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 23E from Chapter 7.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 23E
Chapter:
Problem:
Let Verify that 5 is an eigenvalue of A and v is an eigenvector. Then orthogo¬nally diagonalize A.
Step-by-Step Solution
Given Information
We are given with matrix and vector: \[ A = \left[ \begin{array} { r r r } { 4 } & { - 1 } & { - 1 } \\ { - 1 } & { 4 } & { - 1 } \\ { - 1 } & { - 1 } & { 4 } \end{array} \right] \text { and } \mathbf { v } = \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \end{array} \right] \] We have to verify that 5 is an eigenvalue of A and v is an eigenvector. Then we also have to orthogonally diagonalize A.
Step-1: Eigenvalues
Check if 5 is an eigenvalue: \[\begin{array}{l} |A - 5\cdot I| = \left| {\begin{array}{*{20}{c}} {4 - 5}&{ - 1}&{ - 1}\\ { - 1}&{4 - 5}&{ - 1}\\ { - 1}&{ - 1}&{4 - 5} \end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}} { - 1}&{ - 1}&{ - 1}\\ { - 1}&{ - 1}&{ - 1}\\ { - 1}&{ - 1}&{ - 1} \end{array}} \right|\\ = 0 \end{array}\] Hence, 5 is an eigenvalue of matrix A.
Step-2: Check for eigenvector
Compute the product of A and v \[\begin{array}{l} A{\bf{v}} = \left[ {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 1}\\ { - 1}&4&{ - 1}\\ { - 1}&{ - 1}&4 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} 1\\ 1\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {4\cdot1 + ( - 1)\cdot1 + ( - 1)\cdot1}\\ {( - 1)\cdot1 + 4\cdot1 + ( - 1)\cdot1}\\ {( - 1)\cdot1 + ( - 1)\cdot1 + 4\cdot1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {4 - 1 - 1}\\ { - 1 + 4 - 1}\\ { - 1 - 1 + 4} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 2\\ 2\\ 2 \end{array}} \right]\\ = 2\left[ {\begin{array}{*{20}{l}} 1\\ 1\\ 1 \end{array}} \right]\\ = 2{\bf{v}}\\ \\ \end{array}\] Hence, v is an eigenvector of the matrix with eigenvalue 2 \\
Step-3: Third eigenvalue
Sum of eigenvalues is equal to trace of the matrix: So, \[\begin{array}{l} 5 + 2 + {\lambda _3} = 4 + 4 + 4\\ 7 + {\lambda _3} = 12\\ {\lambda _3} = 12 - 7\\ {\lambda _3} = 5 \end{array}\]
Step-4 Eigenvector for $\lambda = 5$:
\[\begin{array}{l} (A - 5\cdot I){{\bf{v}}_1} = 0\\ \left( {\begin{array}{*{20}{c}} {4 - 5}&{ - 1}&{ - 1}\\ { - 1}&{4 - 5}&{ - 1}\\ { - 1}&{ - 1}&{4 - 5} \end{array}} \right)\left( {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right)\\ \left( {\begin{array}{*{20}{c}} { - 1}&{ - 1}&{ - 1}\\ { - 1}&{ - 1}&{ - 1}\\ { - 1}&{ - 1}&{ - 1} \end{array}} \right)\left( {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right) \end{array}\] Row-reduce the matrix: \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} {{{\bf{A}}_1}}&{\bf{0}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1}&{ - 1}&0\\ { - 1}&{ - 1}&{ - 1}&0\\ { - 1}&{ - 1}&{ - 1}&0 \end{array}} \right]\\ \approx \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1}&{ - 1}&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \to {R_2} - {R_1}}\\ {{R_3} \to {R_3} - {R_1}} \end{array}} \right\}\\ \approx \left[ {\begin{array}{*{20}{c}} 1&1&1&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {{R_1} \to ( - 1){R_1}} \right\} \end{array}\] So, the general solution is \[ \begin{aligned} \left( \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right) & = \left( \begin{array} { c } { - s - t } \\ { s } \\ { t } \end{array} \right) \\ & = ( - s ) \left( \begin{array} { c } { 1 } \\ { - 1 } \\ { 0 } \end{array} \right) + ( - t ) \left( \begin{array} { c } { 1 } \\ { 0 } \\ { - 1 } \end{array} \right) \end{aligned} \] Therefore, the eigenvectors are: \[\begin{array}{l} {v_1} = \left[ {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right]\\ {v_2} = \left[ {\begin{array}{*{20}{c}} 1\\ 0\\ { - 1} \end{array}} \right] \end{array}\]
Step-5 Projection:
The projection of $v_2$ on $v_1$: \[ \begin{aligned} \mathbf { z } _ { 2 } & = \mathbf { v } _ { 2 } - \dfrac { \mathbf { v } _ { 2 } \cdot \mathbf { v } _ { 1 } } { \mathbf { v } _ { 1 } \cdot \mathbf { v } _ { 1 } } \mathbf { v } _ { 1 } \\ & = \left[ \begin{array} { c } { 1 } \\ { 0 } \\ { 0 } \\ { - 1 } \end{array} \right] - \dfrac { 1 } { 2 } \left[ \begin{array} { c } { 1 } \\ { - 1 } \\ { 0 } \end{array} \right] \\ & = \left[ \begin{array} { c } { 1 } \\ { 0 } \\ { - 1 } \end{array} \right] - \dfrac { 1 } { 2 } \left[ \begin{array} { c } { 1 } \\ { - 1 } \\ { 0 } \end{array} \right] \\ & = \left[ \begin{array} { c } { 1 / 2 } \\ { 1 / 2 } \\ { - 1 } \end{array} \right] \end{aligned} \]
Step-6: Normalize the vectors:
\[ \begin{aligned} \mathbf { u } _ { 1 } & = \dfrac { \mathbf { v } _ { 1 } } { \left\| \mathbf { v } _ { 1 } \right\| } \\ & = \left( \dfrac { 1 } { \sqrt { 2 } } , - \dfrac { 1 } { \sqrt { 2 } } , 0 \right) ^ { \mathrm { T } } \end{aligned} \] \[ \begin{aligned} \mathbf { u } _ { 2 } & = \dfrac { \mathbf { Z } _ { 2 } } { \left\| \mathbf { z } _ { 2 } \right\| } \\ & = \left( \dfrac { 1 } { \sqrt { 6 } } , \dfrac { 1 } { \sqrt { 6 } } , - \dfrac { 2 } { \sqrt { 6 } } \right) ^ { \mathrm { T } } \end{aligned} \] \[ \begin{aligned} \mathbf { u } _ { 3 } & = \dfrac { \mathbf { v } _ { 3 } } { \left\| \mathbf { v } _ { 3 } \right\| } \\ & = \left( \dfrac { 1 } { \sqrt { 3 } } , \dfrac { 1 } { \sqrt { 3 } } , \dfrac { 1 } { \sqrt { 3 } } \right) ^ { \mathrm { T } } \end{aligned} \]
Step-7: The matrix Q
