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Expert-verified Found in: Page 395 ### 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 # 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}}.

See the step by step solution

## Step 1: Describe the given information

LetA = \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}}. ### Want to see more solutions like these? 