A damped least-squares step solves a scaled trust-region model

For every nonzero $v$, $v^T(J^{T}J+\lambda D^{T}D)v=\mathrm{\parallel }Jv{\mathrm{\parallel }}_2^2+\lambda \mathrm{\parallel }Dv{\mathrm{\parallel }}_2^2>0$ because $D$ is nonsingular. The matrix is symmetric positive definite, so [F2] gives a unique $p_{\lambda }$. With $z=Dp$, the system is equivalent, after left multiplication by $D^{-T}$, to $({\tilde{J}}^T\tilde{J}+\lambda I)z=-{\tilde{J}}^{T}r$. Expanding $q_{\lambda }$ and completing the square around its unique critical point shows $q_{\lambda }(z)-q_{\lambda }(z_{\lambda })=\frac{1}{2}(z-z_{\lambda }{)}^T({\tilde{J}}^T\tilde{J}+\lambda I)(z-z_{\lambda })\ge 0$, with equality only at $z=z_{\lambda }$.

If $\mathrm{\parallel }z{\mathrm{\parallel }}_2\le \mathrm{\parallel }{z}_{\lambda }{\mathrm{\parallel }}_2$, minimality from step 1.1 gives $m(z_{\lambda })+\frac{\lambda }{2}\mathrm{\parallel }{z}_{\lambda }{\mathrm{\parallel }}_2^2\le m(z)+\frac{\lambda }{2}\mathrm{\parallel }z{\mathrm{\parallel }}_2^2\le m(z)+\frac{\lambda }{2}\mathrm{\parallel }{z}_{\lambda }{\mathrm{\parallel }}_2^2$. Subtracting the common penalty proves $m(z_{\lambda })\le m(z)$ for every feasible $z$, so $z_{\lambda }$ globally minimizes $m$ on the stated ball. Expanding $m$ shows its linear coefficient is ${\tilde{J}}^{T}r$ and its symmetric quadratic coefficient is ${\tilde{J}}^T\tilde{J}$, so it is the trust-region model of [F1].

The damped system gives $-g^T{p}_{\lambda }=\mathrm{\parallel }J{p}_{\lambda }{\mathrm{\parallel }}_2^2+\lambda \mathrm{\parallel }D{p}_{\lambda }{\mathrm{\parallel }}_2^2$. Expanding $m(0)-m(D{p}_{\lambda })=-g^T{p}_{\lambda }-\frac{1}{2}\mathrm{\parallel }J{p}_{\lambda }{\mathrm{\parallel }}_2^2$ yields the displayed formula. Since $g\mathrm{\ne }0$, invertibility of the damped matrix gives $p_{\lambda }\mathrm{\ne }0$, and nonsingularity of $D$ makes the displayed reduction strictly positive. Thus the ratio denominator is valid.