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 34E from Chapter 7.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 34E
Chapter:
Problem:
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Step-by-Step Solution
Given Information
We are given with following matrix:\[A = \left[ \begin{array} { r r r } { 3 } & { - 2 } & { 4 } \\ { - 2 } & { 6 } & { 2 } \\ { 4 } & { 2 } & { 3 } \end{array} \right]\]We have to construct spectral decomposition of A
Step-1:
Find the eigenvalues using the characteristic Polynomial:\[\begin{array}{l}|A - \lambda I| = 0\\\left| {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 2}&4\\{ - 2}&{6 - \lambda }&2\\4&2&{3 - \lambda }\end{array}} \right| = 0\\(3 - \lambda )[(6 - \lambda )(3 - \lambda ) - 4] + 2[ - 2(3 - \lambda ) - 8] + 4[ - 4 - 4(6 - \lambda )] = 0\\ - {\lambda ^3} + 12{\lambda ^2} - 21\lambda - 98 = 0\\ - {(\lambda - 7)^2}(\lambda + 2) = 0\\\lambda = 7,7, - 2\end{array}\]
Step-2:
Eigenvector for $\lambda = 7$\[\begin{array} { l } { ( A - 7 I ) \mathbf { v } = \mathbf { 0 } } \\ { \left( \begin{array} { c c c } { - 4 } & { - 2 } & { 4 } \\ { - 2 } & { - 1 } & { 2 } \\ { 4 } & { 2 } & { - 4 } \end{array} \right) \left( \begin{array} { l } { \mathbf { x } _ { 1 } } \\ { \mathbf { x } _ { 2 } } \\ { \mathbf { x } _ { 3 } } \end{array} \right) = \left( \begin{array} { l } { 0 } \\ { 0 } \\ { 0 } \end{array} \right) } \end{array}\]So, the eigenvectors are:\[\mathbf { v } _ { 1 } = \left( \begin{array} { l } { 1 } \\ { 0 } \\ { 1 } \end{array} \right) , \mathbf { v } _ { 2 } = \left( \begin{array} { r } { - \dfrac { 1 } { 2 } } \\ { 1 } \\ { 0 } \end{array} \right)\]
Step-3:
The component of $v_2$ orthogonal to $v_1$ is:\[\begin{array} { l } { \mathbf { v } _ { 2 } - \dfrac { \mathbf { v } _ { 2 } \cdot \mathbf { v } _ { 1 } } { \mathbf { v } _ { 1 } \cdot \mathbf { v } _ { 1 } } \mathbf { v } _ { 1 } } \\ { \left[ - \dfrac { 1 } { 4 } \right] } \\ { = \left[ \begin{array} { c } { 1 } \\ { 1 } \\ { \dfrac { 1 } { 4 } } \end{array} \right] } \end{array}\]
Step-4:
Eigenvector for $\lambda = -2$\[\begin{array} { l } { ( A + 2 I ) \mathbf { v } = \mathbf { 0 } } \\ { \left( \begin{array} { c c c } { 5 } & { - 2 } & { 4 } \\ { - 2 } & { 8 } & { 2 } \\ { 4 } & { 2 } & { 5 } \end{array} \right) \left( \begin{array} { l } { \mathbf { x } _ { 1 } } \\ { \mathbf { x } _ { 2 } } \\ { \mathbf { x } _ { 3 } } \end{array} \right) = \left( \begin{array} { l } { 0 } \\ { 0 } \\ { 0 } \end{array} \right) } \end{array}\]Thus, the eigenvector is:\[\mathbf { v } _ { 3 } = \left( \begin{array} { r } { - 1 } \\ { - \dfrac { 1 } { 2 } } \\ { 1 } \end{array} \right)\]
Step-5:
Normalize the vectors:\[\begin{aligned} \mathbf { u } _ { 1 } & = \dfrac { 1 } { \left\| \mathbf { v } _ { 1 } \right\| } \mathbf { v } _ { 1 } \\ & = \left( \dfrac { 1 } { \sqrt { 2 } } \right) \\ & = \left( \dfrac { 1 } { \sqrt { 2 } } \right) \end{aligned}\]\[\begin{aligned} \mathbf { u } _ { 2 } & = \dfrac { 1 } { \left\| \mathbf { v } _ { 2 } \right\| } \mathbf { v } _ { 2 } \\ & = \left( \begin{array} { c } { - \dfrac { 1 } { \sqrt { 28 } } } \\ { \dfrac { 4 } { \sqrt { 18 } } } \\ { \dfrac { 2 } { \sqrt { 18 } } } \end{array} \right) \end{aligned}\]\[\begin{aligned} \mathbf { u } _ { 3 } & = \dfrac { 1 } { \left\| \mathbf { v } _ { 3 } \right\| } \mathbf { v } _ { 3 } \\ & = \dfrac { 1 } { \sqrt { 3 } } \mathbf { v } _ { 3 } \\ & = \left( \begin{array} { r } { - \dfrac { 2 } { 3 } } \\ { - \dfrac { 1 } { 3 } } \\ { - \dfrac { 1 } { 3 } } \\ { \dfrac { 2 } { 3 } } \end{array} \right) \end{aligned}\]
