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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 $$\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)$$.8. $$\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{1}}\\{\bf{0}}&{\bf{5}}\end{array}} \right)$$

The general solution is $${x_1}\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)$$. Since we cannot generate an eigenvector basis for $${\mathbb{R}^2}$$, $$A$$ is not 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 $$A = \left( {\begin{array}{*{20}{c}}5&1\\0&5\end{array}} \right)$$. 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 $${\lambda _1} = 5$$ and $${\lambda _2} = 5$$.

## Step 3: Find the matrix form of Equation for finding Eigenvector for $$\lambda {\bf{ = 5}}$$

The matrix equation for finding the eigenvector:

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

Thus, the general solution is $${x_1}\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)$$. Since we cannot generate an eigenvector basis for $${\mathbb{R}^2}$$, $$A$$ is not diagonalizable. ### Want to see more solutions like these? 