Question

Show that block upper triangular matrix \(A\) in Example 5is invertible if and only if both \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible. [Hint: If \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible, the formula for \({A^{ - {\bf{1}}}}\) given in Example 5 actually works as the inverse of \(A\).] This fact about \(A\) is an important part of several computer algorithims that estimates eigenvalues of matrices. Eigenvalues are discussed in chapter 5.

Short answer

\(A\) is invertible.

Step-by-step solution

Analyze \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\)

From Example 5, matrix \(A\) is invertible if and only if \({A_{11}}\) and \({A_{12}}\) are invertible. This condition can be used for theif part.

Find the product \(A{A^{ - {\bf{1}}}}\)

Find the product \(A{A^{ - 1}}\).

\(\begin{array}{c}A{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\0&{{A_{22}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{A_{11}^{ - 1}}&{ - A_{11}^{ - 1}{A_{12}}A_{22}^{ - 1}}\\0&{A_{22}^{ - 1}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{{A_{11}}A_{11}^{ - 1}}&{{A_{11}}\left( { - A_{11}^{ - 1}} \right){A_{12}}A_{22}^{ - 1} + {A_{12}}A_{22}^{ - 1}}\\0&{0\left( { - A_{11}^{ - 1}} \right){A_{12}}A_{22}^{ - 1} + {A_{22}}A_{22}^{ - 1}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}I&{ - {A_{12}}A_{22}^{ - 1} + {A_{12}}A_{22}^{ - 1}}\\0&I\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}I&0\\0&I\end{array}} \right]\end{array}\)

As \(A\) is a square matrix, it is invertible.