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Q5.3-9E

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 $$\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)$$.9. $$\left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right)$$

The general solution is $${{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)$$. 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 eigenvalues of the matrix

Consider the given matrix $$A = \left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right)$$. We need to diagonalize the given matrix if possible.

Write the characteristic equation:

$\begin{array}{c}\det \left( {A - \lambda I} \right) = \det \left( {\left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right) - \lambda \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\\ = \det \left( {\left( {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 1}\\1&{5 - \lambda }\end{array}} \right)} \right)\\ = \left( {3 - \lambda } \right)\left( {5 - \lambda } \right) - \left( { - 1} \right)\left( 1 \right)\\ = {\lambda ^2} - 8\lambda + 16\\ = {\left( {\lambda - 4} \right)^2}\end{array}$

Therefore, $$\det \left( {A - \lambda I} \right) = {\left( {\lambda - 4} \right)^2}$$.

## Step 3: Find the eigenvalues

$\begin{array}{c}\det \left( {A - \lambda I} \right) = 0\\{\left( {\lambda - 4} \right)^2} = 0\\\lambda = 4,4\end{array}$

So, the eigenvalue of the matrix is $$4$$ with a multiplicity of $$2$$.

## Step 4: Find the eigenvectors

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

Therefore, the eigenvectors are shown below:

$- {x_1} - {x_2} = 0$

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

So, eigenvector is $${{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)$$.

Since we cannot generate an eigenvector basis for $${\mathbb{R}^2}$$, $$A$$ is not diagonalizable.

Thus, the matrix is not-Diagonalizable.

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