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 7E from Chapter 7.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 7E
Chapter:
Problem:
Find a unit vector x in R3 at which Q(x) is maximized, subject to xTx = 1. [Hint: The eigenvalues of the matrix of the quadratic form Q are 2, –1, and –4]
Step-by-Step Solution
Given Information
We are given with following equation: \[ Q ( \mathbf { x } ) = - 2 x _ { 1 } ^ { 2 } - x _ { 2 } ^ { 2 } + 4 x _ { 1 } x _ { 2 } + 4 x _ { 2 } x _ { 3 } \] We have to find the unit vector for the maximized subject to the constraint $\mathbf { x } ^ { T } \mathbf { x } = 1$.
Step-1: Find the eigenvalues
\[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} { - 2}&2&0\\ 2&{ - 1}&2\\ 0&2&0 \end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] = 0\\ \left| {\begin{array}{*{20}{r}} { - 2 - \lambda }&2&0\\ 2&{ - 1 - \lambda }&2\\ 0&2&{ - \lambda } \end{array}} \right| = 0\\ \left. {\begin{array}{*{20}{l}} {( - 2 - \lambda )[( - 1 - \lambda )( - \lambda ) - (2)(2)]}\\ { - 2[2( - \lambda ) - 0(2)] + 0[2(2) - 0( - 1 - \lambda )]} \end{array}} \right\} = 0\\ {\lambda ^3} + 3{\lambda ^2} - 6\lambda - 8 = 0 \end{array}\] Solve the characteristic equation: \[\begin{array}{l} {\lambda ^3} + 3{\lambda ^2} - 6\lambda - 8 = 0\\ (\lambda + 4)(\lambda + 1)(\lambda - 2) = 0\\ \lambda = - 4, - 1,2 \end{array}\] The maximum value of Q(x) subject to constraint $\mathbf { x } ^ { T } A \mathbf { x } = 1$ is given by the largest eigenvalue. That is 2
Step-2: (b) Eigenvectors
Eigenvector for $\lambda = 2$ \[\left[ {\begin{array}{*{20}{c}} { - 4}&2&0\\ 2&{ - 3}&2\\ 0&2&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{k_1}}\\ {{k_2}}\\ {{k_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\] Row-reduce the matrix: \[ \begin{aligned} \left[ \begin{array} { c c c } { - 4 } & { 2 } & { 0 } \\ { 2 } & { - 3 } & { 2 } \\ { 0 } & { 2 } & { - 2 } \end{array} \right] & - \left[ \begin{array} { c c c } { - 4 } & { 2 } & { 0 } \\ { 0 } & { - 2 } & { 2 } \\ { 0 } & { 0 } & { 0 } \end{array} \right] \\ & \sim \left[ \begin{array} { r r r } { - 2 } & { 1 } & { 0 } \\ { 0 } & { - 1 } & { 1 } \\ { 0 } & { 0 } & { 0 } \end{array} \right] \end{aligned} \] The general solution: \[x = \left[ {\begin{array}{*{20}{c}} { \dfrac{t}{2}}\\ { t}\\ t \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} { \dfrac{1}{2}}\\ { 1}\\ 1 \end{array}} \right]\] Thus the eigenvector is: \[ \mathbf { v } = \left[ \begin{array} { c } { 1 / 2 } \\ { 1 } \\ { 1 } \end{array} \right] \]
Step-3:
The unit vector is: \[ \begin{aligned} \mathrm { x } & = \dfrac { 1 } { \sqrt { ( 1 / 2 ) ^ { 2 } + 1 ^ { 2 } + 1 ^ { 2 } } } \left[ \begin{array} { c } { 1 / 2 } \\ { 1 } \\ { 1 } \end{array} \right] \\ & = \dfrac { 1 } { \sqrt { 9 / 4 } } \left[ \begin{array} { c } { 1 } \\ { 1 } \end{array} \right] \\ & = \left[ \begin{array} { c } { 1 / 3 } \\ { 2 / 3 } \\ { 2 / 3 } \end{array} \right] \end{aligned} \] The unit vector for the maximum value found in part (a) is:
\[ \left[ \begin{array} { c } { 1 / 3 } \\ { 2 / 3 } \\ { 2 / 3 } \end{array} \right] \]
We are given with following equation: \[ Q ( \mathbf { x } ) = - 2 x _ { 1 } ^ { 2 } - x _ { 2 } ^ { 2 } + 4 x _ { 1 } x _ { 2 } + 4 x _ { 2 } x _ { 3 } \] We have to find the unit vector for the maximized subject to the constraint $\mathbf { x } ^ { T } \mathbf { x } = 1$.
