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Chapter 2: Q2.4-15Q (page 93)

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.

Short Answer

Expert verified

\(X = A_{11}^{ - 1}{A_{21}}\), \(Y = A_{11}^{ - 1}{A_{12}}\)

Step by step solution

01

Solve the given equation

Simplify the expression \(\left[ {\begin{array}{*{20}{c}}I&0\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&0\\0&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\0&I\end{array}} \right]\) using matrix multiplication.

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

02

Equate the right and left-hand sides

For the equation \(\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{{A_{21}}}&{{A_{22}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&0\\0&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\0&I\end{array}} \right]\):

\({A_{11}} = {A_{11}}\), \({A_{11}}Y = {A_{12}}\), \(X{A_{11}} = {A_{21}}\) and \(X{A_{11}}Y + S = {A_{22}}\)

03

Find \(X\) and \(Y\)

By the equation \({A_{11}}Y = {A_{12}}\), since \({A_{11}}\) is invertible, \(Y = A_{11}^{ - 1}{A_{12}}\).

Similarly, \(X = A_{11}^{ - 1}{A_{21}}\).

So, \(X = A_{11}^{ - 1}{A_{21}}\) and \(Y = A_{11}^{ - 1}{A_{12}}\).

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Most popular questions from this chapter

Let \(X\) be \(m \times n\) data matrix such that \({X^T}X\) is invertible., and let \(M = {I_m} - X{\left( {{X^T}X} \right)^{ - {\bf{1}}}}{X^T}\). Add a column \({x_{\bf{0}}}\) to the data and form

\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

Compute \({W^T}W\). The \(\left( {{\bf{1}},{\bf{1}}} \right)\) entry is \({X^T}X\). Show that the Schur complement (Exercise 15) of \({X^T}X\) can be written in the form \({\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}\). It can be shown that the quantity \({\left( {{\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}} \right)^{ - {\bf{1}}}}\) is the \(\left( {{\bf{2}},{\bf{2}}} \right)\)-entry in \({\left( {{W^T}W} \right)^{ - {\bf{1}}}}\). This entry has a useful statistical interpretation, under appropriate hypotheses.

In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

\(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the 鈥渋nput鈥� to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the 鈥渙utput鈥� and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the 鈥渟tate鈥� vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)

If Ais an \(n \times n\) matrix and the equation \(A{\bf{x}} = {\bf{b}}\) has more than one solution for some b, then the transformation \({\bf{x}}| \to A{\bf{x}}\) is not one-to-one. What else can you say about this transformation? Justify your answer.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?

Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{4}}}\\{\bf{7}}&{ - {\bf{8}}}\end{aligned}} \right)\).

Suppose Ais an \(n \times n\) matrix with the property that the equation \[A{\mathop{\rm x}\nolimits} = 0\] has at least one solution for each b in \({\mathbb{R}^n}\). Without using Theorem 5 or 8, explain why each equation Ax = b has in fact exactly one solution.
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