Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 15E from Chapter 2.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 15E
Chapter:
Problem:
Let If A11 is invertible, then the matrix is called the Schur complement of A11. Likewise, if A22 is invertible, the matrix is called the Schur complement of A22. Suppose A11 is invertible. Find X and Y such that
Step-by-Step Solution
Given Information
We are given that $A_{11}$ is invertible. We have to find X and Y such that \[ \left[ \begin{array} { c c } { A _ { 11 } } & { A _ { 12 } } \\ { A _ { 21 } } & { A _ { 22 } } \end{array} \right] = \left[ \begin{array} { c c } { I } & { 0 } \\ { X } & { I } \end{array} \right] \left[ \begin{array} { c c } { A _ { 11 } } & { 0 } \\ { 0 } & { S } \end{array} \right] \left[ \begin{array} { c c } { I } & { Y } \\ { 0 } & { I } \end{array} \right] \]
Step-1:
Using the matrix properties \[\begin{array}{*{20}{l}} {I\cdot A = A}\\ {0\cdot A = 0}\\ {A + 0 = A} \end{array}\] The product of matrices on right hand side: \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} I&0\\ X&I \end{array}} \right]\left[ {\begin{array}{*{20}{r}} {{A_{11}}}&0\\ 0&S \end{array}} \right]\left[ {\begin{array}{*{20}{l}} I&Y\\ 0&I \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {I{A_{11}} + 0\cdot0}&{I\cdot0 + 0\cdot S}\\ {X{A_{11}} + I\cdot0}&{X\cdot0 + IS} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} I&Y\\ 0&I \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{A_{11}}}&0\\ {X{A_{11}}}&S \end{array}} \right]\left[ {\begin{array}{*{20}{l}} I&Y\\ 0&I \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{A_{11}}I + 0\cdot0}&{{A_{11}}Y + 0\cdot I}\\ {X{A_{11}}I + S\cdot0}&{X{A_{11}}Y + SI} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{11}}Y}\\ {X{A_{11}}}&{X{A_{11}}Y + S} \end{array}} \right] \end{array}\] Thus we get: \[ \left[ \begin{array} { l l } { A _ { 11 } } & { A _ { 12 } } \\ { A _ { 21 } } & { A _ { 22 } } \end{array} \right] = \left[ \begin{array} { c c } { A _ { 11 } } & { A _ { 11 } Y } \\ { X A _ { 11 } } & { X A _ { 11 } Y + S } \end{array} \right] \] Upon equating the elements, we get: \[ \begin{array} { l } { A _ { 12 } = A _ { 11 } Y } \\ { A _ { 21 } = X A _ { 11 } } \end{array} \]
Step-3: Solve the Equations
From $A _ { 21 } = X A _ { 11 }$,
\[\begin{array}{l} {A_{21}} = X{A_{11}}\\ {A_{21}}A_{11}^{ - 1} = \left( {X{A_{11}}} \right)A_{11}^{ - 1}\\ {A_{21}}A_{11}^{ - 1} = X\left( {{A_{11}}A_{11}^{ - 1}} \right)\\ {A_{21}}A_{11}^{ - 1} = XI\\ {A_1}A_{11}^{ - 1} = X \end{array}\] From $A _ { 12 } = A _ { 11 } Y$
\[\begin{array}{l} {A_{12}} = {A_{11}}Y\\ A_{11}^{ - 1}{A_{12}} = A_{11}^{ - 1}\left( {{A_{11}}Y} \right)\\ A_{11}^{ - 1}{A_{12}} = \left( {A_{11}^{ - 1}{A_{11}}} \right)Y\\ A_{11}^{ - 1}{A_{12}} = IY\\ A_{11}^{ - 1}{A_{12}} = Y \end{array}\] Therefore,
\[ \begin{array} { l } { X = A _ { 21 } A _ { 11 } ^ { - 1 } } \\ { Y = A _ { 11 } ^ { - 1 } A _ { 12 } } \end{array} \]
We are given that $A_{11}$ is invertible. We have to find X and Y such that \[ \left[ \begin{array} { c c } { A _ { 11 } } & { A _ { 12 } } \\ { A _ { 21 } } & { A _ { 22 } } \end{array} \right] = \left[ \begin{array} { c c } { I } & { 0 } \\ { X } & { I } \end{array} \right] \left[ \begin{array} { c c } { A _ { 11 } } & { 0 } \\ { 0 } & { S } \end{array} \right] \left[ \begin{array} { c c } { I } & { Y } \\ { 0 } & { I } \end{array} \right] \]
Step-1:
Using the matrix properties \[\begin{array}{*{20}{l}} {I\cdot A = A}\\ {0\cdot A = 0}\\ {A + 0 = A} \end{array}\] The product of matrices on right hand side: \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} I&0\\ X&I \end{array}} \right]\left[ {\begin{array}{*{20}{r}} {{A_{11}}}&0\\ 0&S \end{array}} \right]\left[ {\begin{array}{*{20}{l}} I&Y\\ 0&I \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {I{A_{11}} + 0\cdot0}&{I\cdot0 + 0\cdot S}\\ {X{A_{11}} + I\cdot0}&{X\cdot0 + IS} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} I&Y\\ 0&I \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{A_{11}}}&0\\ {X{A_{11}}}&S \end{array}} \right]\left[ {\begin{array}{*{20}{l}} I&Y\\ 0&I \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{A_{11}}I + 0\cdot0}&{{A_{11}}Y + 0\cdot I}\\ {X{A_{11}}I + S\cdot0}&{X{A_{11}}Y + SI} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{11}}Y}\\ {X{A_{11}}}&{X{A_{11}}Y + S} \end{array}} \right] \end{array}\] Thus we get: \[ \left[ \begin{array} { l l } { A _ { 11 } } & { A _ { 12 } } \\ { A _ { 21 } } & { A _ { 22 } } \end{array} \right] = \left[ \begin{array} { c c } { A _ { 11 } } & { A _ { 11 } Y } \\ { X A _ { 11 } } & { X A _ { 11 } Y + S } \end{array} \right] \] Upon equating the elements, we get: \[ \begin{array} { l } { A _ { 12 } = A _ { 11 } Y } \\ { A _ { 21 } = X A _ { 11 } } \end{array} \]
Step-3: Solve the Equations
From $A _ { 21 } = X A _ { 11 }$,
\[\begin{array}{l} {A_{21}} = X{A_{11}}\\ {A_{21}}A_{11}^{ - 1} = \left( {X{A_{11}}} \right)A_{11}^{ - 1}\\ {A_{21}}A_{11}^{ - 1} = X\left( {{A_{11}}A_{11}^{ - 1}} \right)\\ {A_{21}}A_{11}^{ - 1} = XI\\ {A_1}A_{11}^{ - 1} = X \end{array}\] From $A _ { 12 } = A _ { 11 } Y$
\[\begin{array}{l} {A_{12}} = {A_{11}}Y\\ A_{11}^{ - 1}{A_{12}} = A_{11}^{ - 1}\left( {{A_{11}}Y} \right)\\ A_{11}^{ - 1}{A_{12}} = \left( {A_{11}^{ - 1}{A_{11}}} \right)Y\\ A_{11}^{ - 1}{A_{12}} = IY\\ A_{11}^{ - 1}{A_{12}} = Y \end{array}\] Therefore,
\[ \begin{array} { l } { X = A _ { 21 } A _ { 11 } ^ { - 1 } } \\ { Y = A _ { 11 } ^ { - 1 } A _ { 12 } } \end{array} \]