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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, 5.17. \left( {\begin{aligned}{{}}1&{ - 6}&4\\{ - 6}&2&{ - 2}\\4&{ - 2}&{ - 3}\end{aligned}} \right)

The orthogonal diagonalization is \left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\\0&{ - 2/\sqrt 6 }&{1/\sqrt 3 }\\{1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 4}&0&0\\0&4&0\\0&0&7\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\\0&{ - 2/\sqrt 6 }&{1/\sqrt 3 }\\{1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\end{aligned}} \right)^{ - 1}}.

See the step by step solution

## Step 1: Describe the given information

Let A = \left( {\begin{aligned}{{}}1&{ - 6}&4\\{ - 6}&2&{ - 2}\\{\,4}&{ - 2}&{ - 3}\end{aligned}} \right).The eigenvalues of $$A$$are given as $$\lambda = - 4,\,\,4$$ and 7.

## 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 $$\lambda = - 4$$, the basis for Eigenspace is obtained by solving the system of equations $$\left( {A + 4I} \right){\bf{x}} = 0$$. On solving, it is obtained as \left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right).

Similarly, for the Eigen value $$\lambda = 4$$, the basis for Eigenspace is obtained by solving the system of equations $$\left( {A - 4I} \right){\bf{x}} = 0$$. On solving, it is obtained as \left( \begin{aligned}{}\,\,1\\ - 2\\\,\,\,1\end{aligned} \right).

Similarly, for the Eigen value $$\lambda = 7$$, the basis for Eigenspace is obtained by solving the system of equations $$\left( {A - 7I} \right){\bf{x}} = 0$$. On solving it is obtained 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) ,{{\bf{u}}_2} = \left( \begin{aligned}{}\,1/\sqrt 6 \\ - 2/\sqrt 6 \\\,1/\sqrt 6 \end{aligned} \right) and {{\bf{u}}_3} = \left( \begin{aligned}{}1/\sqrt 3 \\1/\sqrt 3 \\1/\sqrt 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 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\\0&{ - 2/\sqrt 6 }&{1/\sqrt 3 }\\{1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\end{aligned}} \right)\end{aligned}

Here, the matrix $$D$$ is obtained as D = \left( {\begin{aligned}{{}}{ - 4}&0&0\\0&4&0\\0&0&7\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 6 }&{1/\sqrt 3 }\\0&{ - 2/\sqrt 6 }&{1/\sqrt 3 }\\{1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 4}&0&0\\0&4&0\\0&0&7\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\\0&{ - 2/\sqrt 6 }&{1/\sqrt 3 }\\{1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\end{aligned}} \right)^{ - 1}}\end{aligned}

Thus, the orthogonal diagonalization is \left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\\0&{ - 2/\sqrt 6 }&{1/\sqrt 3 }\\{1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 4}&0&0\\0&4&0\\0&0&7\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\\0&{ - 2/\sqrt 6 }&{1/\sqrt 3 }\\{1/\sqrt 2 }&{1/\sqrt 6 }&{1/\sqrt 3 }\end{aligned}} \right)^{ - 1}}. ### Want to see more solutions like these? 