I'm trying to express a fixed point when x1 and x2 are both not equal to zero for this system of equations. I know that one fixed point is $(0,0)$ when $x_1 = 0$.
$$
\begin{split}
x_1 &= \lambda x_1(1 – x_1) – 0.2x_1x_2 \\
x_2 &= 0.9x_2 + 0.2x_1x_2
\end{split}
$$
Where $\lambda \in (0,3)$ is a scalar. Initially, $x_1^{(0)} = 0$ and $x_2^{(0)} = 1$.
I built the Jacobian matrix A with both equations for $x_1$ and $x_2$. This gives me a $2 \times 2$ matrix
$$A = \begin{bmatrix}-λ-0.2×2&-0.2×1\\0.2×2&0.9+0.2×1\end{bmatrix}$$
When I try finding the eigenvalues of $A$ using $\det(A – \lambda I)$, it doesn't seem to work out.
$$(A – \lambda I) =
\begin{bmatrix}
-\lambda-0.2x_2-\lambda & -0.2 x_1\\
0.2 x_2 & 0.9 + 0.2x_1-\lambda
\end{bmatrix}
$$
Any ideas what I'm doing wrong?
Best Answer
You are confusing the constant in your problem with the matrix eigenvalues. Rename the matrix eigenvalues to $\mu$ to get $$A - \mu I = \begin{bmatrix} -\lambda-0.2x_2-\mu & -0.2 x_1\\ 0.2 x_2 & 0.9 + 0.2x_1-\mu \end{bmatrix} $$ so $$ \begin{split} \det (A-\mu I) &= -(\lambda+0.2 x_2+\mu)(0.9 + 0.2x_1-\mu) + 0.04 x_1 x_2 \\ &= (\mu+ \lambda+0.2 x_2)(\mu - 0.9 - 0.2x_1) + 0.04 x_1 x_2 \\ &= \mu^2 + \mu \left[ \ldots \right] - \ldots \end{split} $$ and use the quadratic formula