Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
We have solutions for your book!
See our solution for Question 1E from Chapter 5.6 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 1E
Chapter:
Problem:
Let A be a 2 × 2 matrix with eigenvalues 3 and 1/3 and Let be a solution of the difference equation a. Compute x1 = Ax0. [Hint: You do not need to know A itself.] b. Find a formula for xk involving k and the eigenvectors v1 and v2.
Step-by-Step Solution
Given Information
We are given that A is a $2 \times 2$ matrix with eigenvalues as 3 and 1/3. The corresponding eigenvectors are \[ \mathbf { v } _ { 1 } = \left[ \begin{array} { l } { 1 } \\ { 1 } \end{array} \right] \text { and } \mathbf { v } _ { 2 } = \left[ \begin{array} { r } { - 1 } \\ { 1 } \end{array} \right] \] If $x_k$ is a solution of the difference equation $\mathbf { x } _ { k + 1 } = A \mathbf { x } _ { k }$, we have to compute $x_1$ and a formula for $x_k$
Step-1:
The initial vector is given by: \[ \mathbf { x } _ { 0 } = \left[ \begin{array} { l } { 9 } \\ { 1 } \end{array} \right] \] Write $x_0$ as combination of vectors $v_1$ and $v_2$ \[\begin{array}{l} {{\bf{x}}_0} = {c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2}\\ \left[ {\begin{array}{*{20}{l}} 9\\ 1 \end{array}} \right] = {c_1}\left[ {\begin{array}{*{20}{l}} 1\\ 1 \end{array}} \right] + {c_2}\left[ {\begin{array}{*{20}{l}} { - 1}\\ 1 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{l}} 9\\ 1 \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {{c_1} - {c_2}}\\ {{c_1} + {c_2}} \end{array}} \right] \end{array}\] Upon solving: $c _ { 1 } - c _ { 2 } = 9 \text { and } c _ { 1 } + c _ { 2 } = 1$ we get, \[ c _ { 1 } = 5 \text { and } c _ { 2 } = - 4 \] Therefore, \[ \mathbf { x } _ { 0 } = 5 \mathbf { v } _ { 1 } - 4 \mathbf { v } _ { 2 } \]
Step-2:
Since $v_1$ and $v_2$ are eigenvalues of A, hence they satisfy the equations \[ [ A - 3 I ] \mathbf { v } _ { 1 } = \mathbf { 0 } \text { and } \left[ A - \dfrac { 1 } { 3 } I \right] \mathbf { v } _ { 2 } = \mathbf { 0 } \] Thus, \[ A \mathbf { v } _ { 1 } = 3 \mathbf { v } _ { 1 } \text { and } A \mathbf { v } _ { 2 } = \dfrac { 1 } { 3 } \mathbf { v } _ { 2 } \]
Step-3:
Multiply $x_0$ by A: \[ \begin{array} { l } { A \mathbf { x } _ { 0 } = 5 A \mathbf { v } _ { 1 } - 4 A \mathbf { v } _ { 2 } } \\ { \mathbf { x } _ { 1 } = 5 \cdot 3 \cdot \mathbf { v } _ { 1 } - 4 \cdot \left( \dfrac { 1 } { 3 } \right) \cdot \mathbf { v } _ { 2 } } \\ { \mathbf { x } _ { 1 } = 15 \mathbf { v } _ { 1 } - \left( \dfrac { 4 } { 3 } \right) \cdot \mathbf { v } _ { 2 } } \end{array} \] Upon substituting $v_1$ and $v_2$: \[\begin{array}{l} {{\bf{x}}_1} = 15\left[ {\begin{array}{*{20}{l}} 1\\ 1 \end{array}} \right] - \left( {\dfrac{4}{3}} \right)\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {15}\\ {15} \end{array}} \right] - \left[ {\begin{array}{*{20}{r}} { - \dfrac{4}{3}}\\ {\dfrac{4}{3}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {15 + \dfrac{4}{3}}\\ {15 - \dfrac{4}{3}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {\dfrac{{49}}{3}}\\ {\dfrac{{41}}{3}} \end{array}} \right] \end{array}\] The required vector $x_1$ is:
