Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
We have solutions for your book!
See our solution for Question 20E from Chapter 7.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 20E
Chapter:
Problem:
0
Step-by-Step Solution
Given Information
We are given with matrix: \[ A = \left[ \begin{array} { c c c } { 5 } & { 8 } & { - 4 } \\ { 8 } & { 5 } & { - 4 } \\ { - 4 } & { - 4 } & { - 1 } \end{array} \right] \] We have to orthogonally diagonalize the matrix given the eigenvalues are -3 and 15
Step-1: Eigenvector for -3
The matrix Equation: \[\begin{array}{l} A - ( - 3)I = A + 3I\\ = \left[ {\begin{array}{*{20}{c}} 5&8&{ - 4}\\ 8&5&{ - 4}\\ { - 4}&{ - 4}&{ - 1} \end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5&8&{ - 4}\\ 8&5&{ - 4}\\ { - 4}&{ - 4}&{ - 1} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3&0&0\\ 0&3&0\\ 0&0&3 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {5 + 3}&{8 + 0}&{ - 4 + 0}\\ {8 + 0}&{5 + 3}&{ - 4 + 0}\\ { - 4 + 0}&{ - 4 + 0}&{ - 1 + 3} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 8&8&{ - 4}\\ 8&8&{ - 4}\\ { - 4}&{ - 4}&2 \end{array}} \right] \end{array}\] Row-reduce the matrix: \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} 8&8&{ - 4}&0\\ 8&8&{ - 4}&0\\ { - 4}&{ - 4}&2&0 \end{array}} \right]\~\left[ {\begin{array}{*{20}{c}} 1&1&{ - 1/2}&0\\ 1&1&{ - 1/2}&0\\ 1&1&{ - 1/2}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_1} \to (1/8){R_1}}\\ {{R_2} \to (1/8){R_2}}\\ {{R_3} \to ( - 1/4){R_3}} \end{array}} \right\}\\ \left[ {\begin{array}{*{20}{r}} 1&1&{ - 1/2}&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \to {R_2} - {R_1}}\\ {{R_3} \to {R_3} - {R_1}} \end{array}} \right\} \end{array}\] The general solution is: \[ \begin{array} { l } { \mathbf { x } = \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right] } \\ { = \left[ \begin{array} { c } { - s + t } \\ { s } \\ { 2 t } \end{array} \right] } \\ { = s \left[ \begin{array} { c } { - 1 } \\ { 1 } \\ { 0 } \end{array} \right] + t \left[ \begin{array} { l } { 1 } \\ { 0 } \\ { 2 } \end{array} \right] } \end{array} \] Thus, the eigenvectors are: \[v = \left\{ {\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right],\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 2 \end{array}} \right]} \right\}\]
Step-2: Eigenvector for $\lambda = 15$
\[\begin{array}{l} A - 15I = \left[ {\begin{array}{*{20}{c}} 5&8&{ - 4}\\ 8&5&{ - 4}\\ { - 4}&{ - 4}&{ - 1} \end{array}} \right] - 15\left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5&8&{ - 4}\\ 8&5&{ - 4}\\ { - 4}&{ - 4}&{ - 1} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {15}&0&0\\ 0&{15}&0\\ 0&0&{15} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {5 - 15}&{8 - 0}&{ - 4 - 0}\\ {8 - 0}&{5 - 15}&{ - 4 - 0}\\ { - 4 - 0}&{ - 4 - 0}&{ - 1 - 15} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 10}&8&{ - 4}\\ 8&{ - 10}&{ - 4}\\ { - 4}&{ - 4}&{ - 16} \end{array}} \right] \end{array}\] On row-reducing the matrix: \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} { - 10}&8&{ - 4}&0\\ 8&{ - 10}&{ - 4}&0\\ { - 4}&{ - 4}&{ - 16}&0 \end{array}} \right]\~\left[ {\begin{array}{*{20}{c}} { - 10}&8&{ - 4}&0\\ 8&{ - 10}&{ - 4}&0\\ { - 