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Let $A \in \mathbb R^{n\times n}$ be an invertible block anti-diagonal matrix (with $d$ blocks), i.e. $$ A = \begin{pmatrix} & & & A_1 \\ & & A_2 & \\ & \cdot^{\textstyle \cdot^{\textstyle \cdot}} & & \\ A_d\end{pmatrix}, $$ with all square blocks $A_1, \ldots, A_d$ invertible. Is there a formula for its inverse?

In the diagonal case, it is just the diagonal block matrix with the inverses of the blocks, is there an equivalent for the anti-diagonal case?

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  • $\begingroup$ Are the blocks $A_i$ the same size matrix or potentially different sizes? $\endgroup$
    – snulty
    Jul 13 '20 at 14:27
  • $\begingroup$ @snulty In my specific case, yes. It would be interesting though to have generic square blocks. $\endgroup$
    – G. Gare
    Jul 13 '20 at 14:28
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    $\begingroup$ Can you do the case where the blocks are $1\times 1$? That gives a pretty clear hint. $\endgroup$ Jul 13 '20 at 14:48
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I think this is the answer with all the blocks invertible. $$ A = \begin{pmatrix} & & & A_1 \\ & & A_2 & \\ & \dots & & \\ A_d\end{pmatrix}, $$

$$ B = \begin{pmatrix} & & & A_d^{-1} \\ & & A_{d-1}^{-1} & \\ & \dots & & \\ A_1^{-1}\end{pmatrix}, $$ we have

$$AB=I$$

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  • $\begingroup$ very neat answer +1 $\endgroup$
    – user808985
    Jul 18 '20 at 0:47
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There exists a permutation matrix $\rm P$ such that

$${\rm A P} = \mbox{diag} \left( {\rm A}_1, {\rm A}_2, \dots, {\rm A}_d \right)$$

Assuming that all the ${\rm A}_i$ blocks are invertible,

$$\left( \rm A P \right)^{-1} = {\rm P}^\top {\rm A}^{-1} = \mbox{diag} \left( {\rm A}_1^{-1}, {\rm A}_2^{-1}, \dots, {\rm A}_d^{-1} \right)$$

and, thus,

$${\rm A}^{-1} = \color{blue}{{\rm P} \, \mbox{diag} \left( {\rm A}_1^{-1}, {\rm A}_2^{-1}, \dots, {\rm A}_d^{-1} \right)}$$

For example, if $d = 3$,

$${\rm A}^{-1} = \begin{bmatrix} & & {\rm I}\\ & {\rm I} & \\ {\rm I} & & \end{bmatrix} \begin{bmatrix} {\rm A}_1^{-1} & & \\ & {\rm A}_2^{-1} & \\ & & {\rm A}_3^{-1}\end{bmatrix} = \begin{bmatrix} & & {\rm A}_3^{-1}\\ & {\rm A}_2^{-1} & \\ {\rm A}_1^{-1} & & \end{bmatrix}$$


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  • $\begingroup$ SW-NE dots using \cdot^{\textstyle \cdot^{\textstyle \cdot}}: $$\begin{bmatrix} & & {\rm I}\\ & \cdot^{\textstyle \cdot^{\textstyle \cdot}} & \\ {\rm I} & & \end{bmatrix}$$ $\endgroup$ Jul 18 '20 at 0:28

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