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Found in: Page 440

### Linear Algebra With Applications

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

# For the linear system$\frac{\mathbf{d}\stackrel{\mathbf{\to }}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}\left[\begin{array}{cc}3& 0\\ -2.5& 0.5\end{array}\right]\stackrel{\mathbf{\to }}{\mathbf{x}}\left(t\right)$ find the matching phase portrait.

The phase portrait corresponds to I.

See the step by step solution

## Step1: To find the eigenvalues

Consider the linear system as follows.

$\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}3& 0\\ -2.5& 0.5\end{array}\right]\stackrel{\to }{x}\left(t\right)$

To find the eigenvalues, evaluate $\left|A-\lambda I\right|=0$ as follows.

$\left|A-\lambda I\right|=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}3-\lambda & 0\\ -2.5& 0.5-\lambda \end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left(3-\lambda \right)\left(0.5-\lambda \right)-0\left(-2.5\right)=0\phantom{\rule{0ex}{0ex}}1.5-3\lambda -0.5\lambda +{\lambda }^{2}-0=0$

Simplify further as follows.

${\lambda }^{2}-3.5\lambda +1.5=0\phantom{\rule{0ex}{0ex}}{\lambda }^{2}-3\lambda -0.5\lambda +1.5=0\phantom{\rule{0ex}{0ex}}\lambda \left(\lambda -3\right)-0.5\left(\lambda -3\right)=0\phantom{\rule{0ex}{0ex}}\left(\lambda -3\right)\left(\lambda -0.5\right)=0\phantom{\rule{0ex}{0ex}}$

Simplify further as follows.

$\lambda -3=0\phantom{\rule{0ex}{0ex}}{\lambda }_{1}=3\phantom{\rule{0ex}{0ex}}\lambda -0.5=0\phantom{\rule{0ex}{0ex}}{\lambda }_{2}=0.5$

Therefore, the eigenvalues are ${\lambda }_{1}=3$ and ${\lambda }_{2}=0.5$.

## To find the eigenvector for λ1=3

Now, to find the corresponding eigenvector for the eigenvalue ${\lambda }_{1}=3$as follows.

localid="1659888265593" $\left[A-3\right]\stackrel{\to }{a}=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}3-3& 0\\ -2.5& 0.5-3\end{array}\right]\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}0& 0\\ -2.5& 2.5\end{array}\right]\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{c}0{a}_{1}+0{a}_{2}\\ -2.5{a}_{1}-2.5{a}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}$

The corresponding equations are follows.

$0{a}_{1}+0{a}_{2}=0\phantom{\rule{0ex}{0ex}}-2.5{a}_{1}-2.5{a}_{2}=0$

Now, solving the second equation as follows.

$-2.5{a}_{1}-2.5{a}_{2}=0\phantom{\rule{0ex}{0ex}}-2.5{a}_{1}=2.5{a}_{2}\phantom{\rule{0ex}{0ex}}{a}_{1}=\frac{-2.5{a}_{2}}{2.5}\phantom{\rule{0ex}{0ex}}{a}_{1}=-{a}_{2}\phantom{\rule{0ex}{0ex}}$

Therefore, the eigenvectors as follows.

$\stackrel{\to }{a}=\left[{a}_{1}\phantom{\rule{0ex}{0ex}}{a}_{2}\right]\phantom{\rule{0ex}{0ex}}=\left[-{a}_{2}\phantom{\rule{0ex}{0ex}}{a}_{2}\right] \left\{Q{a}_{1}=-{a}_{2}\right\}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}={a}_{2}\left[-1\phantom{\rule{0ex}{0ex}}1\right]\phantom{\rule{0ex}{0ex}}$

## Step3: To find the eigenvector for λ1=0.5

Now, to find the corresponding eigenvector for the eigenvalue ${\lambda }_{1}=0.5$ as follows.

$\left[A-3\right]\stackrel{\to }{b}=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}3-0.5& 0\\ -2.5& 0.5-0.5\end{array}\right]\left[\begin{array}{c}{b}_{1}\\ {b}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}2.5& 0\\ -2.5& 0\end{array}\right]\left[\begin{array}{c}{b}_{1}\\ {b}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{c}2.5{b}_{1}+0{b}_{2}\\ -2.5{b}_{1}-0{b}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}$

The corresponding equations are follows.

$2.5{b}_{1}+0{b}_{2}=0\phantom{\rule{0ex}{0ex}}-2.5{b}_{1}-0{b}_{2}=0$

Now, solving the second equation as follows.

$-2.5{b}_{1}-0{b}_{2}=0\phantom{\rule{0ex}{0ex}}-2.5{b}_{1}=0{b}_{2}\phantom{\rule{0ex}{0ex}}{b}_{1}=0\phantom{\rule{0ex}{0ex}}{b}_{1}=0$

Therefore, the eigenvectors as follows.

$\stackrel{\to }{b}=\left[\begin{array}{c}{b}_{1}\\ {b}_{2}\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[0\phantom{\rule{0ex}{0ex}}{b}_{2}\right] \left\{Q{b}_{1}=0\right\}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{b}={b}_{2}\left[0\phantom{\rule{0ex}{0ex}}1\right]\phantom{\rule{0ex}{0ex}}$

The eigenvectors are as follows

$\left[\begin{array}{c}-1\\ 0\end{array}\right]and\left[\begin{array}{c}0\\ 1\end{array}\right]$

The general solution to the system can be represented as follows.

${\stackrel{\to }{x}}_{n}={c}_{1}{\lambda }_{1}^{n}\stackrel{\to }{a}+{c}_{2}{\lambda }_{2}^{n}\stackrel{\to }{b}\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{x}}_{n}={c}_{1}{\left(3\right)}^{n}\left[\begin{array}{c}-1\\ 1\end{array}\right]+{c}_{2}{\left(0.5\right)}^{n}\left[\begin{array}{c}0\\ 1\end{array}\right]$

## Step4: Observation of the phase portrait

Since, the eigenvalues are real and distinct.

Therefore, the phase portrait is in figure 3 as follows.

Hence, the phase portrait to the linear system $\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}3& 0\\ -2.5& 0.5\end{array}\right]\stackrel{\to }{x}\left(t\right)$

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