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 18E from Chapter 5.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 18E
Chapter:
Problem:
0
Step-by-Step Solution
Given Information
Given the matrix: \[A = \left[ {\begin{array}{*{20}{r}} 1&{ - 1}\\ 4&{.6} \end{array}} \right]\] We have find an invertible matrix $P$ and a matrix $C$ of the form $\left[ \begin{array} { c c } { a } & { - b } \\ { b } & { a } \end{array} \right]$ such that the given matrix has the form $A = P C P ^ { - 1 } .$
Step-1:
Write the characteristic equation: \[ \begin{array} { r } { \operatorname { det } ( A - \lambda I ) = 0 } \\ { \left| \begin{array} { c c } { 1 - \lambda } & { - 1 } \\ { 0.4 } & { 0.6 - \lambda } \end{array} \right| = 0 } \\ { \left( 0.6 - 0.6 \lambda - \lambda + \lambda ^ { 2 } \right) + 0.4 = 0 } \\ { \lambda ^ { 2 } - 1.6 \lambda + 1 = 0 } \end{array} \] The eigenvalues of A \[ \begin{aligned} \lambda & = \dfrac { 1.6 \pm \sqrt { 2.56 - 4 } } { 2 } \\ & = 0.8 \pm 0.6 i \end{aligned} \] Hence, by theorem-9, the matrix C is given by
\[ C = \left[ \begin{array} { l l } { 0.8 } & { - 0.6 } \\ { 0.6 } & { 0.8 } \end{array} \right] \]
Step-2: Eigenvector for $\lambda = 0.8 - 0.6i$
Write the matrix equation: \[ \begin{array} { r } { ( A - \lambda I ) \mathbf { v } = \mathbf { 0 } } \\ { [ A - ( 0.8 - 0.6 i ) I ] \mathbf { v } = \mathbf { 0 } } \\ { \left[ \begin{array} { c c } { 0.2 + 0.6 i } & { - 1 } \\ { 0.4 } & { - 0.2 + 0.6 i } \end{array} \right] \left[ \begin{array} { c } { v _ { 1 } } \\ { v _ { 2 } } \end{array} \right] = \mathbf { 0 } } \\ { \left[ \begin{array} { c } { ( 0.2 + 0.6 i ) v _ { 1 } - v _ { 2 } } \\ { 0.4 v _ { 1 } + ( - 0.2 + 0.6 i ) v _ { 2 } } \end{array} \right] = \left[ \begin{array} { l } { 0 } \\ { 0 } \end{array} \right] } \end{array} \] From the equation: $v _ { 2 } = ( .2 + .6 i ) v _ { 1 }$, the eigenvector is:
\[ \mathbf { v } = \left[ \begin{array} { c } { 5 } \\ { 1 + 3 i } \end{array} \right] \]
Step-3: The matrix P
The matrix is formed by the eigenvectors \[ \begin{aligned} P & = \left[ \begin{array} { l l } { \operatorname { Re } \mathbf { v } } & { \operatorname { Im } \mathbf { v } } \end{array} \right] \\ & = \left[ \begin{array} { l l } { 5 } & { 0 } \\ { 1 } & { 3 } \end{array} \right] \end{aligned} \]
Given the matrix: \[A = \left[ {\begin{array}{*{20}{r}} 1&{ - 1}\\ 4&{.6} \end{array}} \right]\] We have find an invertible matrix $P$ and a matrix $C$ of the form $\left[ \begin{array} { c c } { a } & { - b } \\ { b } & { a } \end{array} \right]$ such that the given matrix has the form $A = P C P ^ { - 1 } .$
Step-1:
Write the characteristic equation: \[ \begin{array} { r } { \operatorname { det } ( A - \lambda I ) = 0 } \\ { \left| \begin{array} { c c } { 1 - \lambda } & { - 1 } \\ { 0.4 } & { 0.6 - \lambda } \end{array} \right| = 0 } \\ { \left( 0.6 - 0.6 \lambda - \lambda + \lambda ^ { 2 } \right) + 0.4 = 0 } \\ { \lambda ^ { 2 } - 1.6 \lambda + 1 = 0 } \end{array} \] The eigenvalues of A \[ \begin{aligned} \lambda & = \dfrac { 1.6 \pm \sqrt { 2.56 - 4 } } { 2 } \\ & = 0.8 \pm 0.6 i \end{aligned} \] Hence, by theorem-9, the matrix C is given by
\[ C = \left[ \begin{array} { l l } { 0.8 } & { - 0.6 } \\ { 0.6 } & { 0.8 } \end{array} \right] \]
Step-2: Eigenvector for $\lambda = 0.8 - 0.6i$
Write the matrix equation: \[ \begin{array} { r } { ( A - \lambda I ) \mathbf { v } = \mathbf { 0 } } \\ { [ A - ( 0.8 - 0.6 i ) I ] \mathbf { v } = \mathbf { 0 } } \\ { \left[ \begin{array} { c c } { 0.2 + 0.6 i } & { - 1 } \\ { 0.4 } & { - 0.2 + 0.6 i } \end{array} \right] \left[ \begin{array} { c } { v _ { 1 } } \\ { v _ { 2 } } \end{array} \right] = \mathbf { 0 } } \\ { \left[ \begin{array} { c } { ( 0.2 + 0.6 i ) v _ { 1 } - v _ { 2 } } \\ { 0.4 v _ { 1 } + ( - 0.2 + 0.6 i ) v _ { 2 } } \end{array} \right] = \left[ \begin{array} { l } { 0 } \\ { 0 } \end{array} \right] } \end{array} \] From the equation: $v _ { 2 } = ( .2 + .6 i ) v _ { 1 }$, the eigenvector is:
\[ \mathbf { v } = \left[ \begin{array} { c } { 5 } \\ { 1 + 3 i } \end{array} \right] \]
Step-3: The matrix P
The matrix is formed by the eigenvectors \[ \begin{aligned} P & = \left[ \begin{array} { l l } { \operatorname { Re } \mathbf { v } } & { \operatorname { Im } \mathbf { v } } \end{array} \right] \\ & = \left[ \begin{array} { l l } { 5 } & { 0 } \\ { 1 } & { 3 } \end{array} \right] \end{aligned} \]