Consider the special block matrix
$$A = \begin{pmatrix} T & l \\l^*& S \end{pmatrix},$$ where we assume that each entry is invertible and $l$ and $l^*$ denote a scalar (and its complex conjugate) times the identity, whereas $T$ and $S$ are full matrices of equal size.
Is there a simple formula for the matrix inverse $A^{-1}?$ I am aware there are general formulas for block matrix inverses, but I thought in this case, things should simplify?
Please let me know if you have any questions.
Best Answer
Applying the block-inversion formula linked in the comments, we have $$ \pmatrix{\mathbf{A}^{-1}+\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & -\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \\ -(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & (\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}} = \\ \pmatrix{T^{-1}+|l|^2T^{-1}(S-|l|^2T^{-1})^{-1}T^{-1} & -l\,T^{-1}(S-|l|^2 T^{-1})^{-1} \\ -l^*(S-|l|^2 T^{-1})^{-1}T^{-1} & (S-|l|^2T^{-1})^{-1}} = \\ \pmatrix{T^{-1}[1+|l|^2(TS-|l|^2)^{-1}] & -l\,(ST-|l|^2 )^{-1} \\ -l^*(TS-|l|^2 )^{-1} & T(ST-|l|^2)^{-1}} $$