# Inverse of block anti-diagonal matrix

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?

• Are the blocks $A_i$ the same size matrix or potentially different sizes? Jul 13, 2020 at 14:27
• @snulty In my specific case, yes. It would be interesting though to have generic square blocks. Jul 13, 2020 at 14:28
• Can you do the case where the blocks are $1\times 1$? That gives a pretty clear hint. Jul 13, 2020 at 14:48

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

$$AB=I$$

– user808985
Jul 18, 2020 at 0:47

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 $$d = 3$$,

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

• SW-NE dots using \cdot^{\textstyle \cdot^{\textstyle \cdot}}: $$\begin{bmatrix} & & {\rm I}\\ & \cdot^{\textstyle \cdot^{\textstyle \cdot}} & \\ {\rm I} & & \end{bmatrix}$$ Jul 18, 2020 at 0:28