 Solution

Assume $X,Y$ are continuously differentiable near the equilibrium. Put $\xi =x-a$, $\eta =y-b$ and expand the vector field:This Jacobian matrix defines the linearized system. For distinct real eigenvalues, two negative values give a stable node, two positive values an unstable node, and opposite signs a saddle point of a dynamical system. A complex-conjugate pair with negative real part gives a stable focus, and one with positive real part an unstable focus. Thus a hyperbolic saddle has two real eigenvalues of opposite signs, equivalently $\det M<0$.

The linearization stability theorem transfers these stability conclusions to the nonlinear system when no eigenvalue has zero real part. Distinctness alone is insufficient: a zero eigenvalue or a purely imaginary conjugate pair leaves a nonhyperbolic equilibrium, for which the nonlinear terms must be examined. In the purely imaginary case the linearized trajectories form a centre, but that does not establish a nonlinear centre.