Take a look at this matrix (Jacobian matrix):
$$ A = \frac{\partial f}{\partial x} \Big\vert_{x=x_{e_1}} = \begin{bmatrix} 0 & 1 & 0 \\ \dfrac{2\mu g}{(1+\mu y_0)} & -\dfrac{k}{m} & -\dfrac{\lambda\mu\sqrt{\dfrac{2gm}{\lambda\mu}}}{m(1+\mu y_0)}\\ 0 & \mu \sqrt{\dfrac{2gm}{\lambda\mu}} & -\dfrac{R(\mu y_0 + 1)}{\lambda} \end{bmatrix} $$ Assume all parameters are positive. I can replace it with $$ A = \frac{\partial f}{\partial x} \Big\vert_{x=x_{e_1}} = \begin{bmatrix} 0 & 1 & 0 \\ a & -b & -c\\ 0 & d & -e \end{bmatrix} $$ where $a,b,c,d,e>0$. Eigenvalues are determined via the characteristic equation $\det(A-\lambda I)=0$. Matlab provides me with this values
Is there an alternative to check if the eigenvalues are positive, negative or at least one of them are zero without going through a tedious calculation? I don't care about the actual numerical values. This has something to do with the stability of the system.
Sum of the eigenvalues should be equal to the trace of the matric (sum of diagonal elements). The product of eigenvalues should be equal to the determinant of the matrix.