91Ó°ÊÓ

Consider the linear system as follows.

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

To find the eigenvalues, evaluate $\left|A-\mathrm{λ}I\right|=0$as follows.

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

Simplify further as follows.

$$\begin{gathered} {\mathrm{λ}}^2-3.5\mathrm{λ}+1.5=0 \\ {\mathrm{λ}}^2-3\mathrm{λ}-0.5\mathrm{λ}+1.5=0 \\ \mathrm{λ}\left(\mathrm{λ}-3\right)-0.5\left(\mathrm{λ}-3\right)=0 \\ \left(\mathrm{λ}-3\right)\left(\mathrm{λ}-0.5\right)=0 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{λ}-3=0 \\ {\mathrm{λ}}_1=3 \\ \mathrm{λ}-0.5=0 \\ {\mathrm{λ}}_2=0.5 \end{gathered}$$

Therefore, the eigenvalues are ${\mathrm{λ}}_1=3$and ${\mathrm{λ}}_2=0.5$.

Now, to find the corresponding eigenvector for the eigenvalue ${\mathrm{λ}}_1=3$as follows.

localid="1659888265593" $$\begin{gathered} \left[A-3\right]\stackrel{→}{a}=0 \\ \left[\begin{array}{cc}3-3& 0\\ -2.5& 0.5-3\end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=0 \\ \left[\begin{array}{cc}0& 0\\ -2.5& 2.5\end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=0 \\ \left[\begin{array}{c}0a_1+0a_2\\ -2.5a_1-2.5a_2\end{array}\right]=0 \end{gathered}$$

The corresponding equations are follows.

$$\begin{gathered} 0a_1+0a_2=0 \\ -2.5a_1-2.5a_2=0 \end{gathered}$$

Now, solving the second equation as follows.

$$\begin{gathered} -2.5a_1-2.5a_2=0 \\ -2.5a_1=2.5a_2 \\ {a}_1=\frac{-2.5a_2}{2.5} \\ {a}_1=-a_2 \end{gathered}$$

Therefore, the eigenvectors as follows.

$$\begin{gathered} \stackrel{→}{a}=\left[a_1 \\ {a}_2\right] \\ =\left[-a_2 \\ {a}_2\right]    \left\{Q{a}_1=-a_2\right\} \\ \stackrel{→}{a}=a_2\left[-1 \\ 1\right] \end{gathered}$$

Now, to find the corresponding eigenvector for the eigenvalue ${\mathrm{λ}}_1=0.5$as follows.

$$\begin{gathered} \left[A-3\right]\stackrel{→}{b}=0 \\ \left[\begin{array}{cc}3-0.5& 0\\ -2.5& 0.5-0.5\end{array}\right]\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]=0 \\ \left[\begin{array}{cc}2.5& 0\\ -2.5& 0\end{array}\right]\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]=0 \\ \left[\begin{array}{c}2.5b_1+0b_2\\ -2.5b_1-0b_2\end{array}\right]=0 \end{gathered}$$

The corresponding equations are follows.

$$\begin{gathered} 2.5b_1+0b_2=0 \\ -2.5b_1-0b_2=0 \end{gathered}$$

Now, solving the second equation as follows.

$$\begin{gathered} -2.5b_1-0b_2=0 \\ -2.5b_1=0b_2 \\ {b}_1=0 \\ {b}_1=0 \end{gathered}$$

Therefore, the eigenvectors as follows.

$$\begin{gathered} \stackrel{→}{b}=\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right] \\ =\left[0 \\ {b}_2\right] \left\{Q{b}_1=0\right\} \\ \stackrel{→}{b}=b_2\left[0 \\ 1\right] \end{gathered}$$

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.

$$\begin{gathered} {\stackrel{→}{x}}_n=c_1{\mathrm{λ}}_1^n\stackrel{→}{a}+c_2{\mathrm{λ}}_2^n\stackrel{→}{b} \\ {\stackrel{→}{x}}_n=c_1{\left(3\right)}^n\left[\begin{array}{c}-1\\ 1\end{array}\right]+c_2{\left(0.5\right)}^n\left[\begin{array}{c}0\\ 1\end{array}\right] \end{gathered}$$

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