I have a non linear system to analyse and sketch the phase portrait of.
At one of the equilibria the Jacobian of the linearised system is given by $$\textbf {J}= \begin{pmatrix} 2 & 7\\ 7/2 & 1 \end{pmatrix}$$
Now we can calculate the eigenvalues of such a matrix to determine the classification of the equilibrium point. In particular here the two eigenvalues are $(3/2)\pm (3\sqrt11)/2$ so we have this particular equilibrium is a saddle point now is where my question begins.
We want to find the straight line paths corresponding to these eigenvalues one way we can do this is by looking for elements of the kernel of $~\lambda \bf{I}-\bf{J}$ for each eigenvalue $\lambda$ this method gives the two eigenvectors as:
$(1/7(1\pm3\sqrt11),1)$
now the other method I know of is to say $y=mx$ and substitute into $dy/dx=\dot y/ \dot x=m$ however this yields
$m=(1/14(1\pm3\sqrt11)$ so we have $y=(1/14(1\pm3\sqrt11)x$ as the straight line paths so my question is are these equivalent/equally valid methods? And also does the eigenector $(1/7(1\pm3\sqrt11),1)$ translate to $y=(1/7(1\pm3\sqrt11)x$ or $x=(1/7(1\pm3\sqrt11)y$? And does it matter which way we order them if it does?
Thank you.
The line spanned by the vector $(\frac{1}{7}(1 \pm 3 \sqrt{11}),1)$ can be parametrised as \begin{equation} x(\theta) = \frac{1}{7}(1 \pm 3 \sqrt{11})\,\theta,\quad y(\theta) = \theta. \end{equation} If you want to write this as a function $y(x)$, you have to write $\theta$ in terms of $x$, i.e. \begin{equation} \theta = \frac{7}{1 \pm 3\sqrt{11}}\,x, \end{equation} so \begin{equation} y = \frac{7}{1 \pm 3 \sqrt{11}}\, x. \tag{1} \end{equation} Now, to see that this is equivalent to the answer you've found by substituting $y = m x$, multiply numerator and denominator of $(1)$ by $1 \mp 3 \sqrt{11}$, which yields \begin{equation} y = -\frac{1}{14}(1 \mp 3 \sqrt{11})\,x. \end{equation}