91影视

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.

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\(D = \left( {\begin{aligned}{{}}{ - 4}&0&0\\0&4&0\\0&0&7\end{aligned}} \right)\).

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