Question

Let \(A = \left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&{\,\,\,2}&{ - 1}\\{ - 1}&{ - 1} &{\,\,\,2}\end{aligned}} \right)\),\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}{ - 1}\\{\,\,\,0}\\{\,\,1}\end{aligned}} \right)\) and and\({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{\,\,\,1}\\{\, - 1}\\{\,\,\,\,1}\end{aligned}} \right)\). Verify that\({{\rm{v}}_1}\), \({{\rm{v}}_2}\) an eigenvector of \(A\). Then orthogonally diagonalize \(A\).

Short answer

It is verified that \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\) are the eigenvector of \(A\). The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)\left( {\begin{aligned}{{}}1&0&0\\0&4&0\\0&0&1\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)^{ - 1}}\).

Step-by-step solution

Verify the eigenvectors

For the matrix\(A\), the given eigenvectors\({{\rm{v}}_1}\)and\({{\rm{v}}_2}\)must satisfy the characteristic equation\(A{\bf{x}} = \lambda {\bf{x}}\).On verifying it for the vector\({{\rm{v}}_1}\), we proceed as follows:

\(\begin{aligned}{}\left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&2&{ - 1}\\{ - 1}&{ - 1}&2\end{aligned}} \right)\left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right) = \left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right)\\ = 1\left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right)\end{aligned}\)

Thus,\(\left( \begin{aligned}{} - 1\\\,\,\,0\\\,\,\,1\end{aligned} \right)\)is eigenvector of A for the eigenvalue\(\lambda = 1\). Now, verifying it for the vector\({{\rm{v}}_2}\), we proceed as follows:

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

Thus, \(\left( \begin{aligned}{}\,\,\,1\\ - 1\\\,\,\,\,1\end{aligned} \right)\) is eigenvector of A for the eigenvalue \(\lambda = 4\).

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

For the Eigen value\(\lambda = 4\), the Eigen vectors is given as\(\left( \begin{aligned}{}\,\,\,1\\ - 1\\\,\,\,1\end{aligned} \right)\).

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

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

Here, the matrix \(D\) is obtained as \(D = \left( {\begin{aligned}{{}}1&0&0\\0&4&0\\0&0&1\end{aligned}} \right)\).

Diagonalize matrix \(A\)

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

\(\begin{aligned}{}A &= PD{P^{ - 1}}\\ &= \left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)\left( {\begin{aligned}{{}}1&0&0\\0&4&0\\0&0&1\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)^{ - 1}}\end{aligned}\)

It is verified that \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\) are the eigenvector of \(A\). The orthogonal diagonalization is \(\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)\left( {\begin{aligned}{{}}1&0&0\\0&4&0\\0&0&1\end{aligned}} \right){\left( {\begin{aligned}{{}}{ - 1/\sqrt 2 }&{1/\sqrt 3 }&{1/\sqrt 6 }\\0&{ - 1/\sqrt 3 }&{2/\sqrt 6 }\\{1/\sqrt 2 }&{\,\,\,\,0}&{\,\,\,1/\sqrt 6 }\end{aligned}} \right)^{ - 1}}\).