I have a complex block tridiagonal matrix that I am trying to prove has an inverse. The matrix M here:
$$M = \begin{bmatrix} A' & B' & 0 & \cdots & 0 \\ B & A & B & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & B & A & B \\ 0 & \cdots & 0 & B' & A' \\ \end{bmatrix} \\ $$
The prime notation, $A'$ and $B'$ simply means they are different from $A$ and $B$. All these blocks, $A$, $B$, $A'$ and $B'$ are complex tridiagonal square matrices and each is nonsingular. My first attempt is using induction to eliminate the lower diagonal of the $B$ and $B'$ blocks. That is:
$$ \begin{bmatrix} A' & B' & 0 & \cdots & \cdots & 0 \\ 0 & A - D_2 & B & \ddots & & \vdots \\ 0 & 0 & A - D_3& B & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 \\ \vdots & & \ddots &0 & A - D_{n-1} & B \\ 0 & \cdots & \cdots & 0 & 0 & A'-D_n \\ \end{bmatrix}$$
where $D_2 = B'(A')^{-1}B$, $D_n = B(A-D_{n-1})^{-1}B'$ and $D_i = B(A-D_{i-1})^{-1}B$ for $i=2,\dots,n-1$. Then use the fact that the determinant is nonzero, $\det(M)\ne0$. The problem I now face is showing that $(A-D_{i-1})^{-1}$ exists for $i=2,\dots,n$ and I am unsure how to do so. Is there a better approach? Something I may be missing?