When is the following block matrix invertible?

Question

Let

$$A = \begin{bmatrix} x_{11} A_{11} & x_{12} A_{12} & x_{13} A_{13} & \cdots & x_{1d} A_{1d}\\ x_{21} A_{21} & x_{22} A_{22} & x_{23} A_{23} & \cdots & x_{2d} A_{2d}\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{d1} A_{d1} & x_{d2} A_{d2} & x_{d3} A_{d3} & \cdots & x_{dd} A_{dd}\\ \end{bmatrix}$$

where $A_{ij}$ is an $n \times n $ invertible matrix and $x_{ij} \neq 0$ is a scalar. All the entries in matrices and scalars are from a finite field $\mathbb{F}_q$. Let $X = (x_{ij})$. It can easily be shown that there exists a matrix $X$ such that the matrix $A$ is invertible if $q > 2$. But is there any nice sufficient condition I can put on matrix $X$ or an explicit construction of $X$ to make $A$ invertible other than saying the $\det(A)$ polynomial in $x_{ij}$ must be non-zero?

Answer 1

The only easy case I can think of is where $\mathrm A_{ij} = \mathrm A_0$, where $\mathrm A_0$ is an invertible $n \times n$ matrix, for all $i,j \in [d]$. In this special case,

$$\mathrm A = \begin{bmatrix} x_{11} \mathrm A_{0} & x_{12} \mathrm A_{0} & \cdots & x_{1d} \mathrm A_{0}\\ x_{21} \mathrm A_{0} & x_{22} \mathrm A_{0} & \cdots & x_{2d} \mathrm A_{0}\\ \vdots & \vdots & \ddots & \vdots \\ x_{d1} \mathrm A_{0} & x_{d2} \mathrm A_{0} & \cdots & x_{dd} \mathrm A_{0}\\ \end{bmatrix} = \mathrm X \otimes \mathrm A_0$$

where $\otimes$ denotes the Kronecker product. Hence,

$$\det (\mathrm A) := \left( \det (\mathrm X) \right)^n \cdot \left( \det (\mathrm A_0) \right)^d$$

and, thus, $\mathrm A$ is invertible if and only if $\mathrm X$ is invertible. If that is indeed the case, the inverse is

$$\rm A^{-1} = X^{-1} \otimes A_0^{-1}$$

Answer 2

If the matrix is block upper-triangular (or lower-triangular) then it is invertible.

That is, if $x_ij=0$ whenever $i>j$.