Use the normalized vectors to form Q \[ Q = \left[ \begin{array} { c c c } { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \\ { - \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \\ { 0 } & { - \dfrac { 2 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \end{array} \right] \] Inverse of Q: \[ Q ^ { - 1 } = \left[ \begin{array} { c c c } { \dfrac { 1 } { \sqrt { 2 } } } & { - \dfrac { 1 } { \sqrt { 2 } } } & { 0 } \\ { \dfrac { 1 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 6 } } } & { - \dfrac { 2 } { \sqrt { 6 } } } \\ { \dfrac { 1 } { \sqrt { 3 } } } & { \dfrac { 1 } { \sqrt { 3 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \end{array} \right] \]
Step-8: The Diagonal matrix D
\[\begin{array}{l} {Q^{ - 1}}AQ = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{\sqrt 2 }}}&{ - \dfrac{1}{{\sqrt 2 }}}&0\\ {\dfrac{1}{{\sqrt 6 }}}&{\dfrac{1}{{\sqrt 6 }}}&{ - \dfrac{2}{{\sqrt 6 }}}\\ {\dfrac{1}{{\sqrt 3 }}}&{\dfrac{1}{{\sqrt 3 }}}&{\dfrac{1}{{\sqrt 3 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 1}\\ { - 1}&4&{ - 1}\\ { - 1}&{ - 1}&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{{\sqrt 6 }}}&{\dfrac{1}{{\sqrt 3 }}}\\ { - \dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{{\sqrt 6 }}}&{\dfrac{1}{{\sqrt 3 }}}\\ 0&{ - \dfrac{2}{{\sqrt 6 }}}&{\dfrac{1}{{\sqrt 3 }}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{\sqrt 2 }}}&{ - \dfrac{1}{{\sqrt 2 }}}&0\\ {\dfrac{1}{{\sqrt 6 }}}&{\dfrac{1}{{\sqrt 6 }}}&{ - \dfrac{2}{{\sqrt 6 }}}\\ {\dfrac{1}{{\sqrt 3 }}}&{\dfrac{1}{{\sqrt 3 }}}&{\dfrac{1}{{\sqrt 3 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\dfrac{5}{{\sqrt 2 }}}&{\dfrac{5}{{\sqrt 6 }}}&{\dfrac{2}{{\sqrt 3 }}}\\ { - \dfrac{5}{{\sqrt 2 }}}&{\dfrac{5}{{\sqrt 6 }}}&{\dfrac{2}{{\sqrt 3 }}}\\ 0&{ - \dfrac{{10}}{{\sqrt 6 }}}&{\dfrac{2}{{\sqrt 3 }}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 5&0&0\\ 0&5&0\\ 0&0&2 \end{array}} \right]\\ = D \end{array}\]
The matrix A is orthogonally diagonalizable with \[ Q = \left[ \begin{array} { c c c } { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \\ { - \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \\ { 0 } & { - \dfrac { 2 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \end{array} \right] \]
We are given with matrix and vector: \[ A = \left[ \begin{array} { r r r } { 4 } & { - 1 } & { - 1 } \\ { - 1 } & { 4 } & { - 1 } \\ { - 1 } & { - 1 } & { 4 } \end{array} \right] \text { and } \mathbf { v } = \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \end{array} \right] \] We have to verify that 5 is an eigenvalue of A and v is an eigenvector. Then we also have to orthogonally diagonalize A.
Step-1: Eigenvalues
Check if 5 is an eigenvalue: \[\begin{array}{l} |A - 5\cdot I| = \left| {\begin{array}{*{20}{c}} {4 - 5}&{ - 1}&{ - 1}\\ { - 1}&{4 - 5}&{ - 1}\\ { - 1}&{ - 1}&{4 - 5} \end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}} { - 1}&{ - 1}&{ - 1}\\ { - 1}&{ - 1}&{ - 1}\\ { - 1}&{ - 1}&{ - 1} \end{array}} \right|\\ = 0 \end{array}\] Hence, 5 is an eigenvalue of matrix A.