Step-6:
The matrix P formed by eigenvectors is:\[P = \left[ \begin{array} { l l l } { \mathbf { u } _ { 1 } } & { \mathbf { u } _ { 2 } } & { \mathbf { u } _ { 3 } } \end{array} \right] = \left[ \begin{array} { c c c } { \dfrac { 1 } { \sqrt { 2 } } } & { - \dfrac { 1 } { \sqrt { 18 } } } & { - \dfrac { 2 } { 3 } } \\ { 0 } & { \dfrac { 4 } { \sqrt { 18 } } } & { - \dfrac { 1 } { 3 } } \\ { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 18 } } } & { \dfrac { 2 } { 3 } } \end{array} \right]\]The matrix D formed by eigenvalues is:\[D = \left[ \begin{array} { c c c } { 7 } & { 0 } & { 0 } \\ { 0 } & { 7 } & { 0 } \\ { 0 } & { 0 } & { - 2 } \end{array} \right]\]
Step-7:
The spectral decomposition of A is:\[\begin{array}{l}A = {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + {\lambda _2}{{\bf{u}}_2}{\bf{u}}_2^T + {\lambda _3}{{\bf{u}}_3}{\bf{u}}_3^T\\ = 7{{\bf{u}}_1}B_1^T + 7{{\bf{u}}_2}{\bf{u}}_2^T - 2{{\bf{u}}_3}{\bf{u}}_3^T\\ = 7\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{{\sqrt 2 }}}\\0\\{\dfrac{1}{{\sqrt 2 }}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{{\sqrt 2 }}}&0&{\dfrac{1}{{\sqrt 2 }}}\end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}}{ - \dfrac{1}{{\sqrt {18} }}}\\{\dfrac{4}{{\sqrt {18} }}}\\{\dfrac{1}{{\sqrt {18} }}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\dfrac{{ - 1}}{{\sqrt {18} }}}&{\dfrac{4}{{\sqrt {18} }}}&{\dfrac{1}{{\sqrt {18} }}}\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}{ - \dfrac{2}{3}}\\{ - \dfrac{1}{3}}\\{\dfrac{2}{3}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - \dfrac{2}{3}}&{ - \dfrac{1}{3}}&{\dfrac{2}{3}}\end{array}} \right]\\7\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{2}}&0&{\dfrac{1}{2}}\\0&0&0\\{\dfrac{1}{2}}&0&{\dfrac{1}{2}}\end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{{18}}}&{ - \dfrac{4}{{18}}}&{ - \dfrac{1}{{18}}}\\{ - \dfrac{4}{{18}}}&{\dfrac{{16}}{{18}}}&{\dfrac{4}{{18}}}\\{ - \dfrac{1}{{18}}}&{\dfrac{4}{{18}}}&{\dfrac{1}{{18}}}\end{array}} \right] - 2\left[ {\begin{array}{*{20}{r}}{\dfrac{4}{9}}&{\dfrac{2}{9}}&{ - \dfrac{4}{9}}\\{\dfrac{2}{9}}&{\dfrac{1}{9}}&{ - \dfrac{2}{9}}\\{ - \dfrac{4}{9}}&{ - \dfrac{2}{9}}&{\dfrac{4}{9}}\end{array}} \right]\end{array}\]The spectral decomposition is:
\[A = 7\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{2}}&0&{\dfrac{1}{2}}\\0&0&0\\{\dfrac{1}{2}}&0&{\dfrac{1}{2}}\end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{{18}}}&{ - \dfrac{4}{{18}}}&{ - \dfrac{1}{{18}}}\\{ - \dfrac{4}{{18}}}&{\dfrac{{16}}{{18}}}&{\dfrac{4}{{18}}}\\{ - \dfrac{1}{{18}}}&{\dfrac{4}{{18}}}&{\dfrac{1}{{18}}}\end{array}} \right] - 2\left[ {\begin{array}{*{20}{r}}{\dfrac{4}{9}}&{\dfrac{2}{9}}&{ - \dfrac{4}{9}}\\{\dfrac{2}{9}}&{\dfrac{1}{9}}&{ - \dfrac{2}{9}}\\{ - \dfrac{4}{9}}&{ - \dfrac{2}{9}}&{\dfrac{4}{9}}\end{array}} \right]\]
