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Let

$$A = \begin{bmatrix} A_{11} & \cdots & A_{1m} \\ \vdots & \ddots & \vdots \\ A_{m1} & \cdots & A_{mm} \end{bmatrix}$$ be a block matrix where each matrix $A_{ij} \in \mathbb{R}^{n\times n}$ is diagonal. What is $A^{-1}$?

It seems that it's possible to iteratively apply the usual $2 \times 2$ inverse formula. However, since that seems as though it would produce something very complicated, I'm not sure if there's a more clever way.

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  • $\begingroup$ Why do you suspect that $A$ is invertible? For instance, the $2 \times 2$ zero matrix satisfies this property — each of the four matrices $A_{ij}$ is the $1 \times 1$ zero matrix, and of course, a $1 \times 1$ matrix is diagonal — but it is not invertible. $\endgroup$ Commented Jul 5, 2020 at 4:40
  • $\begingroup$ These blocks, by virtue of being diagonal, will commute multiplicatively. However I suspect we ought to try an arrangement as a block diagonal matrix, rather than a diagonal block matrix. Do you really want to compute the inverse, or will it,suffice to be able to solve linear systems? $\endgroup$
    – hardmath
    Commented Jul 5, 2020 at 5:03

1 Answer 1

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Any matrix with diagonal blocks (assuming the blocks have the same size) can be converted to a block-diagonal matrix. In particular, suppose that $a_{ijk}$ denotes the $k$th diagonal entry of the block $A_{ij}$, so that $$ A_{ij} = \pmatrix{a_{ij1} \\ & \ddots \\ && a_{ijn}}. $$ There exists a permutation matrix $P$ such that $$ P^TAP = \pmatrix{B_1\\ & \ddots \\ && B_n}, $$ where $$ B_k = \pmatrix{ a_{11k} & \cdots & a_{1mk}\\ \vdots & \ddots & \vdots \\ a_{m1k} & \cdots & a_{mmk}}. $$

It follows that the inverse of $A$ (assuming it exists) satisfies $$ A^{-1} = P\pmatrix{B_1^{-1}\\ & \ddots \\ && B_n^{-1}}P^T. $$ In other words, $A^{-1}$ will have the block-structure $$ A^{-1} = \pmatrix{C_{11} & \cdots & C_{1m}\\ \vdots & \ddots & \vdots\\ C_{m1} & \cdots & C_{mm}}, $$ where $C_{ij}$ is a diagonal matrix whose $k$th diagonal entry is the $i,j$ entry of $B_k^{-1}$.


If you're interested in what the matrix $P$ looks like, it can be written as $$ P = \sum_{i,j = 1}^{m,n} (e_{i}^{(m)} \otimes e_j^{(n)})(e_j^{(n)} \otimes e_i^{(m)})^T $$ where $e_i^{(n)}$ denotes the $i$th canonical basis vector of $\Bbb R^n$ (the $i$th column of the size $n$ identity matrix), and $\otimes$ denotes the Kronecker product.

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  • $\begingroup$ I adapted this answer from my earlier post on a similar question $\endgroup$ Commented Jul 5, 2020 at 5:25
  • $\begingroup$ Thank you, this is very helpful! $\endgroup$
    – user257566
    Commented Jul 5, 2020 at 17:20
  • $\begingroup$ @BenGrossmann Can you please provide some material for the step: "There exists a permutation matrix P such that $P^TAP = B$" ? Is it a theorem? $\endgroup$ Commented Jun 6, 2023 at 5:25

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