Step-2: Check for eigenvector
Compute the product of A and v \[\begin{array}{l} A{\bf{v}} = \left[ {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 1}\\ { - 1}&4&{ - 1}\\ { - 1}&{ - 1}&4 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} 1\\ 1\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {4\cdot1 + ( - 1)\cdot1 + ( - 1)\cdot1}\\ {( - 1)\cdot1 + 4\cdot1 + ( - 1)\cdot1}\\ {( - 1)\cdot1 + ( - 1)\cdot1 + 4\cdot1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {4 - 1 - 1}\\ { - 1 + 4 - 1}\\ { - 1 - 1 + 4} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 2\\ 2\\ 2 \end{array}} \right]\\ = 2\left[ {\begin{array}{*{20}{l}} 1\\ 1\\ 1 \end{array}} \right]\\ = 2{\bf{v}}\\ \\ \end{array}\] Hence, v is an eigenvector of the matrix with eigenvalue 2 \\
Step-3: Third eigenvalue
Sum of eigenvalues is equal to trace of the matrix: So, \[\begin{array}{l} 5 + 2 + {\lambda _3} = 4 + 4 + 4\\ 7 + {\lambda _3} = 12\\ {\lambda _3} = 12 - 7\\ {\lambda _3} = 5 \end{array}\]
Step-4 Eigenvector for $\lambda = 5$:
\[\begin{array}{l} (A - 5\cdot I){{\bf{v}}_1} = 0\\ \left( {\begin{array}{*{20}{c}} {4 - 5}&{ - 1}&{ - 1}\\ { - 1}&{4 - 5}&{ - 1}\\ { - 1}&{ - 1}&{4 - 5} \end{array}} \right)\left( {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right)\\ \left( {\begin{array}{*{20}{c}} { - 1}&{ - 1}&{ - 1}\\ { - 1}&{ - 1}&{ - 1}\\ { - 1}&{ - 1}&{ - 1} \end{array}} \right)\left( {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right) \end{array}\] Row-reduce the matrix: \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} {{{\bf{A}}_1}}&{\bf{0}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1}&{ - 1}&0\\ { - 1}&{ - 1}&{ - 1}&0\\ { - 1}&{ - 1}&{ - 1}&0 \end{array}} \right]\\ \approx \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1}&{ - 1}&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \to {R_2} - {R_1}}\\ {{R_3} \to {R_3} - {R_1}} \end{array}} \right\}\\ \approx \left[ {\begin{array}{*{20}{c}} 1&1&1&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {{R_1} \to ( - 1){R_1}} \right\} \end{array}\] So, the general solution is \[ \begin{aligned} \left( \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right) & = \left( \begin{array} { c } { - s - t } \\ { s } \\ { t } \end{array} \right) \\ & = ( - s ) \left( \begin{array} { c } { 1 } \\ { - 1 } \\ { 0 } \end{array} \right) + ( - t ) \left( \begin{array} { c } { 1 } \\ { 0 } \\ { - 1 } \end{array} \right) \end{aligned} \] Therefore, the eigenvectors are: \[\begin{array}{l} {v_1} = \left[ {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right]\\ {v_2} = \left[ {\begin{array}{*{20}{c}} 1\\ 0\\ { - 1} \end{array}} \right] \end{array}\]
Step-5 Projection:
The projection of $v_2$ on $v_1$: \[ \begin{aligned} \mathbf { z } _ { 2 } & = \mathbf { v } _ { 2 } - \dfrac { \mathbf { v } _ { 2 } \cdot \mathbf { v } _ { 1 } } { \mathbf { v } _ { 1 } \cdot \mathbf { v } _ { 1 } } \mathbf { v } _ { 1 } \\ & = \left[ \begin{array} { c } { 1 } \\ { 0 } \\ { 0 } \\ { - 1 } \end{array} \right] - \dfrac { 1 } { 2 } \left[ \begin{array} { c } { 1 } \\ { - 1 } \\ { 0 } \end{array} \right] \\ & = \left[ \begin{array} { c } { 1 } \\ { 0 } \\ { - 1 } \end{array} \right] - \dfrac { 1 } { 2 } \left[ \begin{array} { c } { 1 } \\ { - 1 } \\ { 0 } \end{array} \right] \\ & = \left[ \begin{array} { c } { 1 / 2 } \\ { 1 / 2 } \\ { - 1 } \end{array} \right] \end{aligned} \]
