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Expert-verified Found in: Page 1 ### 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 # Consider a dynamical system $\stackrel{\mathbf{\to }}{\mathbf{x}}\mathbf{\left(}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{\right)}{\mathbf{=}}{\mathbit{A}}\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}$ with two components. The accompanying sketch shows the initial state vector ${\stackrel{\mathbf{\to }}{\mathbf{x}}}_{{\mathbf{0}}}$ and two eigen vectors $\stackrel{\mathbf{\to }}{{\mathbf{\upsilon }}_{\mathbf{1}}}{\mathbf{ }}{\mathbf{ }}{\mathbit{a}}{\mathbit{n}}{\mathbit{d}}{\mathbf{ }}{\mathbf{ }}\stackrel{\mathbf{\to }}{{\mathbf{\upsilon }}_{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}$ of A (with eigen values $\stackrel{\mathbf{\to }}{{\mathbf{\lambda }}_{\mathbf{1}}}{\mathbf{}}{\mathbf{\text{and}}}{\mathbf{}}\stackrel{\mathbf{\to }}{{\mathbf{\lambda }}_{\mathbf{2}}}$ respectively). For the given values of $\stackrel{\mathbf{\to }}{{\mathbf{\lambda }}_{\mathbf{1}}}{\mathbf{}}{\mathbf{\text{and}}}{\mathbf{}}\stackrel{\mathbf{\to }}{{\mathbf{\lambda }}_{\mathbf{2}}}$ , draw a rough trajectory. Consider the future and the past of the system. $\stackrel{\mathbf{\to }}{{\mathbf{\lambda }}_{\mathbf{1}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{,}}\stackrel{\mathbf{\to }}{{\mathbf{\lambda }}_{\mathbf{2}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{9}}$

So, the required solution is ${A}^{t}{x}_{0}=\alpha {\upsilon }_{1}+0.{9}^{t}\beta {\upsilon }_{2}$.

See the step by step solution

## Step 1: Define the eigenvector

Eigenvector: An eigenvector of${\mathbit{A}}$ is a nonzero vector ${\mathbit{v}}$ in ${{\mathbit{R}}}_{{\mathbf{n}}}$ such that ${\mathbit{A}}{\mathbit{v}}{\mathbf{=}}{\mathbit{\lambda }}{\mathbit{v}}{\mathbf{}}$, for some scalar ${\mathbf{\lambda }}$.

## Step 2: Note the given data

It is given that:

$\stackrel{\to }{{\lambda }_{1}}=1,\phantom{\rule{0ex}{0ex}}\stackrel{\to }{{\lambda }_{2}}=0.9$

Given graph is: ## Step 3: Finding the required solution

We have:

$A{\upsilon }_{1}={\upsilon }_{1}\phantom{\rule{0ex}{0ex}}A{\upsilon }_{2}=0.9{\upsilon }_{2}$

For ${x}_{0}=\alpha {\upsilon }_{1}+\beta {\upsilon }_{2}$ ,We have:

role="math" localid="1668079897383" $A{x}_{0}=A\left(\alpha {\upsilon }_{1}+\beta {\upsilon }_{2}\right)\phantom{\rule{0ex}{0ex}}=\alpha A{\upsilon }_{1}+\beta A{\upsilon }_{2}\phantom{\rule{0ex}{0ex}}=\alpha {\upsilon }_{1}+0.9\beta {\upsilon }_{2}$

Therefore, ${A}^{t}{x}_{0}=\alpha {\upsilon }_{1}+0.{9}^{t}\beta {\upsilon }_{2}$.

Hence, the solutions is ${A}^{t}{x}_{0}=\alpha {\upsilon }_{1}+0.{9}^{t}\beta {\upsilon }_{2}$. ### Want to see more solutions like these? 