# 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?

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?

• Permute the rows and columns of $A$ simultaneously to turn $A$ into a block-diagonal matrix. Take the inverse. Then permute back. – user1551 Oct 19 '19 at 15:28
• @user1551 "Permute the rows and columns of $A$ simultaneously to turn $A$ into a block-diagonal matrix." How would you do that? – amsmath Oct 19 '19 at 15:35
• @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. – sonicboom Oct 19 '19 at 15:35
• 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. – Shiranai Oct 19 '19 at 15:35
• @sonicboom Why is $M=1$ trivial? – amsmath Oct 19 '19 at 15:36

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.
• About the 'unperform' procedure: You find permutation matrices $P$ and $Q$ such that $PAQ$ has the above block-diagonal form. So $A^{-1} = Q(PAQ)^{-1}P$. – amsmath Oct 19 '19 at 19:03