# Inverse of this block matrix

I have three square real matrices $$A, B$$ and $$C$$ of the same order, say $$n$$. I know that $$A+B$$ and $$C$$ are invertible. Then I built a new $$nN \times nN$$ big block matrix as follows: $$M = \begin{pmatrix} A+B & C & C &\cdots & C \\ C & A+B & C & \cdots & C \\ C & C & A+B & \cdots & C \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ C & C & C & \cdots & A+B \end{pmatrix}$$ Is there an expression for the inverse $$M^{-1}$$? I tried toeplitz inverse and also block inverse but both attempts did not work.

• You could try the adjugate approach. $M\textrm{adj}(M)=\textrm{adj}(M)M=\det(M)I$ You get some messiness on the determinant side since you may result in a trying to invert a matrix relation that may not be invertible. For example in the $2 \times 2$ case $\det(M)=(A+B)^2-C^2$ if you know this is invertible you get something like $M^{-1}=\begin{bmatrix} \det(M)^{-1}(A+B) & -\det(M)^{-1}C \\ -\det(M)^{-1}C & \det(M)^{-1}(A+B) \end{bmatrix}$ Commented Nov 1, 2022 at 15:32
• I guess I should use a distinct notation since that is more of a block determinant rather than the conventional one ;) Commented Nov 1, 2022 at 15:36
• Can you compute the inverse when $n=1$? Commented Nov 1, 2022 at 18:03
• @InMath There is a nice approach if we are given that $A+B-C$ is invertible Commented Nov 1, 2022 at 19:06

Let $$X = A + B - C$$. Suppose we know a priori that $$X$$ is invertible. The matrix can be expressed in the form $$M = I_N \otimes X + (ee^T) \otimes C$$ where $$\otimes$$ denotes a Kronecker product and $$e \in \Bbb R^N$$ is the vector $$e = (1,\dots,1)^T$$. With the Woodbury matrix identity, we can express $$M^{-1}$$ in the form $$M^{-1} = I_N \otimes X^{-1} - (e \otimes X^{-1}C)(I_n + N \cdot X^{-1}C)^{-1}(e^T \otimes X^{-1}).$$ This requires only the computation of an $$n\times n$$ inverse
• There is no loss in generality to assuming that $X$ is invertible, because this is required for the block matrix to be invertible. If $X$ is singular with null vector $u$ such that $Xu=0$, then direct calculation shows that $(u,-u,u,-u,\dots,u,-u)$ is in the null space of the block matrix for even $n$, and $(u,-u,\dots,u,-u,0)$ is in the null space for odd $n$. Actually, any block vector where you put the same number of $u$ and $-u$ blocks will be a null vector for the block matrix Commented Nov 1, 2022 at 20:46