Eigenvalues and eigenvectors of a tridiagonal block Toeplitz matrix

Let $T$ be the $2 N \times 2 N$ matrix defined by $$T = \begin{pmatrix} A && B\\ -B^* && -A^* \end{pmatrix}$$ where $*$ is entry wise complex conjugation, $A$ is a Hermitian $N \times N$ tridiagonal Toeplitz matrix $$A = \begin{pmatrix} a & \alpha& 0 & \dots & 0\\ \alpha^* & a & \alpha & \vdots & \vdots\\ 0 & \alpha^* & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & a & \alpha\\ 0 & \dots & \dots & \alpha^* & a \end{pmatrix}$$ with $a$ real and $\alpha$ in general complex and $B$ is the symmetric tridiagonal Toeplitz matrix $$B = \begin{pmatrix} b & \beta& 0 & \dots & 0\\ \beta & b & \beta & \vdots & \vdots\\ 0 & \beta & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & b & \beta\\ 0 & \dots & \dots & \beta & b \end{pmatrix}$$ where $b$ is real and $\beta$ is complex. I want to find the eigenvalues of eigenvectors of $T$. Here's what I have so far.

If $\alpha$ is real, then the solution is quite simple since $A$ is symmetric. Thus, $T$ can be written as $$T = i \sigma_x \otimes Im(B)+i\sigma_y \otimes Re(B)+\sigma_z \otimes A$$ where $\sigma_i$ are the usual Pauli matrices. It is well known that symmetric tridiagonal Toeplitz matrices all the same eigenvectors and their eigenvalues are particularly simple. We can simultaneously diagonalize $Re(B), Im(B),$ and $A$ and rewrite $T$ as a sum of $2 \times 2$ block matrices $$T = \sum_{n =1}^N \begin{pmatrix} a+ 2 \alpha \cos(\frac{\pi n}{(N+1)}) && b+\beta \cos(\frac{\pi n}{(N+1)})\\ -(b+\beta^* \cos(\frac{\pi n}{(N+1)})) && -(a+ 2 \alpha \cos(\frac{\pi n}{(N+1)})) \end{pmatrix}$$ And finding the eigenvalues and eigenvectors becomes diagonalizing a $2 \times 2$ matrix. Now if $\alpha$ is complex, we encounter a problem. $T$ can now be written as $$T = Id \otimes Im(A) +i \sigma_x \otimes Im(B)+i\sigma_y \otimes Re(B)+\sigma_z \otimes Re(A)$$ by hermicity of $A$, $Im(A)$ is an antisymmetric tridiagonal Toeplitz matrix which doesn't commute with $Re(A), Re(B), Im(B)$ (but it almost commutes in that the only non zero elements of the commutator are at the top left and bottom right of the matrix).

So I tried something else. In a different basis (well really just swapping the tensor product) we get $$T = Im(A) \otimes Id + Im(B) \otimes i \sigma_x + Re(B) \otimes i\sigma_y+ Re(A) \otimes \sigma_z$$ Which we can write as a block tridiagonal Toeplitz matrix $$T = \begin{pmatrix} C & D& 0 & \dots & 0\\ E & C & D & \vdots & \vdots\\ 0 & E & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & C & D\\ 0 & \dots & \dots & E & C \end{pmatrix}$$ with $C, D, E$ the $2 \times 2$ matrices. $$C = \begin{pmatrix} a && b\\ -b && -a \end{pmatrix} \: \: \: D = \begin{pmatrix} \alpha^* && \beta\\ -\beta^* && -\alpha \end{pmatrix} \: \: \: E = \begin{pmatrix} \alpha && \beta\\ -\beta^* && -\alpha^* \end{pmatrix}$$ And so the eigenvalue problem becomes a three term difference equation $$E\begin{pmatrix}x_{j-1} \\ y_{j-1}\end{pmatrix} + C\begin{pmatrix}x_{j} \\ y_{j}\end{pmatrix} + D\begin{pmatrix}x_{j+1} \\ y_{j+1}\end{pmatrix} = \lambda \begin{pmatrix} x_{j} \\ y_{j} \end{pmatrix}$$ with boundary conditions $x_0 = x_{N+1} = y_0 = y_{N+1}$. Now, like in the regular tridiagonal Topeplitz matrix case, I make an ansatz of $x_j = x r^j$ and $y_j = y r^j$ the difference equation becomes $$(rE+(C-\lambda)+r^{-1}D ) \begin{pmatrix} x \\y \end{pmatrix} = 0$$ For a non trivial solution to our equation we need the determinant to vanish, which is a polynomial of degree four in $r$ and we assume for the moment that there are four distinct roots $r_k$, $k = {1,2,3,4}$. Thus the general solution is of the form $$\begin{pmatrix} x_j \\ y_j \end{pmatrix} = \sum_{k = 1}^4 c_k \lambda^j_k \begin{pmatrix} x_k \\ y_k \end{pmatrix}$$ where $(x_k,y_k)$ is in the kernel of the difference equation above. We need to find the $c_k$ which satisfy the boundary conditions which equivalent to finding a non trivial solution to $$\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \\ y_1 & y_2 & y_3 & y_4 \\ \lambda_1^{N+1} x_1 & \lambda_2^{N+1} x_2 & \lambda_3^{N+1} x_3 & \lambda_4^{N+1} x_4 \\ \lambda_1^{N+1} y_1 & \lambda_2^{N+1} y_2 & \lambda_3^{N+1} y_3 & \lambda_4^{N+1} y_4 \end{pmatrix} \begin{pmatrix} c_1\\ c_2\\ c_3\\ c_4 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ 0 \end{pmatrix}$$ For a nontrivial solution, we need the determinant to vanish, which imposes a conditions on the roots $\lambda_k$. It is at this point that I'm stuck and not sure what to do, because the determinant involves the eigenvectors $(x_k,y_k)$ which is involved and confusing. Any help would be greatly appreciated!

I've found a partial solution, namely when $a = b = 0$. The thing to realize is that $D$ and $E$ actually commute so they share a common set of eigenvectors. We can then make a change of basis to this eigenbasis $(x,y) \to (X,Y)$ so that the difference equation becomes, in this basis, $$\begin{pmatrix} r E_1-\lambda+r^{-1}D_1 & 0 \\ 0 & r E_1-\lambda+r^{-1}D_2 \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 0 \\0 \end{pmatrix}$$ where $E_i, D_i$ are the eigenvalues of $E, D$ with common eigenvector $(x_i, y_i)$. Our difference equation then "uncouples" into two regular difference equations. The boundary conditions are still the same $X_0 = Y_0 = X_{N+1} = Y_{N+1} = 0$ and so we really have two copies of the regular tridiagonal Toeplitz matrix problem. If $a \neq 0, b \neq 0$ then we can't apply the same trick, because $C$ doesn't commute with $D$ or $E$.