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 29E from Chapter 5.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 29E
Chapter:
Problem:
A factorization A D PDP–1 is not unique. Demonstrate this for the matrix A in Example 2. With Use the information in Example 2 to find a matrix P1 such that Reference: Find a formula for Ak, given that A D PDP–1, where
Step-by-Step Solution
Given Information
We have to explain that the factorization $A = P D P ^ { - 1 } $A D PDP 1 is not unique by using the matrices given below: \[ A = \left[ \begin{array} { c c } { 7 } & { 2 } \\ { - 4 } & { 1 } \end{array} \right] , P = \left[ \begin{array} { c c } { 1 } & { 1 } \\ { - 1 } & { - 2 } \end{array} \right] \text { and } D = \left[ \begin{array} { c c } { 5 } & { 0 } \\ { 0 } & { 3 } \end{array} \right] \]
Step-1:
For eigenvalue of 5, the eigenvector is $\left[ \begin{array} { c } { 1 } \\ { - 1 } \end{array} \right]$ and for eigenvalue of 3, the eigenvector is $\left[ \begin{array} { l } { 1 } \\ { - 2 } \end{array} \right]$.
We have to find another matrix $P_1$ such that $A = P _ { 1 } D _ { 1 } P _ { 1 } ^ { - 1 }$. By reversing the diagonal entris of D, we get another matrix $D_1$ as: \[ D _ { 1 } = \left[ \begin{array} { l l } { 3 } & { 0 } \\ { 0 } & { 5 } \end{array} \right] \] Also, interchange the columns of P to get another matrix $P_1$ that has eigenvectors corresponding to eigenvalues of D. \[ P _ { 1 } = \left[ \begin{array} { c c } { 1 } & { 1 } \\ { - 2 } & { - 1 } \end{array} \right] \]
Step-2:
Find the product $P _ { 1 } D _ { 1 } P _ { 1 } ^ { - 1 }$ \[\begin{array}{l} {P_1}{D_1}P_1^{ - 1} = \left[ {\begin{array}{*{20}{c}} 1&1\\ { - 2}&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&0\\ 0&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1}\\ 2&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 7&2\\ { - 4}&1 \end{array}} \right] \end{array}\] Therefore,
\[ P _ { 1 } = \left[ \begin{array} { c c } { 1 } & { 1 } \\ { - 2 } & { - 1 } \end{array} \right] \]
We have to explain that the factorization $A = P D P ^ { - 1 } $A D PDP 1 is not unique by using the matrices given below: \[ A = \left[ \begin{array} { c c } { 7 } & { 2 } \\ { - 4 } & { 1 } \end{array} \right] , P = \left[ \begin{array} { c c } { 1 } & { 1 } \\ { - 1 } & { - 2 } \end{array} \right] \text { and } D = \left[ \begin{array} { c c } { 5 } & { 0 } \\ { 0 } & { 3 } \end{array} \right] \]
Step-1:
For eigenvalue of 5, the eigenvector is $\left[ \begin{array} { c } { 1 } \\ { - 1 } \end{array} \right]$ and for eigenvalue of 3, the eigenvector is $\left[ \begin{array} { l } { 1 } \\ { - 2 } \end{array} \right]$.
We have to find another matrix $P_1$ such that $A = P _ { 1 } D _ { 1 } P _ { 1 } ^ { - 1 }$. By reversing the diagonal entris of D, we get another matrix $D_1$ as: \[ D _ { 1 } = \left[ \begin{array} { l l } { 3 } & { 0 } \\ { 0 } & { 5 } \end{array} \right] \] Also, interchange the columns of P to get another matrix $P_1$ that has eigenvectors corresponding to eigenvalues of D. \[ P _ { 1 } = \left[ \begin{array} { c c } { 1 } & { 1 } \\ { - 2 } & { - 1 } \end{array} \right] \]
Step-2:
Find the product $P _ { 1 } D _ { 1 } P _ { 1 } ^ { - 1 }$ \[\begin{array}{l} {P_1}{D_1}P_1^{ - 1} = \left[ {\begin{array}{*{20}{c}} 1&1\\ { - 2}&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&0\\ 0&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1}\\ 2&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 7&2\\ { - 4}&1 \end{array}} \right] \end{array}\] Therefore,
\[ P _ { 1 } = \left[ \begin{array} { c c } { 1 } & { 1 } \\ { - 2 } & { - 1 } \end{array} \right] \]