Step-1: Find the eigenvalues
\[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} { - 2}&2&0\\ 2&{ - 1}&2\\ 0&2&0 \end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] = 0\\ \left| {\begin{array}{*{20}{r}} { - 2 - \lambda }&2&0\\ 2&{ - 1 - \lambda }&2\\ 0&2&{ - \lambda } \end{array}} \right| = 0\\ \left. {\begin{array}{*{20}{l}} {( - 2 - \lambda )[( - 1 - \lambda )( - \lambda ) - (2)(2)]}\\ { - 2[2( - \lambda ) - 0(2)] + 0[2(2) - 0( - 1 - \lambda )]} \end{array}} \right\} = 0\\ {\lambda ^3} + 3{\lambda ^2} - 6\lambda - 8 = 0 \end{array}\] Solve the characteristic equation: \[\begin{array}{l} {\lambda ^3} + 3{\lambda ^2} - 6\lambda - 8 = 0\\ (\lambda + 4)(\lambda + 1)(\lambda - 2) = 0\\ \lambda = - 4, - 1,2 \end{array}\] The maximum value of Q(x) subject to constraint $\mathbf { x } ^ { T } A \mathbf { x } = 1$ is given by the largest eigenvalue. That is 2
Step-2: (b) Eigenvectors
Eigenvector for $\lambda = 2$ \[\left[ {\begin{array}{*{20}{c}} { - 4}&2&0\\ 2&{ - 3}&2\\ 0&2&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{k_1}}\\ {{k_2}}\\ {{k_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\] Row-reduce the matrix: \[ \begin{aligned} \left[ \begin{array} { c c c } { - 4 } & { 2 } & { 0 } \\ { 2 } & { - 3 } & { 2 } \\ { 0 } & { 2 } & { - 2 } \end{array} \right] & - \left[ \begin{array} { c c c } { - 4 } & { 2 } & { 0 } \\ { 0 } & { - 2 } & { 2 } \\ { 0 } & { 0 } & { 0 } \end{array} \right] \\ & \sim \left[ \begin{array} { r r r } { - 2 } & { 1 } & { 0 } \\ { 0 } & { - 1 } & { 1 } \\ { 0 } & { 0 } & { 0 } \end{array} \right] \end{aligned} \] The general solution: \[x = \left[ {\begin{array}{*{20}{c}} { \dfrac{t}{2}}\\ { t}\\ t \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} { \dfrac{1}{2}}\\ { 1}\\ 1 \end{array}} \right]\] Thus the eigenvector is: \[ \mathbf { v } = \left[ \begin{array} { c } { 1 / 2 } \\ { 1 } \\ { 1 } \end{array} \right] \]
Step-3:
The unit vector is: \[ \begin{aligned} \mathrm { x } & = \dfrac { 1 } { \sqrt { ( 1 / 2 ) ^ { 2 } + 1 ^ { 2 } + 1 ^ { 2 } } } \left[ \begin{array} { c } { 1 / 2 } \\ { 1 } \\ { 1 } \end{array} \right] \\ & = \dfrac { 1 } { \sqrt { 9 / 4 } } \left[ \begin{array} { c } { 1 } \\ { 1 } \end{array} \right] \\ & = \left[ \begin{array} { c } { 1 / 3 } \\ { 2 / 3 } \\ { 2 / 3 } \end{array} \right] \end{aligned} \] The unit vector for the maximum value found in part (a) is:
\[ \left[ \begin{array} { c } { 1 / 3 } \\ { 2 / 3 } \\ { 2 / 3 } \end{array} \right] \]