Question

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.

22. \(\left( {\begin{aligned}{{}}4&0&1&0\\0&4&0&1\\1&0&4&0\\0&1&0&4\end{aligned}} \right)\)

Short answer

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

Step-by-step solution

Describe the given information

Let\(A = \left( {\begin{aligned}{{}}4&0&1&0\\0&4&0&1\\1&0&4&0\\0&1&0&4\end{aligned}} \right)\).

The eigenvalues of \(A\)are given as \(\lambda = 3\) and 5.

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\(\left\{ {\left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\\\,\,\,0\end{aligned} \right),\left( \begin{aligned}{}\,\,\,0\\ - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right)} \right\}\). This set of vectors is orthogonal as the dot product of elements is 0.

For the Eigen value\(\lambda = 5\), the basis for Eigenspace is obtained by solving the system of equations\(\left( {A - 5I} \right){\bf{x}} = 0\). On solving,, two Eigen vectors are obtained as\(\left( \begin{aligned}{}1\\0\\1\\1\end{aligned} \right)\left( \begin{aligned}{}0\\1\\0\\1\end{aligned} \right)\).This set of vectors is also orthogonal as the dot product of elements is 0.

The normalized vectors are\(\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\} = \left\{ {\left( \begin{aligned}{} - 1/\sqrt 2 \\\,\,\,\,\,\,0\\\,1/\sqrt 2 \\\,\,\,\,\,\,0\end{aligned} \right)\left( \begin{aligned}{}\,\,\,\,\,\,0\\ - 1/\sqrt 2 \\\,\,\,\,\,\,0\\1/\sqrt 2 \end{aligned} \right)} \right\}\)and\(\left\{ {{{\bf{u}}_3},{{\bf{u}}_4}} \right\} = \left\{ {\left( \begin{aligned}{}1/\sqrt 2 \\\,\,\,\,0\\1/\sqrt 2 \\\,\,\,\,0\end{aligned} \right)\left( \begin{aligned}{}\,\,\,\,0\\1/\sqrt 2 \\\,\,\,\,0\\1/\sqrt 2 \end{aligned} \right)} \right\}\). So, the normalized matrix\(P\)is given as;

\(\begin{aligned}{}P &= \left( {{{\bf{u}}_1}\,\,{{\bf{u}}_2}\,{{\bf{u}}_3}\,{{\bf{u}}_4}} \right)\\ &= \left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&0&{1/\sqrt 2 }&0\\0&{ - 1/\sqrt 2 }&0&{1/\sqrt 2 }\\{1/\sqrt 2 }&0&{1/\sqrt 2 }&0\\0&{\,\,\,\,1/\sqrt 2 }&0&{1/\sqrt 2 }\end{aligned}} \right)\end{aligned}\)

Here, the matrix\(D\)is obtained as\(D = \left( {\begin{aligned}{{}}3&0&0&0\\0&3&0&0\\0&0&5&0\\0&0&0&5\end{aligned}} \right)\).

Diagonalize matrix \(A\)

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

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

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