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Related to a previous question: Suppose I want to invert a (sparse) matrix written in block form as \begin{array}{cccc} A_{11} & A_{12} & \ldots & A_{1n}\\ A_{21} & A_{22} & & \vdots\\ \vdots & & \ddots\\ A_{n1} & A_{n2} & \ldots & A_{nn} \end{array}

where all the $A_{i,j}$'s are diagonal. Is the best way to do this just repeated application of the partitioned matrix inverse formula? Also, do such matrices have a name?

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Yes, they are called matrices. Isn't every matrix of that form, with the submatrices $1\times1$? –  Gerry Myerson Apr 30 '12 at 5:46
    
Let me add the qualifier of sparseness then. –  jsouth Apr 30 '12 at 5:50
    
If they don't have a name yet, I'd call them "striped matrices"... –  J. M. Apr 30 '12 at 6:10

1 Answer 1

If the $A_{ij}$ are $m \times m$ diagonal blocks, then your matrix rearranges into $m$ diagonal blocks, each $n \times n$, by the permutation of row and column indices $1 + a + b m \mapsto 1 + b + a n$, $a= 0 \ldots m-1,\ b = 0 \ldots n-1$. So just invert each diagonal block and permute back. For example $$ \left[ \begin {array}{cccccc} a_{{1,1}}&0&0&a_{{1,4}}&0&0\\ 0 &a_{{2,2}}&0&0&a_{{2,5}}&0\\ 0 &0&a_{{3,3}}&0&0&a_{{3,6}}\\a_{{4,1}}&0&0&a_{{4,4}} &0&0\\ 0&a_{{5,2}}&0&0&a_{{5,5}}&0 \\ 0&0&a_{{6,3}}&0&0&a_{{6,6}}\end {array} \right]^{-1} = \left[ \begin {array}{cccccc} b_{{1,1}}&0&0&b_{{1,4}}&0&0\\ 0&b_{{2,2}}&0&0&b_{{2,5}}&0\\ 0 &0&b_{{3,3}}&0&0&b_{{3,6}}\\b_{{4,1}}&0&0&b_{{4,4}} &0&0\\ 0&b_{{5,2}}&0&0&b_{{5,5}}&0 \\ 0&0&b_{{6,3}}&0&0&b_{{6,6}}\end {array} \right]$$ where $$\left[\matrix{b_{11} & b_{14}\cr b_{41} & b_{44}\cr}\right] = \left[\matrix{a_{11} & a_{14}\cr a_{41} & a_{44}\cr}\right]^{-1},\ \left[\matrix{b_{22} & b_{25}\cr b_{52} & b_{55}\cr}\right] = \left[\matrix{a_{22} & a_{25}\cr a_{52} & a_{55}\cr}\right]^{-1},\ \left[\matrix{b_{33} & b_{36}\cr b_{63} & b_{66}\cr}\right] = \left[\matrix{a_{33} & a_{36}\cr a_{63} & a_{66}\cr}\right]^{-1}$$

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