Let $M in mathbb{R}^{N times N}$ be a block-tridiagonal matrix:

$$M = begin{bmatrix} B_1 & C_1 & 0 & 0 & cdots & 0 \ A_2 & B_2 & C_2 & 0 & cdots & 0 \ 0 & A_3 & B_3 & C_3 & cdots & 0 \ 0 & 0 & A_4 & B_4 & cdots & 0 \ vdots & vdots & vdots & vdots & ddots & vdots \ 0 & 0 & cdots & 0 & A_n & B_n end{bmatrix} $$

where each $B_i$ is square and invertible, with varying sizes.

## Problem

What are sufficient conditions on $A_i$, $B_i$ and $C_i$ for showing that $M$ is invertible?

For example, a sufficient, but weak condition is that $C_i = 0$ for each $i$.

### Reformulation by Schur complements

Let

$$D_i = B_i – C_i D_{i + 1}^{-1} A_{i + 1},$$

$$D_n = B_n.$$

Supposing each $D_i$ is invertible, by elimination $M$ reduces to

$$M sim begin{bmatrix} D_1 & 0 & 0 & 0 & cdots & 0 \ A_2 & D_2 & 0 & 0 & cdots & 0 \ 0 & A_3 & D_3 & 0& cdots & 0 \ 0 & 0 & A_4 & D_4 & cdots & 0 \ vdots & vdots & vdots & vdots & ddots & vdots \ 0 & 0 & cdots & 0 & A_n & D_n end{bmatrix} $$

Then

$$det(M) = prod_{i = 1}^n det(D_i).$$

Hence invertibility of each $D_i$ is invertible implies $M$ is invertible. The converse also holds.

Hence sufficient conditions for the invertibility of Schur complements are also useful.

### Reformulation by 2×2 block matrices

Let

$$E_i = begin{bmatrix} B_i & C_i \ A_i & D_{i + 1} end{bmatrix},$$

$$E_n = B_n$$

Then

$$det(E_i) = det(D_{i + 1}) det(D_i) = prod_{j = i}^n det(D_i),$$

$$det(E_1) = det(M).$$

Hence sufficient conditions for a 2×2 block matrix with square invertible diagonal blocks being invertible are also useful.