We are given with following matrix:\[A = \left[ \begin{array} { r r r } { 3 } & { - 2 } & { 4 } \\ { - 2 } & { 6 } & { 2 } \\ { 4 } & { 2 } & { 3 } \end{array} \right]\]We have to construct spectral decomposition of A
Step-1:
Find the eigenvalues using the characteristic Polynomial:\[\begin{array}{l}|A - \lambda I| = 0\\\left| {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 2}&4\\{ - 2}&{6 - \lambda }&2\\4&2&{3 - \lambda }\end{array}} \right| = 0\\(3 - \lambda )[(6 - \lambda )(3 - \lambda ) - 4] + 2[ - 2(3 - \lambda ) - 8] + 4[ - 4 - 4(6 - \lambda )] = 0\\ - {\lambda ^3} + 12{\lambda ^2} - 21\lambda - 98 = 0\\ - {(\lambda - 7)^2}(\lambda + 2) = 0\\\lambda = 7,7, - 2\end{array}\]
Step-2:
Eigenvector for $\lambda = 7$\[\begin{array} { l } { ( A - 7 I ) \mathbf { v } = \mathbf { 0 } } \\ { \left( \begin{array} { c c c } { - 4 } & { - 2 } & { 4 } \\ { - 2 } & { - 1 } & { 2 } \\ { 4 } & { 2 } & { - 4 } \end{array} \right) \left( \begin{array} { l } { \mathbf { x } _ { 1 } } \\ { \mathbf { x } _ { 2 } } \\ { \mathbf { x } _ { 3 } } \end{array} \right) = \left( \begin{array} { l } { 0 } \\ { 0 } \\ { 0 } \end{array} \right) } \end{array}\]So, the eigenvectors are:\[\mathbf { v } _ { 1 } = \left( \begin{array} { l } { 1 } \\ { 0 } \\ { 1 } \end{array} \right) , \mathbf { v } _ { 2 } = \left( \begin{array} { r } { - \dfrac { 1 } { 2 } } \\ { 1 } \\ { 0 } \end{array} \right)\]
Step-3:
The component of $v_2$ orthogonal to $v_1$ is:\[\begin{array} { l } { \mathbf { v } _ { 2 } - \dfrac { \mathbf { v } _ { 2 } \cdot \mathbf { v } _ { 1 } } { \mathbf { v } _ { 1 } \cdot \mathbf { v } _ { 1 } } \mathbf { v } _ { 1 } } \\ { \left[ - \dfrac { 1 } { 4 } \right] } \\ { = \left[ \begin{array} { c } { 1 } \\ { 1 } \\ { \dfrac { 1 } { 4 } } \end{array} \right] } \end{array}\]
Step-4:
Eigenvector for $\lambda = -2$\[\begin{array} { l } { ( A + 2 I ) \mathbf { v } = \mathbf { 0 } } \\ { \left( \begin{array} { c c c } { 5 } & { - 2 } & { 4 } \\ { - 2 } & { 8 } & { 2 } \\ { 4 } & { 2 } & { 5 } \end{array} \right) \left( \begin{array} { l } { \mathbf { x } _ { 1 } } \\ { \mathbf { x } _ { 2 } } \\ { \mathbf { x } _ { 3 } } \end{array} \right) = \left( \begin{array} { l } { 0 } \\ { 0 } \\ { 0 } \end{array} \right) } \end{array}\]Thus, the eigenvector is:\[\mathbf { v } _ { 3 } = \left( \begin{array} { r } { - 1 } \\ { - \dfrac { 1 } { 2 } } \\ { 1 } \end{array} \right)\]
Step-5:
Normalize the vectors:\[\begin{aligned} \mathbf { u } _ { 1 } & = \dfrac { 1 } { \left\| \mathbf { v } _ { 1 } \right\| } \mathbf { v } _ { 1 } \\ & = \left( \dfrac { 1 } { \sqrt { 2 } } \right) \\ & = \left( \dfrac { 1 } { \sqrt { 2 } } \right) \end{aligned}\]\[\begin{aligned} \mathbf { u } _ { 2 } & = \dfrac { 1 } { \left\| \mathbf { v } _ { 2 } \right\| } \mathbf { v } _ { 2 } \\ & = \left( \begin{array} { c } { - \dfrac { 1 } { \sqrt { 28 } } } \\ { \dfrac { 4 } { \sqrt { 18 } } } \\ { \dfrac { 2 } { \sqrt { 18 } } } \end{array} \right) \end{aligned}\]\[\begin{aligned} \mathbf { u } _ { 3 } & = \dfrac { 1 } { \left\| \mathbf { v } _ { 3 } \right\| } \mathbf { v } _ { 3 } \\ & = \dfrac { 1 } { \sqrt { 3 } } \mathbf { v } _ { 3 } \\ & = \left( \begin{array} { r } { - \dfrac { 2 } { 3 } } \\ { - \dfrac { 1 } { 3 } } \\ { - \dfrac { 1 } { 3 } } \\ { \dfrac { 2 } { 3 } } \end{array} \right) \end{aligned}\]
