Suppose block matrix \(A\) on the left side of (7) is invertible and \({A_{{\bf{11}}}}\) is invertible. Show that the Schur component \(S\) of \({A_{{\bf{11}}}}\) is invertible. [Hint: The outside factors on the right side of (7) are always invertible. Verify this.] When \(A\) and \({A_{{\bf{11}}}}\) are invertible, (7) leads to a formula for \({A^{ - {\bf{1}}}}\), using \({S^{ - {\bf{1}}}}\) \(A_{{\bf{11}}}^{ - {\bf{1}}}\), and the other entries in \(A\).

As \(A\) and \({A_{11}}\) are invertible, (7)

\(\left[ {\begin{array}{*{20}{c}}I&0\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&0\\{ - X}&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0\\0&I\end{array}} \right]\).

And

\(\left[ {\begin{array}{*{20}{c}}I&Y\\0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&{ - Y}\\0&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0\\0&I\end{array}} \right]\)

As the matrices \(\left[ {\begin{array}{*{20}{c}}I&0\\X&I\end{array}} \right]\) and \(\left[ {\begin{array}{*{20}{c}}I&Y\\0&I\end{array}} \right]\) are square, they are invertible.

Multiply the equation \(\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{{A_{21}}}&{{A_{22}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&0\\0&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\0&I\end{array}} \right]\) by \({\left[ {\begin{array}{*{20}{c}}I&0\\X&I\end{array}} \right]^{ - 1}}\) and \({\left[ {\begin{array}{*{20}{c}}I&Y\\0&I\end{array}} \right]^{ - 1}}\) on both sides.

\({\left[ {\begin{array}{*{20}{c}}I&0\\X&I\end{array}} \right]^{ - 1}}A{\left[ {\begin{array}{*{20}{c}}I&Y\\0&I\end{array}} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&0\\0&S\end{array}} \right]\)

Since the matrix \(\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&0\\0&S\end{array}} \right]\) is the product of inverse matrices, the Schur complement \(S\) is invertible.