One-step Newton solution of a quadratic stationary equation ($q(x)=\frac{1}{2}{x}^{T}Hx+c^{T}x+d⟹x_1=-H^{-1}c$)

For a real symmetric invertible matrix $H$, the full undamped Newton step for the displayed quadratic function sends every starting point to the unique stationary point:This follows because the gradient is affine and its first-order expansion is exact. If $H$ is a positive-definite matrix the result is the unique global minimum; a negative-definite matrix gives a global maximum, and an indefinite matrix $H$ gives a saddle point. Thus exact arrival does not by itself show that a minimization method has minimized anything. A line search, damping, or a positive-definite matrix replacing the Hessian matrix changes this exact one-step assertion.