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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

Expert verified

\(A\) is invertible.

Step by step solution

01

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.

02

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.

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Most popular questions from this chapter

Suppose \({A_{{\bf{11}}}}\) is invertible. Find \(X\) and \(Y\) such that

\[\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{\bf{0}}\\{\bf{0}}&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\{\bf{0}}&I\end{array}} \right]\]

Where \(S = {A_{{\bf{22}}}} - {A_{21}}A_{{\bf{11}}}^{ - {\bf{1}}}{A_{{\bf{12}}}}\). The matrix \(S\) is called the Schur complement of \({A_{{\bf{11}}}}\). Likewise, if \({A_{{\bf{22}}}}\) is invertible, the matrix \({A_{{\bf{11}}}} - {A_{{\bf{12}}}}A_{{\bf{22}}}^{ - {\bf{1}}}{A_{{\bf{21}}}}\) is called the Schur complement of \({A_{{\bf{22}}}}\). Such expressions occur frequently in the theory of systems engineering, and elsewhere.

Let \(A = \left( {\begin{aligned}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{aligned}} \right)\). Construct a \({\bf{2}} \times {\bf{2}}\) matrix Bsuch that ABis the zero matrix. Use two different nonzero columns for B.

Explain why the columns of an \(n \times n\) matrix Aare linearly independent when Ais invertible.

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

34. \(T\left( {{x_1},{x_2}} \right) = \left( {6{x_1} - 8{x_2}, - 5{x_1} + 7{x_2}} \right)\)

Suppose a linear transformation \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) has the property that \(T\left( {\mathop{\rm u}\nolimits} \right) = T\left( {\mathop{\rm v}\nolimits} \right)\) for some pair of distinct vectors u and v in \({\mathbb{R}^n}\). Can Tmap \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\)? Why or why not?

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