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Found in: Page 395

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# 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 $$A$$.

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}}.

See the step by step solution

## Step 1: 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 $$\lambda = 4$$.

## Step 2: 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).

## Step 3: 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}}.