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 19E from Chapter 7.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 19E
Chapter:
Problem:
Orthogonally diagonalize the matrices in Exercises 13–22, giving an orthogonal matrix P and a diagonal matrix D. To save you time, the eigenvalues in Exercises 17–22 are:
Step-by-Step Solution
Given Information
We are given with a matrix: \[ A = \left[ \begin{array} { c c c } { 3 } & { - 2 } & { 4 } \\ { - 2 } & { 6 } & { 2 } \\ { 4 } & { 2 } & { 3 } \end{array} \right] \] We have to orthogonally diagonalize the matrix if the eigenvalues are 7 and -2
Step-1: Eigenvector for eigenvalue of 7
The matrix form: \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {3 - \lambda }&{ - 2}&4\\ { - 2}&{6 - \lambda }&2\\ 4&2&{3 - \lambda } \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} {3 - 7}&{ - 2}&4\\ { - 2}&{6 - 7}&2\\ 4&2&{3 - 7} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} { - 4}&{ - 2}&4\\ { - 2}&{ - 1}&2\\ 4&2&{ - 4} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right] \end{array}\] The row-reduced augmented matrix \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} { - 4}&{ - 2}&4&0\\ { - 2}&{ - 1}&2&0\\ 4&2&{ - 4}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{1/2}&{ - 1}&0\\ 1&{1/2}&{ - 1}&0\\ 1&{1/2}&{ - 1}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_1} \times ( - 1/4) \to {R_1}}\\ {{R_2} \times ( - 1/2) \to {R_2}}\\ {{R_3} \times (1/4) \to {R_3}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{1/2}&{ - 1}&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} - {R_1} \to {R_2}}\\ {{R_3} - {R_1} \to {R_3}} \end{array}} \right\} \end{array}\] The system of equations: \[ x _ { 1 } + \dfrac { 1 } { 2 } x _ { 2 } - x _ { 3 } = 0 \] The general solution: \[\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - s + t}\\ {2s}\\ t \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - s}\\ {2s}\\ 0 \end{array}} \right] + \left[ {\begin{array}{*{20}{l}} t\\ 0\\ t \end{array}} \right] = s\left[ {\begin{array}{*{20}{c}} { - 1}\\ 2\\ 0 \end{array}} \right] + t\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 1 \end{array}} \right]\] Therefore, the eigenvectors are:
\[\left\{ {\left[ {\begin{array}{*{20}{c}} { - 1}\\ 2\\ 0 \end{array}} \right],\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 1 \end{array}} \right]} \right\}\]
Step-2: Eigenvector for eigenvalue of -2
The matrix form: \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} {3 - \lambda }&{ - 2}&4\\ { - 2}&{6 - \lambda }&2\\ 4&2&{3 - \lambda } \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {3 - ( - 2)}&{ - 2}&4\\ { - 2}&{6 - ( - 2)}&2\\ 4&2&{3 - ( - 2)} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5&{ - 2}&4\\ { - 2}&8&2\\ 4&2&5 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right] \end{array}\] The row-reduced augmented matrix \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 5&{ - 2}&4&0\\ { - 2}&8&2&0\\ 4&2&5&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2/5}&{4/5}&0\\ 1&{ - 4}&{ - 1}&0\\ 1&{1/2}&{5/4}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{c}} {{R_1} \times (1/5) \to {R_1}}\\ {{R_2} \times ( - 1/2) \to {R_2}}\\ {{R_3} \times (1/4) \to {R_3}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2/5}&{4/5}&0\\ 0&{ - 18/5}&{ - 9/5}&0\\ 0&{9/10}&{9/20}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} - {R_1} \to {R_2}}\\ {{R_3} - 4{R_1} \to {R_3}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2/5}&{4/5}&0\\ 0&1&{1/2}&0\\ 0&1&{1/2}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \times ( - 5/18) \to {R_2}}\\ {{R_3} \times (10/9) \to {R_3}} \end{array}} \right\} \end{array}\] The system of equations: \[ x _ { 1 } - ( 2 / 5 ) x _ { 2 } + ( 4 / 5 ) x _ { 3 } = 0 , x _ { 2 } + ( 1 / 2 ) x _ { 3 } = 0 \] The general solution: \[\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2t}\\ { - t}\\ {2t} \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 1}\\ 2 \end{array}} \right]\] Therefore, the eigenvector is:
