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Q20E

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Linear Algebra and its Applications
Found in: Page 395
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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: (17) \( - {\bf{4}}\), 4, 7; (18) \( - {\bf{3}}\), \( - {\bf{6}}\), 9; (19) \( - {\bf{2}}\), 7; (20) \( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

20. \(\left( {\begin{aligned}{{}}5&8&{ - 4}\\8&5&{ - 4}\\{ - 4}&{ - 4}&{ - 1}\end{aligned}} \right)\)

The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{\,\,2/3}&{ - 1/3}&{\,\,2/3}\\{ - 1/3}&{\,\,2/3}&{\,\,\,2/3}\\{\,\,\,2/3}&{\,\,2/3}&{ - 1/3}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 3}&0&0\\0&{ - 3}&0\\0&0&{15}\end{aligned}} \right){\left( {\begin{aligned}{{}}{\,\,2/3}&{ - 1/3}&{\,\,2/3}\\{ - 1/3}&{\,\,2/3}&{\,\,\,2/3}\\{\,\,\,2/3}&{\,\,2/3}&{ - 1/3}\end{aligned}} \right)^{ - 1}}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Describe the given information

Let\(A = \left( {\begin{aligned}{{}}{\,\,5}&8&{\,\,4}\\{\,\,\,8}&5&{ - 4}\\{\, - 4}&{ - 4}&{ - 1}\end{aligned}} \right)\).

The eigenvalues of \(A\)are given as \(\lambda = - 3\) and 15.

Step 2: Find the matrix \(P\) and \(D\)

A matrix \(A\) is diagonalized as \(A = PD{P^{ - 1}}\), where \(P\) is orthogonal matrix of normalized Eigen vectors of matrix \(A\)and \(D\) is a diagonal matrix having Eigen values of matrix \(A\) on its principle diagonal.

For the Eigen value \(\lambda = - 3\), the basis for Eigenspace is obtained by solving the system of equations \(\left( {A + 3I} \right){\bf{x}} = 0\). On solving, two Eigen vectorsare obtained as \(\left\{ {\left( \begin{aligned}{}\,\,\,2\\ - 1\\\,\,\,2\end{aligned} \right),\left( \begin{aligned}{} - 1\\\,\,\,2\\\,\,\,2\end{aligned} \right)} \right\}\). This set of vectors is orthogonal as the dot product of elements is 0.

For the Eigen value \(\lambda = 15\), the basis for Eigenspace is obtained by solving the system of equations \(\left( {A + 15I} \right){\bf{x}} = 0\). On solving, it is obtained as \(\left( \begin{aligned}{}\,\,\,2\\\,\,\,2\\ - 1\end{aligned} \right)\).

The normalized vectors are \({{\bf{u}}_1} = \left( \begin{aligned}{}\,\,2/3\\ - 1/3\\\,\,\,2/3\end{aligned} \right)\) ,\({{\bf{u}}_2} = \left( \begin{aligned}{} - 1/3\\\,\,\,2/3\\\,\,\,2/3\end{aligned} \right)\) and \({{\bf{u}}_3} = \left( \begin{aligned}{}\,\,\,2/3\\\,\,\,2/3\\ - 1/3\end{aligned} \right)\). So, the normalized matrix \(P\) is given as;

\(\begin{aligned}{}P &= \left( {{{\bf{u}}_1}\,\,{{\bf{u}}_2}\,{{\bf{u}}_3}} \right)\\ &= \left( {\begin{aligned}{{}}{\,\,2/3}&{ - 1/3}&{\,\,2/3}\\{ - 1/3}&{\,\,2/3}&{\,\,\,2/3}\\{\,\,\,2/3}&{\,\,2/3}&{ - 1/3}\end{aligned}} \right)\end{aligned}\)

Here, the matrix \(D\) is obtained as \(D = \left( {\begin{aligned}{{}}{ - 3}&0&0\\0&{ - 3}&0\\0&0&{15}\end{aligned}} \right)\).

Step 3: Diagonalize matrix \(A\)

The matrix \(A\) is diagonalized as \(A = PD{P^{ - 1}}\). Thus, the orthogonal diagonalization is as follows:

\(\begin{aligned}{}A &= PD{P^{ - 1}}\\ &= \left( {\begin{aligned}{{}}{\,\,2/3}&{ - 1/3}&{\,\,2/3}\\{ - 1/3}&{\,\,2/3}&{\,\,\,2/3}\\{\,\,\,2/3}&{\,\,2/3}&{ - 1/3}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 3}&0&0\\0&{ - 3}&0\\0&0&{15}\end{aligned}} \right){\left( {\begin{aligned}{{}}{\,\,2/3}&{ - 1/3}&{\,\,2/3}\\{ - 1/3}&{\,\,2/3}&{\,\,\,2/3}\\{\,\,\,2/3}&{\,\,2/3}&{ - 1/3}\end{aligned}} \right)^{ - 1}}\end{aligned}\)

The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{\,\,2/3}&{ - 1/3}&{\,\,2/3}\\{ - 1/3}&{\,\,2/3}&{\,\,\,2/3}\\{\,\,\,2/3}&{\,\,2/3}&{ - 1/3}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 3}&0&0\\0&{ - 3}&0\\0&0&{15}\end{aligned}} \right){\left( {\begin{aligned}{{}}{\,\,2/3}&{ - 1/3}&{\,\,2/3}\\{ - 1/3}&{\,\,2/3}&{\,\,\,2/3}\\{\,\,\,2/3}&{\,\,2/3}&{ - 1/3}\end{aligned}} \right)^{ - 1}}\).

Most popular questions for Math Textbooks

Determine which of the matrices in Exercises 1–6 are symmetric.

\(\left( {\begin{array}{{}{}}1&2&2&1\\2&2&2&1\\2&2&1&2\end{array}} \right)\)

Let A be the matrix of the quadratic form

\({\bf{9}}x_{\bf{1}}^{\bf{2}} + {\bf{7}}x_{\bf{2}}^{\bf{2}} + {\bf{11}}x_{\bf{3}}^{\bf{2}} - {\bf{8}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{8}}{x_{\bf{1}}}{x_{\bf{3}}}\)

It can be shown that the eigenvalues of A are 3,9, and 15. Find an orthogonal matrix P such that the change of variable \({\bf{x}} = P{\bf{y}}\) transforms \({{\bf{x}}^T}A{\bf{x}}\) into a quadratic form which no cross-product term. Give P and the new quadratic form.

Construct a spectral decomposition of A from Example 3.

Determine which of the matrices in Exercises 1–6 are symmetric.

2. \(\left( {\begin{aligned}{{}}3&{\,\, - 5}\\{ - 5}&{ - 3}\end{aligned}} \right)\)

Question: 5. Show that if v is an eigenvector of an \(n \times n\) matrix A and v corresponds to a nonzero eigenvalue of A, then v is in Col A. (Hint: Use the definition of an eigenvector.)
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