Suppose we have:
$\frac{dx}{dt} = 14x-\frac{1}{2}x^2-xy$
$\frac{dy}{dt} = 16y-\frac{1}{2}y^2-xy$
My textbook outlines the following steps:
For the critical point (0,32) of the above system, compute the Jacobian as
$J(0,32) = \left( \begin{array}{ccc} -18 & 0\\ -32 & -16\\ \end{array} \right)$
which corresponds to the linearization
$\frac{du}{dt} = -18u$
$\frac{dv}{dt} = -32u - 16v$
The matrix $J(0,32)$ has characteristic equation $(-18-\lambda)(-16-\lambda)=0$ with eigenvalues $\lambda_1 = -18$ and $\lambda_2 = -16$. Because both eigenvalues are negative, it follows that (0,0) is a nodal sink for the linearized system, and hence (0,32) is also a stable nodal sink for the original system.
Now my question is the part in bold. Why does the Jacobian of the critical point (0,32) tell us anything about (0,0)? In particular, why does knowing anything about the stability of (0,0) help in analyzing the critical point (0,32)?