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Q5.3-12E

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Found in: Page 267

### 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

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# Question: Diagonalize the matrices in Exercises $${\bf{7--20}}$$, if possible. The eigenvalues for Exercises $${\bf{11--16}}$$ are as follows:$$\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}$$; $$\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}$$; $$\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}$$; $$\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}$$; $$\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}$$; $$\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}$$. For exercise $${\bf{18}}$$, one eigenvalue is $$\lambda {\bf{ = 5}}$$ and one eigenvector is $$\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)$$.12. $$\left( {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{4}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{4}}\end{array}} \right)$$

The matrix $$A$$ is diagonalizable with $$P = \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&1\\0&1&1\\1&0&1\end{array}} \right)$$ and $$D = \left( {\begin{array}{*{20}{c}}2&0&0\\0&2&0\\0&0&8\end{array}} \right)$$.

See the step by step solution

## Step 1: Write the Diagonalization Theorem

The Diagonalization Theorem: An $$n \times n$$ matrix $$A$$ is diagonalizable if and only if $$A$$ has $$n$$ linearly independent eigenvectors. As $$A = PD{P^{ - 1}}$$ which has $$D$$ a diagonal matrix if and only if the columns of $$P$$ are $$n$$ linearly independent eigenvectors of $$A$$.

## Step 2: Find eigenvalues of the matrix

Consider the given matrix $$A = \left( {\begin{array}{*{20}{c}}4&2&2\\2&4&2\\2&2&4\end{array}} \right)$$. Since from the given matrix eigenvalues of the matrix are $$2$$ and $$8$$. Since the eigenvalues are distinct, the matrix is diagonalizable.

## Step 3: Find the eigenvalues and eigenvectors

As the sum of all the eigenvalues of $$A$$ is equal to the sum of the diagonal entries of $$A$$, we can find the third eigenvalue:

$$\begin{array}{c}2 + 8 + x = 4 + 4 + 4\\x = 12 - 10\\x = 2\end{array}$$

So, the eigenvalues are $$2,2,8$$.

Find eigenvectors.

Write the matrix form for finding the eigenvector for $$\lambda = 2$$.

$$\begin{array}{c}A - 2I = 0\\\left( {\begin{array}{*{20}{c}}4&2&2\\2&4&2\\2&2&4\end{array}} \right) - 2\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&1&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}4&2&2\\2&4&2\\2&2&4\end{array}} \right) - \left( {\begin{array}{*{20}{c}}2&0&0\\0&2&0\\0&1&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}2&2&2\\2&2&2\\2&2&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\end{array}$$

Write the Row-reduced augmented matrix.

$$\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}2&2&2&0\\2&2&2&0\\2&2&2&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2&2&2&0\\0&0&0&0\\0&0&0&0\end{array}} \right)\;\left\{ \begin{array}{l}{R_3} = {R_3} - {R_1}\\{R_2} = {R_2} - {R_1}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&1&1&0\\0&0&0&0\\0&0&0&0\end{array}} \right)\;\left\{ {{R_1} = \frac{{{R_1}}}{2}} \right\}\end{array}$$

Therefore, the parametric form of the solution is shown below:

$$\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - {x_2} - {x_3}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - {x_2}}\\{{x_2}}\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - {x_3}}\\0\\{{x_3}}\end{array}} \right)\\ = {x_2}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\end{array}$$

Therefore, the eigenvector for $$\lambda = 2$$ are $$\left\{ {{{\rm{v}}_1},{{\rm{v}}_2}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}\begin{array}{l} - 1\\1\end{array}\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}\begin{array}{l} - 1\\0\end{array}\\1\end{array}} \right)} \right\}$$.

Write the matrix form for finding the Eigenvector for $$\lambda = 8$$.

$$\begin{array}{c}A - 8I = 0\\\left( {\begin{array}{*{20}{c}}4&2&2\\2&4&2\\2&2&4\end{array}} \right) - 8\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&1&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}4&2&2\\2&4&2\\2&2&4\end{array}} \right) - \left( {\begin{array}{*{20}{c}}8&0&0\\0&8&0\\0&1&8\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{ - 4}&2&2\\2&{ - 4}&2\\2&2&{ - 4}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\end{array}$$

Write the Row-reduced Augmented matrix.

$$\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}{ - 4}&2&2&0\\2&{ - 4}&2&0\\2&2&{ - 4}&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&{ - \frac{1}{2}}&{ - \frac{1}{2}}&0\\2&{ - 4}&2&0\\2&2&{ - 4}&0\end{array}} \right)\;\left\{ {{R_1} = - \frac{1}{4}{R_1}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&{ - \frac{1}{2}}&{ - \frac{1}{2}}&0\\0&{ - 3}&3&0\\0&3&{ - 3}&0\end{array}} \right)\;\left\{ \begin{array}{l}{R_3} = {R_3} - 2{R_1}\\{R_2} = {R_2} - 2{R_1}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&{ - \frac{1}{2}}&{ - \frac{1}{2}}&0\\0&{ - 3}&3&0\\0&0&0&0\end{array}} \right)\;\left\{ {{R_3} = {R_3} + {R_2}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&{ - 1}&0\\0&1&{ - 1}&0\\0&0&0&0\end{array}} \right)\;\left\{ \begin{array}{l}{R_2} = \frac{1}{6}{R_2}\\{R_1} = {R_1} - {R_2}\\{R_2} = - 2R2\end{array} \right\}\end{array}$$

Therefore, the parametric form of the solution is shown below:

$$\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{x_3}}\\{{x_3}}\\{{x_3}}\end{array}} \right)\\ = {x_3}\left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\end{array}$$

Therefore, the eigenvector for $$\lambda = 8$$ are $$\left\{ {{{\rm{v}}_3}} \right\} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)$$.

The determinant is shown below:

$$\begin{array}{c}{\rm{Det}}\left( {{v_1},{v_2},{v_3}} \right) = \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&1\\0&1&1\\1&0&1\end{array}} \right)\\ = - 1\end{array}$$

## Step 4: Find the matrix $$P$$ and $$D$$

The matrix $$P$$ is formed by eigenvectors

$$P = \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&1\\0&1&1\\1&0&1\end{array}} \right)$$

The matrix $$D$$ is formed by eigenvalues.

$$D = \left( {\begin{array}{*{20}{c}}2&0&0\\0&2&0\\0&0&8\end{array}} \right)$$

## Step 5: Find diagonalizes form of a matrix $$A$$

As the diagonal form of the matrix $$A$$ is $$A = PD{P^{ - 1}}$$.

Where $$P = \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&1\\0&1&1\\1&0&1\end{array}} \right)$$ and $$D = \left( {\begin{array}{*{20}{c}}2&0&0\\0&2&0\\0&0&8\end{array}} \right)$$.

Thus, the matrix $$A$$ is diagonalizable with $$P = \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&1\\0&1&1\\1&0&1\end{array}} \right)$$ and $$D = \left( {\begin{array}{*{20}{c}}2&0&0\\0&2&0\\0&0&8\end{array}} \right)$$.

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