Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 5B / a / Solution

For any $h$,If the normal equation $A^{T}A{c}^{\ast }=A^{T}y$ holds, the last two terms reduce to $\Vert Ah{\Vert }^2\ge 0$, so $c^{\ast }$ minimizes. Conversely, at a minimizer the directional derivative in every $h$ vanishes, forcing $A^T(A{c}^{\ast }-y)=0$.