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.

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

Short answer

The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{ - 2/3}&{\,\,\,\,2/3}&{\,\,2/3}\\{ - 1/3}&{ - 1/3}&{ - 2/3}\\{\,\,2/3}&{\,\,\,2/3}&{\,\,1/3}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 3}&0&0\\0&{ - 6}&0\\0&0&9\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 2/3}&{\,\,\,\,2/3}&{\,\,2/3}\\{ - 1/3}&{ - 1/3}&{ - 2/3}\\{\,\,2/3}&{\,\,\,2/3}&{\,\,1/3}\end{aligned}} \right)^{ - 1}}\).

Step-by-step solution

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 \(\lambda = - 3,\,\, - 6\) and 9.

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, it is obtained as\(\left( \begin{aligned}{}1\\2\\2\end{aligned} \right)\).

Similarly, for the Eigen value\(\lambda = - 6\), the basis for Eigenspace is obtained by solving the system of equations\(\left( {A + 6I} \right){\bf{x}} = 0\). On solving, it is obtained as\(\left( \begin{aligned}{} - 2\\ - 1\\\,\,\,2\end{aligned} \right)\).

Similarly, for the Eigen value\(\lambda = 9\), the basis for Eigenspace is obtained by solving the system of equations\(\left( {A - 9I} \right){\bf{x}} = 0\). On solving it is obtained as\(\left( \begin{aligned}{}\,\,\,2\\ - 2\\\,\,\,1\end{aligned} \right)\).

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

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

Here, the matrix \(D\) is obtained as \(D = \left( {\begin{aligned}{{}}{ - 3}&0&0\\0&{ - 6}&0\\0&0&9\end{aligned}} \right)\).

Diagonalize matrix \(A\)

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

\(\begin{aligned}{}A &= PD{P^{ - 1}}\\ &= \left( {\begin{aligned}{{}}{ - 2/3}&{\,\,\,\,2/3}&{\,\,2/3}\\{ - 1/3}&{ - 1/3}&{ - 2/3}\\{\,\,2/3}&{\,\,\,2/3}&{\,\,1/3}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 3}&0&0\\0&{ - 6}&0\\0&0&9\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 2/3}&{\,\,\,\,2/3}&{\,\,2/3}\\{ - 1/3}&{ - 1/3}&{ - 2/3}\\{\,\,2/3}&{\,\,\,2/3}&{\,\,1/3}\end{aligned}} \right)^{ - 1}}\end{aligned}\)

The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{ - 2/3}&{\,\,\,\,2/3}&{\,\,2/3}\\{ - 1/3}&{ - 1/3}&{ - 2/3}\\{\,\,2/3}&{\,\,\,2/3}&{\,\,1/3}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 3}&0&0\\0&{ - 6}&0\\0&0&9\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 2/3}&{\,\,\,\,2/3}&{\,\,2/3}\\{ - 1/3}&{ - 1/3}&{ - 2/3}\\{\,\,2/3}&{\,\,\,2/3}&{\,\,1/3}\end{aligned}} \right)^{ - 1}}\).