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Q5.3-10E

Expert-verified
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

# 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 $$\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)$$.10. $$\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{3}}\\{\bf{4}}&{\bf{1}}\end{array}} \right)$$

The matrix $$A$$ is diagonalizable with $$P = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)$$, $${P^{ - 1}} = \left( {\begin{array}{*{20}{c}}{ - \frac{4}{7}}&{\frac{4}{7}}\\{\frac{4}{7}}&{\frac{3}{7}}\end{array}} \right)$$ and $$D = \left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&5\end{array}} \right)$$.

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 $$A = \left( {\begin{array}{*{20}{c}}2&3\\4&1\end{array}} \right)$$. We need to diagonalize the given matrix if possible.

Write the characteristic equation:

$\begin{array}{c}\left| {A - \lambda I} \right| = 0\\\left| {\begin{array}{*{20}{c}}{2 - \lambda }&3\\4&{1 - \lambda }\end{array}} \right| = 0\\\left( {\lambda + 2} \right)\left( {\lambda - 5} \right) = 0\\\lambda = - 2,5\end{array}$

## Step 3: Find the eigenvectors

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

$\begin{array}{c}\left( {A - \lambda I} \right){\bf{x}} = 0\\\left( {A - \left( { - 2} \right)I} \right){\bf{x}} = 0\\\left( {\left( {\begin{array}{*{20}{c}}2&3\\4&1\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}4&3\\4&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}$

Therefore, the augmented matrix is shown below:

$\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}4&3&0\\4&3&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4&3&0\\4&3&0\end{array}} \right)\;\left\{ {{R_2} = {R_2} - {R_1}} \right\}\end{array}$

Since there are $$2$$ variables and $$1$$ equitation, consider $${x_2}$$ as free.

$\begin{array}{c}4{x_1} + 3{x_2} = 0\\4{x_1} = - 3{x_2}\\{x_1} = - \frac{3}{4}{x_2}\end{array}$

Write the parametric form of solution.

$\begin{array}{c}{\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\\ = {x_2}\left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}\\1\end{array}} \right)\end{array}$

Therefore, the eigenvector for $$\lambda = - 2$$ is $${\rm{v}} = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}\\1\end{array}} \right)$$.

Write the matrix form for finding the eigenvector for $$\lambda = 5$$.

$\begin{array}{c}\left( {A - \lambda I} \right){\bf{x}} = 0\\\left( {A - \left( 5 \right)I} \right){\bf{x}} = 0\\\left( {\left( {\begin{array}{*{20}{c}}2&3\\4&1\end{array}} \right) - 5\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{ - 3}&3\\4&{ - 4}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}$

Therefore, the augmented matrix is shown below:

$\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}{ - 3}&3&0\\4&{ - 4}&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&{ - 1}&0\\1&{ - 1}&0\end{array}} \right)\;\left\{ {{R_1} = - \frac{{{R_1}}}{3},{R_2} = - \frac{{{R_2}}}{3}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&{ - 1}&0\\0&0&0\end{array}} \right)\;\left\{ {{R_1} = {R_2} - {R_1}} \right\}\end{array}$

Since there are $$2$$ variables and $$1$$ equitation, consider $${x_2}$$ as free.

$\begin{array}{c}{x_1} - {x_2} = 0\\{x_1} = {x_2}\end{array}$

Write the parametric form of solution.

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

Therefore, the eigenvector for $$\lambda = 5$$ is $$v = \left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right)$$.

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

The matrix $$P$$ is formed by eigenvectors

$P = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)$

The inverse of the matrix $$P$$ is shown below:

$\begin{array}{c}{P^{ - 1}} = {\left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)^{ - 1}}\\ = \frac{1}{{1\left( { - \frac{3}{4}} \right) - 1}}\left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&{ - \frac{3}{4}}\end{array}} \right)\\ = - \frac{4}{7}\left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&{ - \frac{3}{4}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - \frac{4}{7}}&{\frac{4}{7}}\\{\frac{4}{7}}&{\frac{3}{7}}\end{array}} \right)\end{array}$

## Step 5: Find the matrix $$D$$

As the matrix that diagonalizes $$A$$is shown below:

$\begin{array}{c}D = {P^{ - 1}}AP\\ = \left( {\begin{array}{*{20}{c}}{ - \frac{4}{7}}&{\frac{4}{7}}\\{\frac{4}{7}}&{\frac{3}{7}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&3\\4&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\frac{8}{7}}&{ - \frac{8}{7}}\\{\frac{{20}}{7}}&{\frac{{15}}{7}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&5\end{array}} \right)\end{array}$

Thus, the matrix $$A$$ is diagonalizable with $$P = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}}&1\\1&1\end{array}} \right)$$, $${P^{ - 1}} = \left( {\begin{array}{*{20}{c}}{ - \frac{4}{7}}&{\frac{4}{7}}\\{\frac{4}{7}}&{\frac{3}{7}}\end{array}} \right)$$ and $$D = \left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&5\end{array}} \right)$$.