\[ \left[ \begin{array} { l } { - 2 } \\ { - 1 } \\ { 2 } \end{array} \right] \]
Step-3: Orthogonal basis for first eigenvector
\[\begin{array}{l} {{\bf{u}}_1} = \dfrac{{{{\bf{v}}_1}}}{{\left\| {{{\bf{v}}_1}} \right\|}}\\ = \dfrac{1}{{\sqrt {1 + 0 + 1} }}\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 1 \end{array}} \right]\\ = \dfrac{{\sqrt 2 }}{2}\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {\sqrt 2 /2}\\ 0\\ {\sqrt 2 /2} \end{array}} \right] \end{array}\]
Step-4: Orthogonal basis for second eigenvector
\[\begin{array}{l} {{\bf{u}}_2} = \dfrac{{{{\bf{z}}_2}}}{{\left\| {{{\bf{z}}_2}} \right\|}}\\ = \dfrac{1}{{\sqrt {\dfrac{1}{4} + 4 + \dfrac{1}{4}} }}\left[ {\begin{array}{*{20}{l}} { - 1/2}\\ 2\\ {1/2} \end{array}} \right]\\ = \dfrac{{2\sqrt 2 }}{6}\left[ {\begin{array}{*{20}{l}} { - 1/2}\\ 2\\ {1/2} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - \sqrt 2 /6}\\ {4\sqrt 2 /6}\\ {\sqrt 2 /6} \end{array}} \right] \end{array}\]
Step-5: Orthogonal basis for third eigenvector
\[\begin{array}{l} {{\bf{u}}_3} = \dfrac{{{{\bf{v}}_3}}}{{\left\| {{{\bf{v}}_3}} \right\|}}\\ = \dfrac{1}{{\sqrt {4 + 1 + 4} }}\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 1}\\ 2 \end{array}} \right]\\ = \dfrac{1}{3}\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 1}\\ 2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} { - 2/3}\\ { - 1/3}\\ {2/3} \end{array}} \right] \end{array}\]
Step-7: Orthogonal Matrix
Required matrix $P$ \[\begin{array}{l} P = \left[ {{{\bf{u}}_1}\,\,{{\bf{u}}_2}\,\,\,{{\bf{u}}_3}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {\sqrt 2 /2}&{ - \sqrt 2 /6}&{ - 2/3}\\ 0&{2\sqrt 2 /3}&{ - 1/3}\\ {\sqrt 2 /2}&{\sqrt 2 /6}&{2/3} \end{array}} \right] \end{array}\] The required diagonal matrix:
\[D = \left[ {\begin{array}{*{20}{r}} 7&0&0\\ 0&7&0\\ 0&0&{ - 2} \end{array}} \right]\]
We are given with a matrix: \[ A = \left[ \begin{array} { c c c } { 3 } & { - 2 } & { 4 } \\ { - 2 } & { 6 } & { 2 } \\ { 4 } & { 2 } & { 3 } \end{array} \right] \] We have to orthogonally diagonalize the matrix if the eigenvalues are 7 and -2
Step-1: Eigenvector for eigenvalue of 7
The matrix form: \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {3 - \lambda }&{ - 2}&4\\ { - 2}&{6 - \lambda }&2\\ 4&2&{3 - \lambda } \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} {3 - 7}&{ - 2}&4\\ { - 2}&{6 - 7}&2\\ 4&2&{3 - 7} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} { - 4}&{ - 2}&4\\ { - 2}&{ - 1}&2\\ 4&2&{ - 4} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right] \end{array}\] The row-reduced augmented matrix \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} { - 4}&{ - 2}&4&0\\ { - 2}&{ - 1}&2&0\\ 4&2&{ - 4}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{1/2}&{ - 1}&0\\ 1&{1/2}&{ - 1}&0\\ 1&{1/2}&{ - 1}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_1} \times ( - 1/4) \to {R_1}}\\ {{R_2} \times ( - 1/2) \to {R_2}}\\ {{R_3} \times (1/4) \to {R_3}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{1/2}&{ - 1}&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} - {R_1} \to {R_2}}\\ {{R_3} - {R_1} \to {R_3}} \end{array}} \right\} \end{array}\] The system of equations: \[ x _ { 1 } + \dfrac { 1 } { 2 } x _ { 2 } - x _ { 3 } = 0 \] The general solution: \[\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - s + t}\\ {2s}\\ t \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - s}\\ {2s}\\ 0 \end{array}} \right] + \left[ {\begin{array}{*{20}{l}} t\\ 0\\ t \end{array}} \right] = s\left[ {\begin{array}{*{20}{c}} { - 1}\\ 2\\ 0 \end{array}} \right] + t\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 1 \end{array}} \right]\] Therefore, the eigenvectors are:
