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Q20E

Expert-verifiedFound in: Page 395

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

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

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.

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 \(D = \left( {\begin{aligned}{{}}{ - 3}&0&0\\0&{ - 3}&0\\0&0&{15}\end{aligned}} \right)\).

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

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