# Schur complement of a matrix $A$

Let $A\in\mathbb{R}^{n\times n}$ and its inverse be partitioned $$A = \begin{pmatrix} A_{11} & A_{12}\\ A_{21} & A_{22}\\ \end{pmatrix},\:\: A^{-1} = \begin{pmatrix} \tilde{A_{11}} & \tilde{A_{12}}\\ \tilde{A_{21}} & \tilde{A_{22}}\\ \end{pmatrix}$$ where $A_{11}\in\mathbb{R}^{k\times k}$ and $\tilde{A_{11}}\in\mathbb{R}^{k\times k}$.

a.) Show that if, $S = A_{22} - A_{21}A_{11}^{-1}A_{12}$, the Schur complement of $A$ with respect to $A_{11}$ exists then $A$ is nonsingular iff $S$ is nonsingular.

b.) Show that $S^{-1} = \tilde{A_{11}}$.

Attempted solution a.) The Schur complement arises as a result of performing a block Gaussian elimination by multiplying the matrix $A$ from the right with the block lower triangular matrix $$L = \begin{pmatrix} I_n & 0\\ -A_{22}^{-1}A_{21} & I_m\\ \end{pmatrix}$$ where $I_n$ is a $n\times n$ identity matrix. \begin{align*} AL &= \begin{pmatrix} A_{11} & A_{12}\\ A_{21} & A_{22}\\ \end{pmatrix} \begin{pmatrix} I_{n} & 0\\ -A_{22}^{-1}A_{21} & I_m\\ \end{pmatrix} = \begin{pmatrix} A_{11} - A_{12}A_{22}^{-1}A_{21} & A_{12}\\ 0 & A_{22}\\ \end{pmatrix}\\ &= \begin{pmatrix} I_n & A_{12}A_{22}^{-1}\\ 0 & I_m\\ \end{pmatrix}\begin{pmatrix} A_{11} - A_{12}A_{22}^{-1}A_{21} & 0\\ 0 & A_{22}\\ \end{pmatrix} \end{align*} As you can see after multiplication with the matrix $L$ the Schur complement appears in the upper $n\times n$ block. I am not really sure where to go from here any suggestions is greatly appreciated.

Take determinants and you have finished (apart from the difference with respect to the problem statement that you are considering the Schur complement of $A_{22}$ instead of that of $A_{11}$).