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?
Best Answer
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
$$AB=I$$