Show that all eigenvalues of this pentadiagonal matrix are double degenerate

Question

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!

Answer 1

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 $a$.

Answer 2

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 $\left(a^4+a^2 ((17-3 \lambda ) \lambda -23)+(\lambda -4) (\lambda -3) (\lambda -2) (\lambda -1)\right)^2\,.$