AHL 5.17—Phase portrait

The system is already decoupled, so the coefficient matrix is $A=\left(\begin{array}{cc}4& 0\\ 0& -2\end{array}\right)$, which is diagonal.

For a diagonal matrix, the eigenvalues are simply the diagonal entries: ${\lambda }_1=4$, ${\lambda }_2=-2$.

The eigenvalues are real with opposite signs ($4>0$, $-2<0$).

Real eigenvalues of opposite sign always give a saddle point, which is unstable.

This would require both eigenvalues negative, but ${\lambda }_1=4>0$.

This would require both eigenvalues positive, but ${\lambda }_2=-2<0$.

A centre requires purely imaginary eigenvalues, but here both eigenvalues are real.

Opposite-sign real eigenvalues confirm the saddle classification.

${\lambda }^2-\text{tr}(A)\lambda +\det (A)=0$ becomes ${\lambda }^2+6\lambda +8=0$.

Factorising: $(\lambda +2)(\lambda +4)=0$, so ${\lambda }_1=-2$, ${\lambda }_2=-4$.

Both eigenvalues are real (discriminant $36-32=4>0$) and both negative.

Two real negative eigenvalues give a stable node.

This requires both eigenvalues positive, but both are negative here.

A saddle needs opposite-sign eigenvalues, but both $-2$ and $-4$ are negative.

A spiral requires complex eigenvalues, but the discriminant is positive so eigenvalues are real.

Real, both-negative eigenvalues confirm the stable node classification.