4}&{ - 4}&{ - 16}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_1} \to (1/2){R_1}}\\ {{R_2} \to (1/2){R_2}}\\ {{R_3} \to ( - 1/4){R_3}} \end{array}} \right\}\\ \left[ {\begin{array}{*{20}{c}} { - 5}&4&{ - 2}&0\\ 4&{ - 5}&{ - 2}&0\\ 1&1&4&0 \end{array}} \right]::\left\{ {{R_1} \leftrightarrow {R_3}} \right\}\\ \left[ {\begin{array}{*{20}{c}} 1&1&4&0\\ 4&{ - 5}&{ - 2}&0\\ { - 5}&4&{ - 2}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{c}} {{R_2} \to {R_2} + ( - 4){R_1}}\\ {{R_3} \to {R_3} + (5){R_1}} \end{array}} \right\}\\ \left[ {\begin{array}{*{20}{c}} 1&1&4&0\\ 0&{ - 9}&{ - 18}&0\\ 0&9&{18}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \to ( - 1/9){R_2}}\\ {{R_3} \to (1/9){R_3}} \end{array}} \right\}\\ \left[ {\begin{array}{*{20}{l}} 1&1&4&0\\ 0&1&2&0\\ 0&1&2&0 \end{array}} \right]::\left\{ {{R_1} \to {R_1} + ( - 1){R_2}} \right\}\\ \left[ {\begin{array}{*{20}{l}} 1&0&2&0\\ 0&1&2&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {{R_3} \to {R_3} + ( - 1){R_2}} \right\} \end{array}\] Thus general solution: \[{\bf{x}} = \left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2t}\\ { - 2t}\\ t \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 2}\\ 1 \end{array}} \right]\] Hence, \[{v_3} = \left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 2}\\ 1 \end{array}} \right]\]
Step-4:
Projection of $v_2$ onto $v_1$: \[\begin{array}{l} \dfrac{{{{\bf{v}}_2}\cdot{{\bf{v}}_1}}}{{{{\bf{v}}_1}\cdot{{\bf{v}}_1}}}{{\bf{v}}_1} = \dfrac{{\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 2 \end{array}} \right]\cdot\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right]}}{{{{( - 1)}^2} + {1^2} + 0}}\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right]\\ = \dfrac{{ - 1}}{2}\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right] \end{array}\] The component of $v_2$ orthogonal to $v_1$ \[\begin{array}{l} {\bf{Z}} = {{\bf{V}}_2} - \dfrac{{{{\bf{V}}_2}\cdot{{\bf{V}}_1}}}{{{{\bf{V}}_1}\cdot{{\bf{v}}_1}}}{{\bf{v}}_1}\\ = \left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 2 \end{array}} \right] - \left( {\dfrac{{ - 1}}{2}\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right]} \right)\\ = \left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 2 \end{array}} \right] + \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 2 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { - 1/2}\\ {1/2}\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right] \end{array}\]
Step-5
Normalize the vector $v_1$: \[ \begin{aligned} \mathbf { u } _ { 1 } & = \dfrac { \mathbf { v } _ { 1 } } { \left\| \mathbf { v } _ { 1 } \right\| } \\ & = \dfrac { 1 } { \sqrt { ( - 1 ) ^ { 2 } + 1 ^ { 2 } + 0 } } \left[ \begin{array} { c } { - 1 } \\ { 1 } \\ { 0 } \end{array} \right] \\ & = \dfrac { 1 } { \sqrt { 2 } } \left[ \begin{array} { c } { - 1 / \sqrt { 2 } } \\ { 0 } \end{array} \right] \\ & = \left[ \begin{array} { c } { - 1 / \sqrt { 2 } } \\ { 1 / \sqrt { 2 } } \\ { 0 } \end{array} \right] \end{aligned} \] Normalize the vector $v_2$: \[\begin{array}{l} {{\bf{u}}_2} = \dfrac{{\bf{z}}}{{{\bf{z}}}}\\ = \dfrac{1}{{\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^2} + {2^2}} }}\left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right]\\ = \dfrac{1}{{\sqrt {\dfrac{{18}}{4}} }}\left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right]\\ = \dfrac{1}{{\sqrt {\dfrac{9}{2}} }}\left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right]\\ = \dfrac{1}{{\dfrac{3}{{\sqrt 2 }}}}\left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right]\\ = \dfrac{{\sqrt 2 }}{3}\left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {\dfrac{1}{6}\sqrt 2 }\\ {\dfrac{1}{6}\sqrt 2 }\\ {\dfrac{2}{3}\sqrt 2 } \end{array}} \right] \end{array}\] Normalize the vector $v_3$: \[\begin{array}{l} {{\bf{u}}_3} = \dfrac{{{{\bf{v}}_3}}}{{\left| {\left| {{{\bf{v}}_3}} \right|} \right|}}\\ = \dfrac{1}{{\sqrt {{{( - 2)}^2} + {{( - 2)}^2} + {1^2}} }}\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 2}\\ 1 \end{array}} \right]\\ = \dfrac{1}{{\sqrt 9 }}\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 2}\\ 1 \end{array}} \right]\\ = \dfrac{1}{3}\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 2}\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 2/3}\\ { - 2/3}\\ {1/3} \end{array}} \right] \end{array}\]
Step-6: The matrix P
\[ \begin{aligned} P & = \left[ \begin{array} { c c c } { \mathbf { u } _ { 1 } } & { \mathbf { u } _ { 2 } } & { \mathbf { u } _ { 3 } } \end{array} \right] \\ & = \left[ \begin{array} { c c c c } { - \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { 6 } \sqrt { 2 } } & { - \dfrac { 2 } { 3 } } \\ { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { 6 } \sqrt { 2 } } & { - \dfrac { 2 } { 3 } } \\ { 0 } & { \dfrac { 2 } { 3 } \sqrt { 2 } } & { \dfrac { 1 } { 3 } } \end{array} \right] \end{aligned} \]
Step-7: The Diagonal matrix D
\[\begin{array}{l} D = {P^T}AP\\ = \left[ {\begin{array}{*{20}{c}} { - \dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{{\sqrt 2 }}}&0\\ {\dfrac{1}{6}\sqrt 2 }&{\dfrac{1}{6}\sqrt 2 }&{\dfrac{2}{3}\sqrt 2 }\\ { - \dfrac{2}{3}}&{ - \dfrac{2}{3}}&{\dfrac{1}{3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&8&{ - 4}\\ 8&5&{ - 4}\\ { - 4}&{ - 4}&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - \dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{6}\sqrt 2 }&{ - \dfrac{2}{3}}\\ {\dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{6}\sqrt 2 }&{ - \dfrac{2}{3}}\\ 0&{\dfrac{2}{3}\sqrt 2 }&{\dfrac{1}{3}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - \dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{{\sqrt 2 }}}&0\\ {\dfrac{1}{6}\sqrt 2 }&{\dfrac{1}{6}\sqrt 2 }&{\dfrac{2}{3}\sqrt 2 }\\ { - \dfrac{2}{3}}&{ - \dfrac{2}{3}}&{\dfrac{1}{3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\dfrac{3}{2}\sqrt 2 }&{ - \dfrac{1}{2}\sqrt 2 }&{ - 10}\\ { - \dfrac{3}{2}\sqrt 2 }&{ - \dfrac{1}{2}\sqrt 2 }&{ - 10}\\ 0&{ - 2\sqrt 2 }&5 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{r}} { - 3}&0&0\\ 0&{ - 3}&0\\ 0&0&{15} \end{array}} \right] \end{array}\]
The matrix A is orthogonally diagonalizable with \[ P = \left[ \begin{array} { c c c } { - \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { 6 } \sqrt { 2 } } & { - \dfrac { 2 } { 3 } } \\ { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { 6 } \sqrt { 2 } } & { - \dfrac { 2 } { 3 } } \\ { 0 } & { \dfrac { 2 } { 3 } \sqrt { 2 } } & { \dfrac { 1 } { 3 } } \end{array} \right] \]