\[\left\{ {\left[ {\begin{array}{*{20}{c}} { - 1}\\ 2\\ 0 \end{array}} \right],\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 1 \end{array}} \right]} \right\}\]
Step-2: Eigenvector for eigenvalue of -2
The matrix form: \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} {3 - \lambda }&{ - 2}&4\\ { - 2}&{6 - \lambda }&2\\ 4&2&{3 - \lambda } \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {3 - ( - 2)}&{ - 2}&4\\ { - 2}&{6 - ( - 2)}&2\\ 4&2&{3 - ( - 2)} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5&{ - 2}&4\\ { - 2}&8&2\\ 4&2&5 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right] \end{array}\] The row-reduced augmented matrix \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 5&{ - 2}&4&0\\ { - 2}&8&2&0\\ 4&2&5&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2/5}&{4/5}&0\\ 1&{ - 4}&{ - 1}&0\\ 1&{1/2}&{5/4}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{c}} {{R_1} \times (1/5) \to {R_1}}\\ {{R_2} \times ( - 1/2) \to {R_2}}\\ {{R_3} \times (1/4) \to {R_3}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2/5}&{4/5}&0\\ 0&{ - 18/5}&{ - 9/5}&0\\ 0&{9/10}&{9/20}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} - {R_1} \to {R_2}}\\ {{R_3} - 4{R_1} \to {R_3}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2/5}&{4/5}&0\\ 0&1&{1/2}&0\\ 0&1&{1/2}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \times ( - 5/18) \to {R_2}}\\ {{R_3} \times (10/9) \to {R_3}} \end{array}} \right\} \end{array}\] The system of equations: \[ x _ { 1 } - ( 2 / 5 ) x _ { 2 } + ( 4 / 5 ) x _ { 3 } = 0 , x _ { 2 } + ( 1 / 2 ) x _ { 3 } = 0 \] The general solution: \[\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2t}\\ { - t}\\ {2t} \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 1}\\ 2 \end{array}} \right]\] Therefore, the eigenvector is:
\[ \left[ \begin{array} { l } { - 2 } \\ { - 1 } \\ { 2 } \end{array} \right] \]
Step-3: Orthogonal basis for first eigenvector
\[\begin{array}{l} {{\bf{u}}_1} = \dfrac{{{{\bf{v}}_1}}}{{\left\| {{{\bf{v}}_1}} \right\|}}\\ = \dfrac{1}{{\sqrt {1 + 0 + 1} }}\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 1 \end{array}} \right]\\ = \dfrac{{\sqrt 2 }}{2}\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {\sqrt 2 /2}\\ 0\\ {\sqrt 2 /2} \end{array}} \right] \end{array}\]
Step-4: Orthogonal basis for second eigenvector
\[\begin{array}{l} {{\bf{u}}_2} = \dfrac{{{{\bf{z}}_2}}}{{\left\| {{{\bf{z}}_2}} \right\|}}\\ = \dfrac{1}{{\sqrt {\dfrac{1}{4} + 4 + \dfrac{1}{4}} }}\left[ {\begin{array}{*{20}{l}} { - 1/2}\\ 2\\ {1/2} \end{array}} \right]\\ = \dfrac{{2\sqrt 2 }}{6}\left[ {\begin{array}{*{20}{l}} { - 1/2}\\ 2\\ {1/2} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - \sqrt 2 /6}\\ {4\sqrt 2 /6}\\ {\sqrt 2 /6} \end{array}} \right] \end{array}\]
Step-5: Orthogonal basis for third eigenvector
\[\begin{array}{l} {{\bf{u}}_3} = \dfrac{{{{\bf{v}}_3}}}{{\left\| {{{\bf{v}}_3}} \right\|}}\\ = \dfrac{1}{{\sqrt {4 + 1 + 4} }}\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 1}\\ 2 \end{array}} \right]\\ = \dfrac{1}{3}\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 1}\\ 2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} { - 2/3}\\ { - 1/3}\\ {2/3} \end{array}} \right] \end{array}\]
Step-7: Orthogonal Matrix
Required matrix $P$ \[\begin{array}{l} P = \left[ {{{\bf{u}}_1}\,\,{{\bf{u}}_2}\,\,\,{{\bf{u}}_3}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {\sqrt 2 /2}&{ - \sqrt 2 /6}&{ - 2/3}\\ 0&{2\sqrt 2 /3}&{ - 1/3}\\ {\sqrt 2 /2}&{\sqrt 2 /6}&{2/3} \end{array}} \right] \end{array}\] The required diagonal matrix:
\[D = \left[ {\begin{array}{*{20}{r}} 7&0&0\\ 0&7&0\\ 0&0&{ - 2} \end{array}} \right]\]