= One-step Newton solution of a quadratic stationary equation
{title2=$q(x)=\tfrac12x^THx+c^Tx+d\ \Longrightarrow\ x_1=-H^{-1}c$}
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>:
$$
x_1=x_0-H^{-1}(Hx_0+c)=-H^{-1}c.
$$
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.
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