\[{{\bf{x}}_1} = \left[ {\begin{array}{*{20}{l}} {\frac{{49}}{3}}\\ {\frac{{41}}{3}} \end{array}} \right]\]
Step-4:
Solve for $x_2$ \[ \begin{array} { l } { \mathbf { x } _ { 2 } = A \mathbf { x } _ { 1 } } \\ { = A \left( 15 \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } \mathbf { v } _ { 2 } \right) } \\ { = 15 A \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } A \mathbf { v } _ { 2 } } \\ { = 15 \cdot 3 \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } \cdot \dfrac { 1 } { 3 } \cdot \mathbf { v } _ { 2 } } \end{array} \] Solve for $x_3$ \[ \begin{array} { l } { \mathbf { x } _ { 3 } = A \mathbf { x } _ { 2 } } \\ { = A \left( 15 \cdot 3 \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } \cdot \dfrac { 1 } { 3 } \cdot \mathbf { v } _ { 2 } \right) } \\ { = 15 \cdot 3 \cdot A \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } \cdot \dfrac { 1 } { 3 } \cdot A \mathbf { v } _ { 2 } } \\ { = 15 \cdot 3 \cdot 3 \cdot \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } \cdot \dfrac { 1 } { 3 } \cdot \dfrac { 1 } { 3 } \cdot \mathbf { v } _ { 2 } } \\ { = 5 \cdot 3 ^ { 3 } \cdot \mathbf { v } _ { 1 } - 4 \cdot \left( \dfrac { 1 } { 3 } \right) ^ { 3 } \cdot \mathbf { v } _ { 2 } } \end{array} \] Thus, the general formula for $x_k$ is:
\[ \mathbf { x } _ { k } = 5 \cdot 3 ^ { k } \cdot \mathbf { v } _ { 1 } - 4 \cdot \left( \dfrac { 1 } { 3 } \right) ^ { k } \cdot \mathbf { v } _ { 2 } \]
We are given that A is a $2 \times 2$ matrix with eigenvalues as 3 and 1/3. The corresponding eigenvectors are \[ \mathbf { v } _ { 1 } = \left[ \begin{array} { l } { 1 } \\ { 1 } \end{array} \right] \text { and } \mathbf { v } _ { 2 } = \left[ \begin{array} { r } { - 1 } \\ { 1 } \end{array} \right] \] If $x_k$ is a solution of the difference equation $\mathbf { x } _ { k + 1 } = A \mathbf { x } _ { k }$, we have to compute $x_1$ and a formula for $x_k$
Step-1:
The initial vector is given by: \[ \mathbf { x } _ { 0 } = \left[ \begin{array} { l } { 9 } \\ { 1 } \end{array} \right] \] Write $x_0$ as combination of vectors $v_1$ and $v_2$ \[\begin{array}{l} {{\bf{x}}_0} = {c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2}\\ \left[ {\begin{array}{*{20}{l}} 9\\ 1 \end{array}} \right] = {c_1}\left[ {\begin{array}{*{20}{l}} 1\\ 1 \end{array}} \right] + {c_2}\left[ {\begin{array}{*{20}{l}} { - 1}\\ 1 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{l}} 9\\ 1 \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {{c_1} - {c_2}}\\ {{c_1} + {c_2}} \end{array}} \right] \end{array}\] Upon solving: $c _ { 1 } - c _ { 2 } = 9 \text { and } c _ { 1 } + c _ { 2 } = 1$ we get, \[ c _ { 1 } = 5 \text { and } c _ { 2 } = - 4 \] Therefore, \[ \mathbf { x } _ { 0 } = 5 \mathbf { v } _ { 1 } - 4 \mathbf { v } _ { 2 } \]
