 Solution

The least-squares normal equations areThe full column rank of $G$ gives $z^{T}Hz=\Vert Gz{\Vert }_2^2>0$ for nonzero $z$, so $H$ is a symmetric positive-definite matrix. Apply the Conjugate gradient method to this linear system. To verify that its solution is the desired one, let $x_{\ast }$ solve the displayed equations. Then $G^T(G{x}_{\ast }-b)=0$, and for any $x$ the Pythagorean identity givesThus $x_{\ast }$ is the unique minimizer of the linear least-squares problem.

One can apply $H$ without forming the Gram matrix: compute $Hp=G^T(Gp)$. More explicitly, initialize $x_0$, $r_0=c-H{x}_0$, $p_0=r_0$; until the linear-system residual vanishes, use the Conjugate gradient method recurrenceEach dense application of $H$ this way costs $O(mn)$ operations. The condition number satisfies ${\kappa }_2(H)={\kappa }_2(G{)}^2$, where ${\kappa }_2(G)={\sigma }_{\max }(G)/{\sigma }_{\min }(G)$; this is the conditioning cost of the least-squares normal equations.