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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.

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  • $\begingroup$ 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 $M^{-1}=\begin{bmatrix} \det(M)^{-1}(A+B) & -\det(M)^{-1}C \\ -\det(M)^{-1}C & \det(M)^{-1}(A+B) \end{bmatrix}$ $\endgroup$
    – N8tron
    Commented Nov 1, 2022 at 15:32
  • $\begingroup$ I guess I should use a distinct notation since that is more of a block determinant rather than the conventional one ;) $\endgroup$
    – N8tron
    Commented Nov 1, 2022 at 15:36
  • $\begingroup$ Can you compute the inverse when $n=1$? $\endgroup$ Commented Nov 1, 2022 at 18:03
  • $\begingroup$ @InMath There is a nice approach if we are given that $A+B-C$ is invertible $\endgroup$ Commented Nov 1, 2022 at 19:06

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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

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    $\begingroup$ 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 $\endgroup$
    – Nick Alger
    Commented Nov 1, 2022 at 20:46
  • $\begingroup$ @Nick Nice observation! $\endgroup$ Commented Nov 1, 2022 at 22:36
  • $\begingroup$ Nice answer! Thanks! $\endgroup$ Commented Nov 2, 2022 at 13:58

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