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Q24E

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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

Let \(A = \left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&{\,\,\,2}&{ - 1}\\{ - 1}&{ - 1} &{\,\,\,2}\end{aligned}} \right)\),\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}{ - 1}\\{\,\,\,0}\\{\,\,1}\end{aligned}} \right)\) and and\({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{\,\,\,1}\\{\, - 1}\\{\,\,\,\,1}\end{aligned}} \right)\). Verify that\({{\rm{v}}_1}\), \({{\rm{v}}_2}\) an eigenvector of \(A\). Then orthogonally diagonalize \(A\).

It is verified that \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\) are the eigenvector of \(A\). The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)\left( {\begin{aligned}{{}}1&0&0\\0&4&0\\0&0&1\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)^{ - 1}}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Verify the eigenvectors

For the matrix \(A\), the given eigenvectors \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\) must satisfy the characteristic equation \(A{\bf{x}} = \lambda {\bf{x}}\). On verifying it for the vector \({{\rm{v}}_1}\), we proceed as follows:

\(\begin{aligned}{}\left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&2&{ - 1}\\{ - 1}&{ - 1}&2\end{aligned}} \right)\left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right) = \left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right)\\ = 1\left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right)\end{aligned}\)

Thus, \(\left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right)\) is eigenvector of A for the eigenvalue \(\lambda = 1\). Now, verifying it for the vector \({{\rm{v}}_2}\), we proceed as follows:

\(\begin{aligned}{}\left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&2&{ - 1}\\{ - 1}&{ - 1}&2\end{aligned}} \right)\left( \begin{aligned}{}\,\,\,1\\ - 1\\\,\,\,1\end{aligned} \right) = \left( \begin{aligned}{}\,\,\,4\\ - 4\\\,\,\,4\end{aligned} \right)\\ = 4\left( \begin{aligned}{}\,\,\,1\\ - 1\\\,\,\,\,1\end{aligned} \right)\end{aligned}\)

Thus, \(\left( \begin{aligned}{}\,\,\,1\\ - 1\\\,\,\,\,1\end{aligned} \right)\) is eigenvector of A for the eigenvalue \(\lambda = 4\).

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 \(\lambda = 1\), the basis for its Eigenspace is obtained by solving the system of equations \(\left( {A - I} \right){\bf{x}} = 0\). On solving, two Eigen vectorsare obtained as \(\left\{ {\left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right),\left( \begin{aligned}{}1\\1\\0\end{aligned} \right)} \right\}\).This set of vectors is converted to an orthogonal basis via orthogonal projection to obtain it as\(\left\{ {\left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right),\left( \begin{aligned}{}\,\,1\\\,\,2\\\,\,\,1\end{aligned} \right)} \right\}\).

For the Eigen value\(\lambda = 4\), the Eigen vectors is given as \(\left( \begin{aligned}{}\,\,\,1\\ - 1\\\,\,\,1\end{aligned} \right)\).

The normalized vectors are \({{\bf{u}}_1} = \left( \begin{aligned}{} - 1/\sqrt 2 \\\,\,\,\,\,\,\,0\\\,\,\,1/\sqrt 2 \end{aligned} \right)\)and \({{\bf{u}}_2} = \left( \begin{aligned}{}\,\,\,1/\sqrt 3 \\ - 1/\sqrt 3 \\\,\,\,1/\sqrt 3 \end{aligned} \right)\), and\({{\bf{u}}_2} = \left( \begin{aligned}{}1/\sqrt 6 \\\,2/\sqrt 6 \\\,1/\sqrt 6 \end{aligned} \right)\).So, the normalized matrix \(P\) is given as;

\(\begin{aligned}{}P &= \left( {{{\bf{u}}_1}\,\,{{\bf{u}}_2}\,{{\bf{u}}_3}} \right)\\ &= \left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)\end{aligned}\)

Here, the matrix \(D\) is obtained as \(D = \left( {\begin{aligned}{{}}1&0&0\\0&4&0\\0&0&1\end{aligned}} \right)\).

Step 3: Diagonalize matrix \(A\)

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

\(\begin{aligned}{}A &= PD{P^{ - 1}}\\ &= \left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)\left( {\begin{aligned}{{}}1&0&0\\0&4&0\\0&0&1\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)^{ - 1}}\end{aligned}\)

It is verified that \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\) are the eigenvector of \(A\). The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)\left( {\begin{aligned}{{}}1&0&0\\0&4&0\\0&0&1\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)^{ - 1}}\).

Most popular questions for Math Textbooks

Orhtogonally diagonalize the matrices in Exercises 37-40. To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and for each eigenvalue \(\lambda \), find an orthogonal basis for \({\bf{Nul}}\left( {A - \lambda I} \right)\), as in Examples 2 and 3.

37. \(\left( {\begin{aligned}{{}}{\bf{6}}&{\bf{2}}&{\bf{9}}&{ - {\bf{6}}}\\{\bf{2}}&{\bf{6}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{9}}&{ - {\bf{6}}}&{\bf{6}}&{\bf{2}}\\{\bf{6}}&{\bf{9}}&{\bf{2}}&{\bf{6}}\end{aligned}} \right)\)

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

5. \(\left( {\begin{aligned}{{}{}}{ - 6}&2&0\\2&{ - 6}&2\\0&2&{ - 6}\end{aligned}} \right)\)

Question: 3. Let A be an \(n \times n\) symmetric matrix of rank r. Explain why the spectral decomposition of A represents A as the sum of r rank 1 matrices.

Find the matrix of the quadratic form. Assume x is in \({\mathbb{R}^2}\).

a. \(5x_1^2 + 16{x_1}{x_2} - 5x_2^2\)

b. \(2{x_1}{x_2}\)

Question: 2. Let \(\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},....,{{\bf{u}}_n}} \right\}\) be an orthonormal basis for \({\mathbb{R}_n}\) , and let \({\lambda _1},....{\lambda _n}\) be any real scalars. Define

\(A = {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + ..... + {\lambda _n}{\bf{u}}_n^T\)

a. Show that A is symmetric.

b. Show that \({\lambda _1},....{\lambda _n}\) are the eigenvalues of A
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