Step-2:
Since $v_1$ and $v_2$ are eigenvalues of A, hence they satisfy the equations \[ [ A - 3 I ] \mathbf { v } _ { 1 } = \mathbf { 0 } \text { and } \left[ A - \dfrac { 1 } { 3 } I \right] \mathbf { v } _ { 2 } = \mathbf { 0 } \] Thus, \[ A \mathbf { v } _ { 1 } = 3 \mathbf { v } _ { 1 } \text { and } A \mathbf { v } _ { 2 } = \dfrac { 1 } { 3 } \mathbf { v } _ { 2 } \]
Step-3:
Multiply $x_0$ by A: \[ \begin{array} { l } { A \mathbf { x } _ { 0 } = 5 A \mathbf { v } _ { 1 } - 4 A \mathbf { v } _ { 2 } } \\ { \mathbf { x } _ { 1 } = 5 \cdot 3 \cdot \mathbf { v } _ { 1 } - 4 \cdot \left( \dfrac { 1 } { 3 } \right) \cdot \mathbf { v } _ { 2 } } \\ { \mathbf { x } _ { 1 } = 15 \mathbf { v } _ { 1 } - \left( \dfrac { 4 } { 3 } \right) \cdot \mathbf { v } _ { 2 } } \end{array} \] Upon substituting $v_1$ and $v_2$: \[\begin{array}{l} {{\bf{x}}_1} = 15\left[ {\begin{array}{*{20}{l}} 1\\ 1 \end{array}} \right] - \left( {\dfrac{4}{3}} \right)\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {15}\\ {15} \end{array}} \right] - \left[ {\begin{array}{*{20}{r}} { - \dfrac{4}{3}}\\ {\dfrac{4}{3}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {15 + \dfrac{4}{3}}\\ {15 - \dfrac{4}{3}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {\dfrac{{49}}{3}}\\ {\dfrac{{41}}{3}} \end{array}} \right] \end{array}\] The required vector $x_1$ is:
\[{{\bf{x}}_1} = \left[ {\begin{array}{*{20}{l}} {\frac{{49}}{3}}\\ {\frac{{41}}{3}} \end{array}} \right]\]
Step-4:
Solve for $x_2$ \[ \begin{array} { l } { \mathbf { x } _ { 2 } = A \mathbf { x } _ { 1 } } \\ { = A \left( 15 \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } \mathbf { v } _ { 2 } \right) } \\ { = 15 A \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } A \mathbf { v } _ { 2 } } \\ { = 15 \cdot 3 \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } \cdot \dfrac { 1 } { 3 } \cdot \mathbf { v } _ { 2 } } \end{array} \] Solve for $x_3$ \[ \begin{array} { l } { \mathbf { x } _ { 3 } = A \mathbf { x } _ { 2 } } \\ { = A \left( 15 \cdot 3 \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } \cdot \dfrac { 1 } { 3 } \cdot \mathbf { v } _ { 2 } \right) } \\ { = 15 \cdot 3 \cdot A \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } \cdot \dfrac { 1 } { 3 } \cdot A \mathbf { v } _ { 2 } } \\ { = 15 \cdot 3 \cdot 3 \cdot \mathbf { v } _ { 1 } - \dfrac { 4 } { 3 } \cdot \dfrac { 1 } { 3 } \cdot \dfrac { 1 } { 3 } \cdot \mathbf { v } _ { 2 } } \\ { = 5 \cdot 3 ^ { 3 } \cdot \mathbf { v } _ { 1 } - 4 \cdot \left( \dfrac { 1 } { 3 } \right) ^ { 3 } \cdot \mathbf { v } _ { 2 } } \end{array} \] Thus, the general formula for $x_k$ is:
\[ \mathbf { x } _ { k } = 5 \cdot 3 ^ { k } \cdot \mathbf { v } _ { 1 } - 4 \cdot \left( \dfrac { 1 } { 3 } \right) ^ { k } \cdot \mathbf { v } _ { 2 } \]