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 13E from Chapter 7.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 13E
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: \[ A = \left[ \begin{array} { l l } { 3 } & { 1 } \\ { 1 } & { 3 } \end{array} \right] \] We have to check whether P orthogonally diagonalizes A or not.
Step-1: The characteristic polynomial
\[\begin{array}{l} det (A - \lambda I) = 0\\ \left| {\begin{array}{*{20}{c}} {3 - \lambda }&1\\ 1&{3 - \lambda } \end{array}} \right| = 0\\ {(3 - \lambda )^2} - 1 = 0\\ (\lambda - 4)(\lambda - 2) = 0 \end{array}\] So, the eigenvalues are 4 and 2
Step-2: The eigenvector for $\lambda = 4$
\[ \begin{aligned} ( A - \lambda I ) X & = 0 \\ \left( \begin{array} { c c } { - 1 } & { 1 } \\ { 1 } & { - 1 } \end{array} \right) \left( \begin{array} { c } { x } \\ { y } \end{array} \right) & = \left( \begin{array} { l } { 0 } \\ { 0 } \end{array} \right) \\ x & = y \\ \left( \begin{array} { l } { x } \\ { y } \end{array} \right) & = x \left( \begin{array} { l } { 1 } \\ { 1 } \end{array} \right) \end{aligned} \] Thus, eigenvector is:
\[{V_1} = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]\]
Step-3: The normalized eigenvector for $\lambda = 4$
\[{{\bf{u}}_1} = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{\sqrt {{1^2} + {{(1)}^2}} }}}\\ {\dfrac{1}{{\sqrt {{1^2} + {{(1)}^2}} }}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{\sqrt 2 }}}\\ {\dfrac{1}{{\sqrt 2 }}} \end{array}} \right]\]
Step-4: The eigenvector for $\lambda = 2$
\[ \begin{aligned} ( A - \lambda I ) X & = 0 \\ \left( \begin{array} { l l } { 1 } & { 1 } \\ { 1 } & { 1 } \end{array} \right) \left( \begin{array} { l } { x } \\ { y } \end{array} \right) & = \left( \begin{array} { l } { 0 } \\ { 0 } \end{array} \right) \\ y & = - x \\ \left( \begin{array} { l } { x } \\ { y } \end{array} \right) & = x \left( \begin{array} { c } { 1 } \\ { - 1 } \end{array} \right) \end{aligned} \] Thus, eigenvector is:
\[{V_2} = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]\]
Step-4: The normalized eigenvector for $\lambda = 2$
\[{{\bf{u}}_2} = \left[ {\begin{array}{*{20}{c}} 1\\ {\sqrt {{1^2} + {{( - 1)}^2}} }\\ {\dfrac{1}{{\sqrt {{1^2} + {{( - 1)}^2}} }}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \dfrac{1}{{\sqrt 2 }}}\\ {\dfrac{1}{{\sqrt 2 }}} \end{array}} \right]\]
Step-5: The P matrix
\[ \begin{aligned} P & = \left[ \mathbf { u } _ { 1 } \ \ \mathbf { u } _ { 2 } \right] \\ & = \left[ \begin{array} { c c } { \dfrac { 1 } { \sqrt { 2 } } } & { - \dfrac { 1 } { \sqrt { 2 } } } \\ { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 2 } } } \end{array} \right] \end{aligned} \]
Step-6: The D matrix
The D matrix is formed by eigenvalues \[ D = \left[ \begin{array} { l l } { 4 } & { 0 } \\ { 0 } & { 2 } \end{array} \right] \]
Step-7: Check for orthogonality
\[ \begin{aligned} P D P ^ { - 1 } & = P D P ^ { T } \\ & = \left[ \begin{array} { c c } { \dfrac { 1 } { \sqrt { 2 } } } & { - \dfrac { 1 } { \sqrt { 2 } } } \\ { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 2 } } } \end{array} \right] \left[ \begin{array} { c c } { 4 } & { 0 } \\ { 0 } & { 2 } \end{array} \right] \left[ \begin{array} { c c } { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 2 } } } \\ { - \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 2 } } } \end{array} \right] \\ & = \left[ \begin{array} { c c } { 3 } & { 1 } \\ { 1 } & { 3 } \end{array} \right] \\ & = A \end{aligned} \] Therefore,
Therefore, the matrix P orthogonally diagonalizes A .
