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I want to invert a non-singular $N\times N$ block matrix $A$ where each $M\times M$ block in $A$ is a diagonal matrix. That is, I have an $N\times N$ block matrix $$ A = \begin{pmatrix} A^{11} & A^{12} & \cdots & \cdots & A^{1N} \\ A^{21} & A^{22} & & & \\ \vdots & & \ddots & & \\ \vdots & & & \ddots & \\ A^{N1} & & & & A^{NN} \end{pmatrix}, $$ where for $i,j \in \{1,\dots,N\}$, each $M\times M$ block $A_{ij}$ is of the form $$ A_{ij} = \begin{pmatrix} a_{11} & 0 & \cdots & \cdots & 0 \\ 0 & a_{22} & & & \\ \vdots & & \ddots & & \\ \vdots & & & \ddots & \\ 0 & & & & a_{MM}. \end{pmatrix} $$

A very simple explicit example where $N=3$ and $M=3$ (with the non-zero diagonals in red for clarity) would be $$ A = \begin{pmatrix} \color{red}{1} & 0 & 0 & \color{red}{2} & 0 & 0 & \color{red}{3} & 0 & 0 \\ 0 & \color{red}{2} & 0 & 0 & \color{red}{3} & 0 & 0 & \color{red}{4} & 0 \\ 0 & 0 & \color{red}{3} & 0 & 0 & \color{red}{4} & 0 & 0 & \color{red}{5} \\ \color{red}{4} & 0 & 0 & \color{red}{5} & 0 & 0 & \color{red}{6} & 0 & 0 \\ 0 & \color{red}{5} & 0 & 0 & \color{red}{6} & 0 & 0 & \color{red}{7} & 0 \\ 0 & 0 & \color{red}{6} & 0 & 0 & \color{red}{7} & 0 & 0 & \color{red}{8} \\ \color{red}{7} & 0 & 0 & \color{red}{8} & 0 & 0 & \color{red}{9} & 0 & 0 \\ 0 & \color{red}{8} & 0 & 0 & \color{red}{9} & 0 & 0 & \color{red}{0} & 0 \\ 0 & 0 & \color{red}{9} & 0 & 0 & \color{red}{0} & 0 & 0 & \color{red}{1} \\ \end{pmatrix}. $$

Is there a special formula for the inverse a matrix like this, i.e. is it possible to obtain a special explicit expression for the inverse of an $N\times N$ block matrix $A$ where each $M\times M$ block in $A$ is a diagonal matrix?

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    $\begingroup$ Permute the rows and columns of $A$ simultaneously to turn $A$ into a block-diagonal matrix. Take the inverse. Then permute back. $\endgroup$
    – user1551
    Commented Oct 19, 2019 at 15:28
  • $\begingroup$ @user1551 "Permute the rows and columns of $A$ simultaneously to turn $A$ into a block-diagonal matrix." How would you do that? $\endgroup$
    – amsmath
    Commented Oct 19, 2019 at 15:35
  • $\begingroup$ @amsmath $N=2, M\in \mathbb{N}$ is straightforward using the Schur complement approach to the inverse of a $2\times 2$ block matrix. How it generalize beyond these cases is what I am unsure of. $\endgroup$
    – sonicboom
    Commented Oct 19, 2019 at 15:35
  • $\begingroup$ It should be computationally easy to solve, using permutations and such, but it seems there will not usually be nice and simple representation. For $M=2$, link shows that it won't have a nice form even with all the $A^{ij}$ being diagonal. $\endgroup$ Commented Oct 19, 2019 at 15:35
  • $\begingroup$ @sonicboom Why is $M=1$ trivial? $\endgroup$
    – amsmath
    Commented Oct 19, 2019 at 15:36

1 Answer 1

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Since user1551 doesn't seem to show up anymore, let me make precise their idea here. Each row of your matrix follows a specific pattern. The first one starts with some entry and then $M-1$ zeros, followed by some entry and then $M-1$ zeros and so on. Let's call this Pattern(1). The next row follows Pattern(2) which is just a shifted version of Pattern(1). And so on until Pattern(M) in the $M$-th row. Then we start again with Pattern(1) in the $(M+1)$-th row. Let's now permute the rows such that we sort by patterns. So, first $N$ times Patten(1), then $N$ times Pattern(2) and so on. Then we obtain a $M\times N$ block matrix with blocks of size $N\times M$: $$ B = \begin{pmatrix} B^{11} & B^{12} & \cdots & \cdots & B^{1N} \\ B^{21} & B^{22} & & & \\ \vdots & & \ddots & & \\ \vdots & & & \ddots & \\ B^{M1} & & & & B^{MN} \end{pmatrix}. $$ Here, for any block-row $k$, $B^{k1},\ldots,B^{kN}$ all have the same form, namely each column being zero except the $k$-th. Now, let us consider the columns of $B$. They are all of the form $$ \begin{bmatrix}v_1\\ \vdots\\ v_M\end{bmatrix} $$ with each $v_j$ being an $N\times 1$-vector and only one of them is not the zero vector. Sorting the columns by pattern will then result in an $M\times M$ block-diagonal matrix with $N\times N$-matrices as entries.

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  • $\begingroup$ Yes this seems correct. So we perform this process, invert the resulting block-diagonal matrix and 'unperform' the process. Recursive Schur complements could also be used (arxiv.org/pdf/1612.00001.pdf). $\endgroup$
    – sonicboom
    Commented Oct 19, 2019 at 17:41
  • $\begingroup$ About the 'unperform' procedure: You find permutation matrices $P$ and $Q$ such that $PAQ$ has the above block-diagonal form. So $A^{-1} = Q(PAQ)^{-1}P$. $\endgroup$
    – amsmath
    Commented Oct 19, 2019 at 19:03

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