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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 $$\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)$$.7. $$\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{6}}&{{\bf{ - 1}}}\end{array}} \right)$$

As the matrix product $$AP$$ and $$PD$$ are same, that is, the matrix $$A$$ is diagonalizable.

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 the inverse of the invertible matrix

Consider the given matrix $$A = \left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right)$$. We need to diagonalize the given matrix if possible.

As the given matrix is an upper triangular matrix. Therefore, the eigenvalues of $$A$$ are its diagonal values.

The eigenvalues are $${\lambda _1} = 1$$ and $${\lambda _2} = - 1$$.

Thus, the diagonal matrix is $$D = \left( {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right)$$.

## Step 3: Find Eigenvectors

Eigenvectors for $${\lambda _1} = 1$$,

$\begin{array}{c}A{\bf{x}} = \lambda {\bf{x}}\\A{\bf{x}} = {\bf{x}}\\\left( {A - I} \right){\bf{x}} = 0\\\left( {\left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}0&0\\6&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}$

Assume $${x_2} = t$$ then we get:

$\begin{array}{l}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}t\\{3t}\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1\\3\end{array}} \right)t\end{array}$

Eigenvectors for $${\lambda _1} = - 1$$,

$\begin{array}{c}A{\bf{x}} = \lambda {\bf{x}}\\A{\bf{x}} = {\bf{x}}\\\left( {A - I} \right){\bf{x}} = 0\\\left( {\left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}2&0\\6&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}$

Assume $${x_2} = t$$ then we get:

$\begin{array}{l}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\t\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)t\end{array}$

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

$\begin{array}{c}P = \left\{ {{v_1},{v_2}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\end{array}$

## Step 5: Find the matrix $${P^{ - {\bf{1}}}}$$

$\begin{array}{l}{P^{ - 1}} = \frac{1}{{1\left( 1 \right) - 0}}\left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&3\end{array}} \right)\\{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&3\end{array}} \right)\end{array}$

## Step 6: Find the matrix product $$AP$$

$\begin{array}{c}AP = \left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right)\end{array}$

## Step 7: Find the matrix product $$PD$$

$\begin{array}{c}PD = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right)\end{array}$

Thus, as matrix product $$AP$$ and $$PD$$ are same, that is, the matrix $$A$$ is diagonalizable. ### Want to see more solutions like these? 