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.


1 Answer 1


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 $A_{11}$).


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