Assume that we are given a block matrix of the form:$$M = \left[\begin{array}{cc}A   & b \\b^T & c \\\end{array}\right]$$where $b$ is a column vector. and $c$ is a scalar.Schur's complement of $A$ in $M$ is given by:$$ s = c - b^T A^{-1} b $$Assume that we know $A$ is invertible and we want to check if $M$ is invertible. Is it true that $M$ is invertible if, and only if $s$ is different than zero?Thank you
Schur's Complement and Inverse of Block Matrices
linear algebra
Yes.Under the assumption that $A^{-1}$ exists, then $\det (M) = \det (A) s.$Since $\det(A) \neq 0$, $\det (M) \neq 0$ if and only if $s \neq 0$.  Thus $M$ is inveritible if and only if $s \neq 0$.    This theorem on block determinants is standard material that can be found in many linear algebra textbooks.  See for example Carl Meyer's textbook, Matrix Analysis and Linear Algebra.  The proof is pretty simple- write $M$ as $M=\left[ \begin{array}{cc}I & 0 \\b^{T}A^{-1} & 1 \\\end{array}\right]\left[ \begin{array}{cc}A & b \\0 & c-b^{T}A^{-1}b \\\end{array}\right]$Then apply the product rule and the rule for determinants of (block) diagonal matrices.