We are given with matrix: \[ A = \left[ \begin{array} { c c c } { 5 } & { 8 } & { - 4 } \\ { 8 } & { 5 } & { - 4 } \\ { - 4 } & { - 4 } & { - 1 } \end{array} \right] \] We have to orthogonally diagonalize the matrix given the eigenvalues are -3 and 15
Step-1: Eigenvector for -3
The matrix Equation: \[\begin{array}{l} A - ( - 3)I = A + 3I\\ = \left[ {\begin{array}{*{20}{c}} 5&8&{ - 4}\\ 8&5&{ - 4}\\ { - 4}&{ - 4}&{ - 1} \end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5&8&{ - 4}\\ 8&5&{ - 4}\\ { - 4}&{ - 4}&{ - 1} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3&0&0\\ 0&3&0\\ 0&0&3 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {5 + 3}&{8 + 0}&{ - 4 + 0}\\ {8 + 0}&{5 + 3}&{ - 4 + 0}\\ { - 4 + 0}&{ - 4 + 0}&{ - 1 + 3} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 8&8&{ - 4}\\ 8&8&{ - 4}\\ { - 4}&{ - 4}&2 \end{array}} \right] \end{array}\] Row-reduce the matrix: \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} 8&8&{ - 4}&0\\ 8&8&{ - 4}&0\\ { - 4}&{ - 4}&2&0 \end{array}} \right]\~\left[ {\begin{array}{*{20}{c}} 1&1&{ - 1/2}&0\\ 1&1&{ - 1/2}&0\\ 1&1&{ - 1/2}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_1} \to (1/8){R_1}}\\ {{R_2} \to (1/8){R_2}}\\ {{R_3} \to ( - 1/4){R_3}} \end{array}} \right\}\\ \left[ {\begin{array}{*{20}{r}} 1&1&{ - 1/2}&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \to {R_2} - {R_1}}\\ {{R_3} \to {R_3} - {R_1}} \end{array}} \right\} \end{array}\] The general solution is: \[ \begin{array} { l } { \mathbf { x } = \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right] } \\ { = \left[ \begin{array} { c } { - s + t } \\ { s } \\ { 2 t } \end{array} \right] } \\ { = s \left[ \begin{array} { c } { - 1 } \\ { 1 } \\ { 0 } \end{array} \right] + t \left[ \begin{array} { l } { 1 } \\ { 0 } \\ { 2 } \end{array} \right] } \end{array} \] Thus, the eigenvectors are: \[v = \left\{ {\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right],\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 2 \end{array}} \right]} \right\}\]
Step-2: Eigenvector for $\lambda = 15$
\[\begin{array}{l} A - 15I = \left[ {\begin{array}{*{20}{c}} 5&8&{ - 4}\\ 8&5&{ - 4}\\ { - 4}&{ - 4}&{ - 1} \end{array}} \right] - 15\left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5&8&{ - 4}\\ 8&5&{ - 4}\\ { - 4}&{ - 4}&{ - 1} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {15}&0&0\\ 0&{15}&0\\ 0&0&{15} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {5 - 15}&{8 - 0}&{ - 4 - 0}\\ {8 - 0}&{5 - 15}&{ - 4 - 0}\\ { - 4 - 0}&{ - 4 - 0}&{ - 1 - 15} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 10}&8&{ - 4}\\ 8&{ - 10}&{ - 4}\\ { - 4}&{ - 4}&{ - 16} \end{array}} \right] \end{array}\] On row-reducing the matrix: \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} { - 10}&8&{ - 4}&0\\ 8&{ - 10}&{ - 4}&0\\ { - 4}&{ - 4}&{ - 16}&0 \end{array}} \right]\~\left[ {\begin{array}{*{20}{c}} { - 10}&8&{ - 4}&0\\ 8&{ - 10}&{ - 4}&0\\ { - 4}&{ - 4}&{ - 16}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_1} \to (1/2){R_1}}\\ {{R_2} \to (1/2){R_2}}\\ {{R_3} \to ( - 1/4){R_3}} \end{array}} \right\}\\ \left[ {\begin{array}{*{20}{c}} { - 5}&4&{ - 2}&0\\ 4&{ - 5}&{ - 2}&0\\ 1&1&4&0 \end{array}} \right]::\left\{ {{R_1} \leftrightarrow {R_3}} \right\}\\ \left[ {\begin{array}{*{20}{c}} 1&1&4&0\\ 4&{ - 5}&{ - 2}&0\\ { - 5}&4&{ - 2}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{c}} {{R_2} \to {R_2} + ( - 4){R_1}}\\ {{R_3} \to {R_3} + (5){R_1}} \end{array}} \right\}\\ \left[ {\begin{array}{*{20}{c}} 1&1&4&0\\ 0&{ - 9}&{ - 18}&0\\ 0&9&{18}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \to ( - 1/9){R_2}}\\ {{R_3} \to (1/9){R_3}} \end{array}} \right\}\\ \left[ {\begin{array}{*{20}{l}} 1&1&4&0\\ 0&1&2&0\\ 0&1&2&0 \end{array}} \right]::\left\{ {{R_1} \to {R_1} + ( - 1){R_2}} \right\}\\ \left[ {\begin{array}{*{20}{l}} 1&0&2&0\\ 0&1&2&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {{R_3} \to {R_3} + ( - 1){R_2}} \right\} \end{array}\] Thus general solution: \[{\bf{x}} = \left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2t}\\ { - 2t}\\ t \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 2}\\ 1 \end{array}} \right]\] Hence, \[{v_3} = \left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 2}\\ 1 \end{array}} \right]\]
Step-4:
Projection of $v_2$ onto $v_1$: \[\begin{array}{l} \dfrac{{{{\bf{v}}_2}\cdot{{\bf{v}}_1}}}{{{{\bf{v}}_1}\cdot{{\bf{v}}_1}}}{{\bf{v}}_1} = \dfrac{{\left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 2 \end{array}} \right]\cdot\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right]}}{{{{( - 1)}^2} + {1^2} + 0}}\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right]\\ = \dfrac{{ - 1}}{2}\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right] \end{array}\] The component of $v_2$ orthogonal to $v_1$ \[\begin{array}{l} {\bf{Z}} = {{\bf{V}}_2} - \dfrac{{{{\bf{V}}_2}\cdot{{\bf{V}}_1}}}{{{{\bf{V}}_1}\cdot{{\bf{v}}_1}}}{{\bf{v}}_1}\\ = \left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 2 \end{array}} \right] - \left( {\dfrac{{ - 1}}{2}\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right]} \right)\\ = \left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 2 \end{array}} \right] + \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 1\\ 0\\ 2 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { - 1/2}\\ {1/2}\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right] \end{array}\]
Step-5
Normalize the vector $v_1$: \[ \begin{aligned} \mathbf { u } _ { 1 } & = \dfrac { \mathbf { v } _ { 1 } } { \left\| \mathbf { v } _ { 1 } \right\| } \\ & = \dfrac { 1 } { \sqrt { ( - 1 ) ^ { 2 } + 1 ^ { 2 } + 0 } } \left[ \begin{array} { c } { - 1 } \\ { 1 } \\ { 0 } \end{array} \right] \\ & = \dfrac { 1 } { \sqrt { 2 } } \left[ \begin{array} { c } { - 1 / \sqrt { 2 } } \\ { 0 } \end{array} \right] \\ & = \left[ \begin{array} { c } { - 1 / \sqrt { 2 } } \\ { 1 / \sqrt { 2 } } \\ { 0 } \end{array} \right] \end{aligned} \] Normalize the vector $v_2$: \[\begin{array}{l} {{\bf{u}}_2} = \dfrac{{\bf{z}}}{{{\bf{z}}}}\\ = \dfrac{1}{{\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^2} + {2^2}} }}\left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right]\\ = \dfrac{1}{{\sqrt {\dfrac{{18}}{4}} }}\left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right]\\ = \dfrac{1}{{\sqrt {\dfrac{9}{2}} }}\left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right]\\ = \dfrac{1}{{\dfrac{3}{{\sqrt 2 }}}}\left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right]\\ = \dfrac{{\sqrt 2 }}{3}\left[ {\begin{array}{*{20}{c}} {1/2}\\ {1/2}\\ 2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {\dfrac{1}{6}\sqrt 2 }\\ {\dfrac{1}{6}\sqrt 2 }\\ {\dfrac{2}{3}\sqrt 2 } \end{array}} \right] \end{array}\] Normalize the vector $v_3$: \[\begin{array}{l} {{\bf{u}}_3} = \dfrac{{{{\bf{v}}_3}}}{{\left| {\left| {{{\bf{v}}_3}} \right|} \right|}}\\ = \dfrac{1}{{\sqrt {{{( - 2)}^2} + {{( - 2)}^2} + {1^2}} }}\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 2}\\ 1 \end{array}} \right]\\ = \dfrac{1}{{\sqrt 9 }}\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 2}\\ 1 \end{array}} \right]\\ = \dfrac{1}{3}\left[ {\begin{array}{*{20}{c}} { - 2}\\ { - 2}\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 2/3}\\ { - 2/3}\\ {1/3} \end{array}} \right] \end{array}\]
Step-6: The matrix P
\[ \begin{aligned} P & = \left[ \begin{array} { c c c } { \mathbf { u } _ { 1 } } & { \mathbf { u } _ { 2 } } & { \mathbf { u } _ { 3 } } \end{array} \right] \\ & = \left[ \begin{array} { c c c c } { - \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { 6 } \sqrt { 2 } } & { - \dfrac { 2 } { 3 } } \\ { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { 6 } \sqrt { 2 } } & { - \dfrac { 2 } { 3 } } \\ { 0 } & { \dfrac { 2 } { 3 } \sqrt { 2 } } & { \dfrac { 1 } { 3 } } \end{array} \right] \end{aligned} \]
Step-7: The Diagonal matrix D
\[\begin{array}{l} D = {P^T}AP\\ = \left[ {\begin{array}{*{20}{c}} { - \dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{{\sqrt 2 }}}&0\\ {\dfrac{1}{6}\sqrt 2 }&{\dfrac{1}{6}\sqrt 2 }&{\dfrac{2}{3}\sqrt 2 }\\ { - \dfrac{2}{3}}&{ - \dfrac{2}{3}}&{\dfrac{1}{3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&8&{ - 4}\\ 8&5&{ - 4}\\ { - 4}&{ - 4}&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - \dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{6}\sqrt 2 }&{ - \dfrac{2}{3}}\\ {\dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{6}\sqrt 2 }&{ - \dfrac{2}{3}}\\ 0&{\dfrac{2}{3}\sqrt 2 }&{\dfrac{1}{3}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - \dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{{\sqrt 2 }}}&0\\ {\dfrac{1}{6}\sqrt 2 }&{\dfrac{1}{6}\sqrt 2 }&{\dfrac{2}{3}\sqrt 2 }\\ { - \dfrac{2}{3}}&{ - \dfrac{2}{3}}&{\dfrac{1}{3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\dfrac{3}{2}\sqrt 2 }&{ - \dfrac{1}{2}\sqrt 2 }&{ - 10}\\ { - \dfrac{3}{2}\sqrt 2 }&{ - \dfrac{1}{2}\sqrt 2 }&{ - 10}\\ 0&{ - 2\sqrt 2 }&5 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{r}} { - 3}&0&0\\ 0&{ - 3}&0\\ 0&0&{15} \end{array}} \right] \end{array}\]
The matrix A is orthogonally diagonalizable with \[ P = \left[ \begin{array} { c c c } { - \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { 6 } \sqrt { 2 } } & { - \dfrac { 2 } { 3 } } \\ { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { 6 } \sqrt { 2 } } & { - \dfrac { 2 } { 3 } } \\ { 0 } & { \dfrac { 2 } { 3 } \sqrt { 2 } } & { \dfrac { 1 } { 3 } } \end{array} \right] \]