Step-6: Normalize the vectors:
\[ \begin{aligned} \mathbf { u } _ { 1 } & = \dfrac { \mathbf { v } _ { 1 } } { \left\| \mathbf { v } _ { 1 } \right\| } \\ & = \left( \dfrac { 1 } { \sqrt { 2 } } , - \dfrac { 1 } { \sqrt { 2 } } , 0 \right) ^ { \mathrm { T } } \end{aligned} \] \[ \begin{aligned} \mathbf { u } _ { 2 } & = \dfrac { \mathbf { Z } _ { 2 } } { \left\| \mathbf { z } _ { 2 } \right\| } \\ & = \left( \dfrac { 1 } { \sqrt { 6 } } , \dfrac { 1 } { \sqrt { 6 } } , - \dfrac { 2 } { \sqrt { 6 } } \right) ^ { \mathrm { T } } \end{aligned} \] \[ \begin{aligned} \mathbf { u } _ { 3 } & = \dfrac { \mathbf { v } _ { 3 } } { \left\| \mathbf { v } _ { 3 } \right\| } \\ & = \left( \dfrac { 1 } { \sqrt { 3 } } , \dfrac { 1 } { \sqrt { 3 } } , \dfrac { 1 } { \sqrt { 3 } } \right) ^ { \mathrm { T } } \end{aligned} \]
Step-7: The matrix Q
Use the normalized vectors to form Q \[ Q = \left[ \begin{array} { c c c } { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \\ { - \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \\ { 0 } & { - \dfrac { 2 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \end{array} \right] \] Inverse of Q: \[ Q ^ { - 1 } = \left[ \begin{array} { c c c } { \dfrac { 1 } { \sqrt { 2 } } } & { - \dfrac { 1 } { \sqrt { 2 } } } & { 0 } \\ { \dfrac { 1 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 6 } } } & { - \dfrac { 2 } { \sqrt { 6 } } } \\ { \dfrac { 1 } { \sqrt { 3 } } } & { \dfrac { 1 } { \sqrt { 3 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \end{array} \right] \]
Step-8: The Diagonal matrix D
\[\begin{array}{l} {Q^{ - 1}}AQ = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{\sqrt 2 }}}&{ - \dfrac{1}{{\sqrt 2 }}}&0\\ {\dfrac{1}{{\sqrt 6 }}}&{\dfrac{1}{{\sqrt 6 }}}&{ - \dfrac{2}{{\sqrt 6 }}}\\ {\dfrac{1}{{\sqrt 3 }}}&{\dfrac{1}{{\sqrt 3 }}}&{\dfrac{1}{{\sqrt 3 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 1}\\ { - 1}&4&{ - 1}\\ { - 1}&{ - 1}&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{{\sqrt 6 }}}&{\dfrac{1}{{\sqrt 3 }}}\\ { - \dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{{\sqrt 6 }}}&{\dfrac{1}{{\sqrt 3 }}}\\ 0&{ - \dfrac{2}{{\sqrt 6 }}}&{\dfrac{1}{{\sqrt 3 }}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{\sqrt 2 }}}&{ - \dfrac{1}{{\sqrt 2 }}}&0\\ {\dfrac{1}{{\sqrt 6 }}}&{\dfrac{1}{{\sqrt 6 }}}&{ - \dfrac{2}{{\sqrt 6 }}}\\ {\dfrac{1}{{\sqrt 3 }}}&{\dfrac{1}{{\sqrt 3 }}}&{\dfrac{1}{{\sqrt 3 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\dfrac{5}{{\sqrt 2 }}}&{\dfrac{5}{{\sqrt 6 }}}&{\dfrac{2}{{\sqrt 3 }}}\\ { - \dfrac{5}{{\sqrt 2 }}}&{\dfrac{5}{{\sqrt 6 }}}&{\dfrac{2}{{\sqrt 3 }}}\\ 0&{ - \dfrac{{10}}{{\sqrt 6 }}}&{\dfrac{2}{{\sqrt 3 }}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 5&0&0\\ 0&5&0\\ 0&0&2 \end{array}} \right]\\ = D \end{array}\]
The matrix A is orthogonally diagonalizable with \[ Q = \left[ \begin{array} { c c c } { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \\ { - \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \\ { 0 } & { - \dfrac { 2 } { \sqrt { 6 } } } & { \dfrac { 1 } { \sqrt { 3 } } } \end{array} \right] \]