So I am attempting to find the equilibrium points of a nonlinear system and I am getting the following jacobian matrix:
$$ \begin{pmatrix} 1 & 0 \\ 0 & 1.5 \\ \end{pmatrix} $$
I have found the eigenvalues as 1 and 1.5 but am at a loss when attempting to find the eigenvectors for each. For eigenvalue equal to 1 i get the following matrix:
$$ \begin{pmatrix} 0 & 0 \\ 0 & 0.5 \\ \end{pmatrix} \begin{pmatrix} v1 \\ v2 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix} $$
Which using gauss jordan becomes:
$$ \begin{pmatrix} 0 & 1 & | & 0 \\ 0 & 0 & | & 0\\ \end{pmatrix} $$
From here I am lost as what to do because every answer becomes zero. When throwing the jacobian matrix into wolfram alpha, the vector for eigenvalue equals to 1 is:
$$ \begin{pmatrix} -1 \\ 0 \\ \end{pmatrix} $$
Is there some sort of trick I am missing?
For $\lambda = 1$, we have the RREF of $[A -I]v_1 = 0$ as:
$$ \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right)v_1 = \left( \begin{array}{cc} 0 \\ 0 \\ \end{array} \right)$$
Lets write this out to understand what it is telling us. We have:
$$0a + 1 b = 0 \\ 0a + 0b = 0$$
If we choose $b = 1$, can we satisfy this system? No. If we choose $b = 0$, this would give us a zero eigenvector, which is not allowed. What is we chose $a = 1$ and $b=0$. Hey, that satisfies the system and gives us a non-zero eigenvector. This is also called the null space of the matrix.
This means we have $a = 1 $, so we can choose $b = 0$, so $v_1 = (1, 0)$. We could have also chosen $v_1 = (-1, 0)$. Try using $3/2$ instead of $1.5$ in WA and see what happens. Regardless, both answers are acceptable as eigenvectors are not unique.
Do you see how that vector satisfies the system and is not a zero eigenvector?