Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 7E from Chapter 5.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 7E
Chapter:
Problem:
Diagonalize the matrices in Exercises 7-20, if possible. The real eigenvalues for Exercises 11-16 and 18 are included below the matrix...
Step-by-Step Solution
Step 1
Given matrix\[A = \left[ {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right]\]We have to diagonalize the given matrix if possible.
Step 2: possibility of Diagonalization
A $ n \times n $ matrix A is diagonalizable if and only if A has $n$ linearly independent eigenvectors.
The given matrix is an upper triangular matrix. Hence, the eigenvalues of A are its diagonal values.
Step 3: Diagonal Matrix
The eigenvalues are: \[{\lambda _1} = 1,\,\,\,{\lambda _2} = - 1\]So, the diagonal matrix is:\[D = \left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right]\]
Step 4: Eigen vector for ${\lambda _1} = 1$
\[\begin{array}{l}A{\bf{x}} = \lambda {\bf{x}}\\A{\bf{x}} = {\bf{x}}\\\left( {A - I} \right){\bf{x}} = 0\\\left( {\left[ {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]} \right)\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}0&0\\6&{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\end{array}\]Choose $x_2 = t$, \[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}t\\{3t}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\3\end{array}} \right]t\end{array}\]
Step 5: Eigen vector for ${\lambda _1} = -1$
\[\begin{array}{l}A{\bf{x}} = \lambda {\bf{x}}\\A{\bf{x}} = {\bf{x}}\\\left( {A - I} \right){\bf{x}} = 0\\\left( {\left[ {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]} \right)\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}2&0\\6&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\end{array}\]Choose $x_2 = t$, \[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\t\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]t\end{array}\]
Step 6: The matrix $P$
\[\begin{array}{l}P = \left\{ {{v_1},{v_2}} \right\}\\P = \left[ {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right]\end{array}\]
Step 6: The matrix $P^{-1}$
\[\begin{array}{l}{P^{ - 1}} = \dfrac{1}{{1 \times 1 - 0}}\left[ {\begin{array}{*{20}{c}}1&0\\{ - 3}&3\end{array}} \right]\\{P^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0\\{ - 3}&3\end{array}} \right]\end{array}\]
Step 7: The matrix product $AP$
\[\begin{array}{l}AP = \left[ {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right]\\AP = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\end{array}\]
Step 8: The matrix product $PD$
\[\begin{array}{l}PD = \left[ {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right]\\PD = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\end{array}\]As matrix product AP and PD are same, the matrix A is diagonalizable
Given matrix\[A = \left[ {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right]\]We have to diagonalize the given matrix if possible.
Step 2: possibility of Diagonalization
A $ n \times n $ matrix A is diagonalizable if and only if A has $n$ linearly independent eigenvectors.
The given matrix is an upper triangular matrix. Hence, the eigenvalues of A are its diagonal values.
Step 3: Diagonal Matrix
The eigenvalues are: \[{\lambda _1} = 1,\,\,\,{\lambda _2} = - 1\]So, the diagonal matrix is:\[D = \left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right]\]
Step 4: Eigen vector for ${\lambda _1} = 1$
\[\begin{array}{l}A{\bf{x}} = \lambda {\bf{x}}\\A{\bf{x}} = {\bf{x}}\\\left( {A - I} \right){\bf{x}} = 0\\\left( {\left[ {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]} \right)\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}0&0\\6&{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\end{array}\]Choose $x_2 = t$, \[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}t\\{3t}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\3\end{array}} \right]t\end{array}\]
Step 5: Eigen vector for ${\lambda _1} = -1$
\[\begin{array}{l}A{\bf{x}} = \lambda {\bf{x}}\\A{\bf{x}} = {\bf{x}}\\\left( {A - I} \right){\bf{x}} = 0\\\left( {\left[ {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]} \right)\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}2&0\\6&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\end{array}\]Choose $x_2 = t$, \[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\t\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]t\end{array}\]
Step 6: The matrix $P$
\[\begin{array}{l}P = \left\{ {{v_1},{v_2}} \right\}\\P = \left[ {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right]\end{array}\]
Step 6: The matrix $P^{-1}$
\[\begin{array}{l}{P^{ - 1}} = \dfrac{1}{{1 \times 1 - 0}}\left[ {\begin{array}{*{20}{c}}1&0\\{ - 3}&3\end{array}} \right]\\{P^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0\\{ - 3}&3\end{array}} \right]\end{array}\]
Step 7: The matrix product $AP$
\[\begin{array}{l}AP = \left[ {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right]\\AP = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\end{array}\]
Step 8: The matrix product $PD$
\[\begin{array}{l}PD = \left[ {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right]\\PD = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\end{array}\]As matrix product AP and PD are same, the matrix A is diagonalizable