I am having difficulties in proving the closed form of matrix $(I-(1- α)P)^{-1}$ into the equation below. I got this equation from a research paper and i failed to understand how they got there.
If $I$ is an $m × m$ identity matrix. According to matrix
they said a closed form of $(I-(1- α)P)^{-1}$ is equals to below, but I failed to understand the process of getting here
Any help would be really appreciated, thank you..
Let $a$ be a scalar, $b$ a row and $c$ a column vector, $D$ an invertible matrices, all of compatible dimensions to form a block matrix. Then $$ \begin{bmatrix} a&b\\c&D \end{bmatrix}⋅ \begin{bmatrix} 1&0\\ -D^{-1}c&I \end{bmatrix} = \begin{bmatrix} a-bD^{-1}c&b\\0&D \end{bmatrix} = \begin{bmatrix} 1&bD^{-1}\\ 0&I \end{bmatrix}⋅ \begin{bmatrix} a-bD^{-1}c&0\\0&D \end{bmatrix} $$ so that $$ \begin{bmatrix} a&b\\c&D \end{bmatrix}^{-1} = \begin{bmatrix} 1&0\\ -D^{-1}c&I \end{bmatrix}⋅ \begin{bmatrix} (a-bD^{-1}c)^{-1}&0\\0&D^{-1} \end{bmatrix}⋅ \begin{bmatrix} 1&-bD^{-1}\\ 0&I \end{bmatrix} $$ and so on...