Question

Consider a noninvertible $\mathbf{2}\mathbf{\times }\mathbf{2}$matrix A with a positive trace. What does the phase portrait of the system$\mathit{d}\overrightarrow{\mathbf{x}}\mathbf{/}\mathit{d}\mathit{y}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$look like?

Short answer

The phase portrait of the system$d\overrightarrow{x}/dy=A\overrightarrow{x}$ look like asymptotically stable

Step-by-step solution

Step 1:Explanation for the phase of the system dx→/dy=Ax→

Consider a noninvertible $2\times 2$matrix A with a positive trace.

Where A is a real$2\times 2$matrix.

Then the zero state is an asymptotically stable equilibrium solution if and only if $trA<0$and $\det A>0$

Explanation for the phase of the system dx→/dy=Ax→

Here is the phase looks like asymptotically stable.

${\lambda }_{1,2}=p\pm iq$

$$\begin{gathered} p^2+q^2=1 \\ {\lambda }_{1,2}=\pm iq \end{gathered}$$

The phase portrait of the system $d\overrightarrow{x}/dy=A\overrightarrow{x}$look like asymptotically stable.