91Ó°ÊÓ

In the given matrix\(A\), each row sums to 2.So, \(\lambda = 2\)must be one of the Eigenvalues of the matrix\(A\). The basis for the Eigenspace is obtained by solving the system of equations\(\left( {A - 2I} \right){\bf{x}} = 0\).

On solving, the corresponding Eigenvectorsare obtained as\(\left( \begin{aligned}{}1\\1\\1\end{aligned} \right)\). So,\(\left( \begin{aligned}{}1\\1\\1\end{aligned} \right)\)is one of the Eigenvectors of the matrix\(A\). It can also be verified by the equation\(A{\bf{x}} = \lambda \), as follows:

\(\left( {\begin{aligned}{{}}{\,\,\,4}&{ - 1}&{ - 1}\\{ - 1}&4&{ - 1}\\{ - 1}&{ - 1}&4\end{aligned}} \right)\left( \begin{aligned}{}1\\1\\1\end{aligned} \right) = \left( \begin{aligned}{}2\\2\\2\end{aligned} \right)\)

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.

As the trace of matrix\(A\)is 12. So, it must have the Eigen value\(\lambda = 5\)with the multiplicity of 2. So, the basis for its Eigenspace is obtained by solving the system of equations\(\left( {A - 5I} \right){\bf{x}} = 0\). On solving, two Eigen vectorsare obtained as\(\left\{ {\left( \begin{aligned}{} - 1\\\,\,\,1\\\,\,\,0\end{aligned} \right),\left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right)} \right\}\).This set of vectors is converted to an orthogonal basis via orthogonal projection to obtain it as\(\left\{ {\left( \begin{aligned}{} - 1\\\,\,\,1\\\,\,\,0\end{aligned} \right),\left( \begin{aligned}{} - 1\\\,\,\,1\\\,\,\,2\end{aligned} \right)} \right\}\).

The normalized vectors are\({{\bf{u}}_1} = \left( \begin{aligned}{}1/\sqrt 3 \\1/\sqrt 3 \\1/\sqrt 3 \end{aligned} \right)\), and\({{\bf{u}}_2} = \left( \begin{aligned}{} - 1/\sqrt 2 \\\,\,\,1/\sqrt 2 \\\,\,\,\,\,\,\,0\end{aligned} \right)\), and\({{\bf{u}}_2} = \left( \begin{aligned}{} - 1/\sqrt 6 \\\,\,\,1/\sqrt 6 \\\,\,\,\,2/\sqrt 6 \end{aligned} \right)\).So, the normalized matrix\(P\)is given as;

\(\begin{aligned}{}P &= \left( {{{\bf{u}}_1}\,\,{{\bf{u}}_2}\,{{\bf{u}}_3}} \right)\\ &= \left( {\begin{aligned}{{}}{1/\sqrt 3 }&{ - 1/\sqrt 2 }&{ - 1/\sqrt 6 }\\{1/\sqrt 3 }&{\,\,\,1/\sqrt 2 }&{ - 1/\sqrt 6 }\\{1/\sqrt 3 }&{\,\,\,\,0}&{\,\,\,2/\sqrt 6 }\end{aligned}} \right)\end{aligned}\)

Here, the matrix \(D\) is obtained as \(D = \left( {\begin{aligned}{{}}2&0&0\\0&5&0\\0&0&5\end{aligned}} \right)\).

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

\(\begin{aligned}{}A &= PD{P^{ - 1}}\\ &= \left( {\begin{aligned}{{}}{1/\sqrt 3 }&{ - 1/\sqrt 2 }&{ - 1/\sqrt 6 }\\{1/\sqrt 3 }&{\,\,\,1/\sqrt 2 }&{ - 1/\sqrt 6 }\\{1/\sqrt 3 }&{\,\,\,\,0}&{\,\,\,2/\sqrt 6 }\end{aligned}} \right)\left( {\begin{aligned}{{}}2&0&0\\0&5&0\\0&0&5\end{aligned}} \right){\left( {\begin{aligned}{{}}{1/\sqrt 3 }&{ - 1/\sqrt 2 }&{ - 1/\sqrt 6 }\\{1/\sqrt 3 }&{\,\,\,1/\sqrt 2 }&{ - 1/\sqrt 6 }\\{1/\sqrt 3 }&{\,\,\,\,0}&{\,\,\,2/\sqrt 6 }\end{aligned}} \right)^{ - 1}}\end{aligned}\)

It is verified that 5 is an eigenvalue of\(A\)and\({\rm{v}}\)is an eigenvector.

The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{1/\sqrt 3 }&{ - 1/\sqrt 2 }&{ - 1/\sqrt 6 }\\{1/\sqrt 3 }&{\,\,\,1/\sqrt 2 }&{ - 1/\sqrt 6 }\\{1/\sqrt 3 }&{\,\,\,\,0}&{\,\,\,2/\sqrt 6 }\end{aligned}} \right)\left( {\begin{aligned}{{}}2&0&0\\0&5&0\\0&0&5\end{aligned}} \right){\left( {\begin{aligned}{{}}{1/\sqrt 3 }&{ - 1/\sqrt 2 }&{ - 1/\sqrt 6 }\\{1/\sqrt 3 }&{\,\,\,1/\sqrt 2 }&{ - 1/\sqrt 6 }\\{1/\sqrt 3 }&{\,\,\,\,0}&{\,\,\,2/\sqrt 6 }\end{aligned}} \right)^{ - 1}}\).