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Q26E

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Found in: Page 336

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# 26: Based on your answers in Exercises 24 and 25, sketch a phase portrait of the dynamical system$\overline{\mathbf{x}}\left(t+1\right){\mathbf{=}}\left[\begin{array}{cc}0.5& 0.25\\ 0.5& 0.75\end{array}\right]\overline{\mathbf{x}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}$

Phase portrait of a dynamical system is: ${A}^{t}{x}_{0}=\left[\begin{array}{c}0.{25}^{t}\alpha +0.25\beta \\ -0.{25}^{t}\alpha +0.5\beta \end{array}\right]$

See the step by step solution

## Step 1: Transition Matrix

Transition matrix may refer to: The matrix associated with a change of basis for a vector space. Stochastic matrix, a square matrix used to describe the transitions of a Markov chain. State-transition matrix, a matrix whose product with the state vector at an initial time gives state vector at that time.

## Step 2: Sketching a phase portrait of the dynamical system:

As we clearly now that,

${v}_{1}=\left[\begin{array}{c}1\\ -1\end{array}\right]\phantom{\rule{0ex}{0ex}}{v}_{2}=\left[\begin{array}{c}0.25\\ 0.5\end{array}\right]$

role="math" localid="1659584611493" ${x}_{0}=\alpha \left[\begin{array}{c}1\\ -1\end{array}\right]+\beta \left[\begin{array}{c}0.25\\ 0.5\end{array}\right]\phantom{\rule{0ex}{0ex}}{A}^{t}{x}_{0}=0.{25}^{t}\alpha \left[\begin{array}{c}1\\ -1\end{array}\right]+\beta \left[\begin{array}{c}0.25\\ 0.5\end{array}\right]=\left[\begin{array}{c}0.{25}^{t}\alpha +0.25\beta \\ -0.{25}^{t}\alpha +0.5\beta \end{array}\right]$

Hence, the final answer is: ${A}^{t}{x}_{0}=\left[\begin{array}{c}0.{25}^{t}\alpha +0.25\beta \\ -0.{25}^{t}\alpha +0.5\beta \end{array}\right]$