# Eigenvalues and Eigenvectors of a block tridiagonal block 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 $x_0 = x_{N+1} = y_0 = y_{N+1} = 0$. 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 $\lambda_k$.

So I tried something else. The difference equation above implies that once we know $(x_1, y_1)$, we know can recursively find the value of $(x_j, y_j)$ for any $j$. So defined the $2 \times 2$ matrix $T_j$ by $$\begin{pmatrix} u_j \\ v_j \end{pmatrix} = T_j \begin{pmatrix} u_1 \\ v_1 \end{pmatrix}$$ Then this sequence of matrices satisfies a second order matrix difference equation $$E T_{j-1} + (C-\lambda)T_j+D T_{j+1} = 0$$ with boundary conditions $T_0 = 0$ and $T_1 = 1$ (the two $\times$ two identity matrix). Again if $\alpha$ is real, then $E = D$ and we can multiply both sides of the difference equation by $E^{-1}$ to obtain $$T_{j-1}+E^{-1} (C-\lambda) T_{j} + T_{j+1} = 0$$
This difference equation is solved by the matrix $$T_j = \dfrac{\sin(\beta j)}{\sin(\beta)}$$ where the matrix $\beta$ is defined by $\cos(\beta) = - \frac{1}{2}E^{-1}(C-\lambda)$. The boundary condition for $x_{N+1} = y{N+1} = 0$ means that $(x_1, y_1)$ must be in the kernel of $T_{N+1}$ which can only occur if $\det(T_{N+1}) = 0$. This conditions then translates to a condition on the eigenvalues $\lambda$ and we do in fact recover what we obtained before.

But when $\alpha$ is complex, $E \neq D$ and we can't do as we did before since we have to worry about matrices which don't commute. The only time we can in fact do that is if $a = b = 0$ so that our difference equation becomes $$E T_{j-1}-\lambda T_j + D T_{j+1}$$ Since $E$ and $D$ commute, we can work in the basis which diagonalizes both simultaneously, and our matrix equation just becomes two uncoupled scalar second order difference equations which are easy to solve.

All this to say that I'm quite stuck. My other idea was to first find the eigenvalues of $T$ directly by computing the determinant (the determinant must also satisfies some recursion relation, right?) Anyway, any help would be greatly appreciated!

• Here's an idea: start by finding the unitary $U$ such that $UMU^\dagger$ is diagonal whenever $M$ is symmetric and tridiagonal. Then, compute $$(Id \otimes U) T (Id \otimes U)^\dagger = \pmatrix{UAU^\dagger & UBU^\dagger\\ UB^*U^\dagger & -UA^*U^\dagger }$$ – Omnomnomnom Aug 14 '17 at 14:29
• Right, this is exactly the approach I took when $\alpha$ is real. In that case, $A, B$ and $B^*$ commute (in fact they are symmetric tridiagonal Toeplitz matrices) and so their eigenvalues and eigenvectors are well known. But when $\alpha$ is complex, $A$ and $A^*$ are no longer symmetric and they don't commute with $B$ or $B^*$ anymore. – PhysKid Aug 14 '17 at 14:35
• Sure, they don't commute. The point, however, is that if you go through the computation, you might find that $UAU^\dagger$ has a convenient (ableit non-diagonal) form. – Omnomnomnom Aug 14 '17 at 14:37
• Ultimately, it gives me something quite messy unfortunately. The matrix is no longer sparse. – PhysKid Aug 14 '17 at 14:51
• All right, good to know. – Omnomnomnom Aug 14 '17 at 14:55