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Q19E

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.**

**19. \(\left( {\begin{aligned}{{}}3&{ - 2}&4\\{ - 2}&6&2\\4&2&3\end{aligned}} \right)\)**

The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{ - 1/\sqrt 5 }&{4/\sqrt {45} }&{\,\,2/3}\\{2/\sqrt 5 }&{2/\sqrt {45} }&{ - 2/3}\\0&{5/\sqrt {45} }&{\,\,1/3}\end{aligned}} \right)\left( {\begin{aligned}{{}}7&0&0\\0&7&0\\0&0&{ - 2}\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 5 }&{4/\sqrt {45} }&{\,\,2/3}\\{2/\sqrt 5 }&{2/\sqrt {45} }&{ - 2/3}\\0&{5/\sqrt {45} }&{\,\,1/3}\end{aligned}} \right)^{ - 1}}\).

Let\(A = \left( {\begin{aligned}{{}}3&{ - 2}&4\\{ - 2}&6&2\\{\,4}&2&3\end{aligned}} \right)\).The eigenvalues of \(A\)are given as \(\lambda = - 2\) and 7.

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 \(\lambda = 7\), 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 \(\left\{ {\left( \begin{aligned}{} - 1\\\,\,\,2\\\,\,\,0\end{aligned} \right),\left( \begin{aligned}{}1\\0\\1\end{aligned} \right)} \right\}\). This set is converted to an orthogonal basis, via orthogonal projection, to obtain it as \(\left\{ {\left( \begin{aligned}{} - 1\\\,\,\,2\\\,\,\,0\end{aligned} \right),\left( \begin{aligned}{}4\\2\\5\end{aligned} \right)} \right\}\).

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

The normalized vectors are \({{\bf{u}}_1} = \left( \begin{aligned}{}1/\sqrt 5 \\2/\sqrt 5 \\\,\,0\end{aligned} \right)\) ,\({{\bf{u}}_2} = \left( \begin{aligned}{}4/\sqrt {45} \\2/\sqrt {45} \\5/\sqrt {45} \end{aligned} \right)\) and \({{\bf{u}}_3} = \left( \begin{aligned}{}\,\,\,2/3\\ - 2/3\\\,\,\,\,1/3\end{aligned} \right)\). So, the normalized matrix \(P\) is given as

\(\begin{aligned}{}P &= \left( {{{\bf{u}}_1}\,\,{{\bf{u}}_2}\,{{\bf{u}}_3}} \right)\\ &= \left( {\begin{aligned}{{}}{ - 1/\sqrt 5 }&{4/\sqrt {45} }&{\,\,2/3}\\{2/\sqrt 5 }&{2/\sqrt {45} }&{ - 2/3}\\0&{5/\sqrt {45} }&{\,\,1/3}\end{aligned}} \right)\end{aligned}\),

whereas the matrix \(D\) is obtained as \(D = \left( {\begin{aligned}{{}}7&0&0\\0&7&0\\0&0&{ - 2}\end{aligned}} \right)\).

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

\(\begin{aligned}{}A &= PD{P^{ - 1}}\\ &= \left( {\begin{aligned}{{}}{ - 1/\sqrt 5 }&{4/\sqrt {45} }&{\,\,2/3}\\{2/\sqrt 5 }&{2/\sqrt {45} }&{ - 2/3}\\0&{5/\sqrt {45} }&{\,\,1/3}\end{aligned}} \right)\left( {\begin{aligned}{{}}7&0&0\\0&7&0\\0&0&{ - 2}\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 5 }&{4/\sqrt {45} }&{\,\,2/3}\\{2/\sqrt 5 }&{2/\sqrt {45} }&{ - 2/3}\\0&{5/\sqrt {45} }&{\,\,1/3}\end{aligned}} \right)^{ - 1}}\end{aligned}\)

The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{ - 1/\sqrt 5 }&{4/\sqrt {45} }&{\,\,2/3}\\{2/\sqrt 5 }&{2/\sqrt {45} }&{ - 2/3}\\0&{5/\sqrt {45} }&{\,\,1/3}\end{aligned}} \right)\left( {\begin{aligned}{{}}7&0&0\\0&7&0\\0&0&{ - 2}\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 5 }&{4/\sqrt {45} }&{\,\,2/3}\\{2/\sqrt 5 }&{2/\sqrt {45} }&{ - 2/3}\\0&{5/\sqrt {45} }&{\,\,1/3}\end{aligned}} \right)^{ - 1}}\).

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