We are given with a matrix: \[ A = \left[ \begin{array} { l l } { 3 } & { 1 } \\ { 1 } & { 3 } \end{array} \right] \] We have to check whether P orthogonally diagonalizes A or not.
Step-1: The characteristic polynomial
\[\begin{array}{l} det (A - \lambda I) = 0\\ \left| {\begin{array}{*{20}{c}} {3 - \lambda }&1\\ 1&{3 - \lambda } \end{array}} \right| = 0\\ {(3 - \lambda )^2} - 1 = 0\\ (\lambda - 4)(\lambda - 2) = 0 \end{array}\] So, the eigenvalues are 4 and 2
Step-2: The eigenvector for $\lambda = 4$
\[ \begin{aligned} ( A - \lambda I ) X & = 0 \\ \left( \begin{array} { c c } { - 1 } & { 1 } \\ { 1 } & { - 1 } \end{array} \right) \left( \begin{array} { c } { x } \\ { y } \end{array} \right) & = \left( \begin{array} { l } { 0 } \\ { 0 } \end{array} \right) \\ x & = y \\ \left( \begin{array} { l } { x } \\ { y } \end{array} \right) & = x \left( \begin{array} { l } { 1 } \\ { 1 } \end{array} \right) \end{aligned} \] Thus, eigenvector is:
\[{V_1} = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]\]
Step-3: The normalized eigenvector for $\lambda = 4$
\[{{\bf{u}}_1} = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{\sqrt {{1^2} + {{(1)}^2}} }}}\\ {\dfrac{1}{{\sqrt {{1^2} + {{(1)}^2}} }}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{\sqrt 2 }}}\\ {\dfrac{1}{{\sqrt 2 }}} \end{array}} \right]\]
Step-4: The eigenvector for $\lambda = 2$
\[ \begin{aligned} ( A - \lambda I ) X & = 0 \\ \left( \begin{array} { l l } { 1 } & { 1 } \\ { 1 } & { 1 } \end{array} \right) \left( \begin{array} { l } { x } \\ { y } \end{array} \right) & = \left( \begin{array} { l } { 0 } \\ { 0 } \end{array} \right) \\ y & = - x \\ \left( \begin{array} { l } { x } \\ { y } \end{array} \right) & = x \left( \begin{array} { c } { 1 } \\ { - 1 } \end{array} \right) \end{aligned} \] Thus, eigenvector is:
\[{V_2} = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]\]
Step-4: The normalized eigenvector for $\lambda = 2$
\[{{\bf{u}}_2} = \left[ {\begin{array}{*{20}{c}} 1\\ {\sqrt {{1^2} + {{( - 1)}^2}} }\\ {\dfrac{1}{{\sqrt {{1^2} + {{( - 1)}^2}} }}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \dfrac{1}{{\sqrt 2 }}}\\ {\dfrac{1}{{\sqrt 2 }}} \end{array}} \right]\]
Step-5: The P matrix
\[ \begin{aligned} P & = \left[ \mathbf { u } _ { 1 } \ \ \mathbf { u } _ { 2 } \right] \\ & = \left[ \begin{array} { c c } { \dfrac { 1 } { \sqrt { 2 } } } & { - \dfrac { 1 } { \sqrt { 2 } } } \\ { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 2 } } } \end{array} \right] \end{aligned} \]
Step-6: The D matrix
The D matrix is formed by eigenvalues \[ D = \left[ \begin{array} { l l } { 4 } & { 0 } \\ { 0 } & { 2 } \end{array} \right] \]
Step-7: Check for orthogonality
\[ \begin{aligned} P D P ^ { - 1 } & = P D P ^ { T } \\ & = \left[ \begin{array} { c c } { \dfrac { 1 } { \sqrt { 2 } } } & { - \dfrac { 1 } { \sqrt { 2 } } } \\ { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 2 } } } \end{array} \right] \left[ \begin{array} { c c } { 4 } & { 0 } \\ { 0 } & { 2 } \end{array} \right] \left[ \begin{array} { c c } { \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 2 } } } \\ { - \dfrac { 1 } { \sqrt { 2 } } } & { \dfrac { 1 } { \sqrt { 2 } } } \end{array} \right] \\ & = \left[ \begin{array} { c c } { 3 } & { 1 } \\ { 1 } & { 3 } \end{array} \right] \\ & = A \end{aligned} \] Therefore,
Therefore, the matrix P orthogonally diagonalizes A .