Step-6:
The matrix P formed by eigenvectors is:\[P = \left[ \begin{array} { l l l } { \mathbf { u } _ { 1 } } & { \mathbf { u } _ { 2 } } & { \mathbf { u } _ { 3 } } \end{array} \right] = \left[ \begin{array} { c c c } { \dfrac { 1 } { \sqrt { 2 } } } & { - \dfrac { 1 } { \sqrt { 18 } } } & { - \dfrac { 2 } { 3 } } \\ { 0 } & { \dfrac { 4 } { \sqrt { 18 } } } & { - \dfrac { 1 } { 3 } } \\ { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 18 } } } & { \dfrac { 2 } { 3 } } \end{array} \right]\]The matrix D formed by eigenvalues is:\[D = \left[ \begin{array} { c c c } { 7 } & { 0 } & { 0 } \\ { 0 } & { 7 } & { 0 } \\ { 0 } & { 0 } & { - 2 } \end{array} \right]\]
Step-7:
The spectral decomposition of A is:\[\begin{array}{l}A = {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + {\lambda _2}{{\bf{u}}_2}{\bf{u}}_2^T + {\lambda _3}{{\bf{u}}_3}{\bf{u}}_3^T\\ = 7{{\bf{u}}_1}B_1^T + 7{{\bf{u}}_2}{\bf{u}}_2^T - 2{{\bf{u}}_3}{\bf{u}}_3^T\\ = 7\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{{\sqrt 2 }}}\\0\\{\dfrac{1}{{\sqrt 2 }}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{{\sqrt 2 }}}&0&{\dfrac{1}{{\sqrt 2 }}}\end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}}{ - \dfrac{1}{{\sqrt {18} }}}\\{\dfrac{4}{{\sqrt {18} }}}\\{\dfrac{1}{{\sqrt {18} }}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\dfrac{{ - 1}}{{\sqrt {18} }}}&{\dfrac{4}{{\sqrt {18} }}}&{\dfrac{1}{{\sqrt {18} }}}\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}{ - \dfrac{2}{3}}\\{ - \dfrac{1}{3}}\\{\dfrac{2}{3}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - \dfrac{2}{3}}&{ - \dfrac{1}{3}}&{\dfrac{2}{3}}\end{array}} \right]\\7\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{2}}&0&{\dfrac{1}{2}}\\0&0&0\\{\dfrac{1}{2}}&0&{\dfrac{1}{2}}\end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{{18}}}&{ - \dfrac{4}{{18}}}&{ - \dfrac{1}{{18}}}\\{ - \dfrac{4}{{18}}}&{\dfrac{{16}}{{18}}}&{\dfrac{4}{{18}}}\\{ - \dfrac{1}{{18}}}&{\dfrac{4}{{18}}}&{\dfrac{1}{{18}}}\end{array}} \right] - 2\left[ {\begin{array}{*{20}{r}}{\dfrac{4}{9}}&{\dfrac{2}{9}}&{ - \dfrac{4}{9}}\\{\dfrac{2}{9}}&{\dfrac{1}{9}}&{ - \dfrac{2}{9}}\\{ - \dfrac{4}{9}}&{ - \dfrac{2}{9}}&{\dfrac{4}{9}}\end{array}} \right]\end{array}\]The spectral decomposition is:
\[A = 7\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{2}}&0&{\dfrac{1}{2}}\\0&0&0\\{\dfrac{1}{2}}&0&{\dfrac{1}{2}}\end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{{18}}}&{ - \dfrac{4}{{18}}}&{ - \dfrac{1}{{18}}}\\{ - \dfrac{4}{{18}}}&{\dfrac{{16}}{{18}}}&{\dfrac{4}{{18}}}\\{ - \dfrac{1}{{18}}}&{\dfrac{4}{{18}}}&{\dfrac{1}{{18}}}\end{array}} \right] - 2\left[ {\begin{array}{*{20}{r}}{\dfrac{4}{9}}&{\dfrac{2}{9}}&{ - \dfrac{4}{9}}\\{\dfrac{2}{9}}&{\dfrac{1}{9}}&{ - \dfrac{2}{9}}\\{ - \dfrac{4}{9}}&{ - \dfrac{2}{9}}&{\dfrac{4}{9}}\end{array}} \right]\]