11 a^4 \lambda ^4-122 a^4
   \lambda ^3+497 a^4 \lambda ^2-882 a^4 \lambda +577 a^4-6 a^2 \lambda
   ^6+94 a^2 \lambda ^5-596 a^2 \lambda ^4+1950 a^2 \lambda ^3-3454 a^2
   \lambda ^2+3116 a^2 \lambda -1104 a^2+\lambda ^8-20 \lambda ^7+170
   \lambda ^6-800 \lambda ^5+2273 \lambda ^4-3980 \lambda ^3+4180 \lambda
   ^2-2400 \lambda +576\,,$$
which simplifies to $\left(a^4+a^2 ((17-3 \lambda ) \lambda -23)+(\lambda -4) (\lambda -3)
   (\lambda -2) (\lambda -1)\right)^2\,.$
