The minimum-norm IRLS step is well-defined for every design rank

Each predictor is a finite sum of products of real design and parameter entries, and each mean is a unique finite real value by [F2]. The observed response coordinates are finite real numbers by [F1]. Thus $z_i={\eta }_i+(y_i-{\mu }_i)\mathrm{/}{\dot{\mu }}_i$ is finite and uniquely determined because its denominator is nonzero. Likewise $w_i={\dot{\mu }}_i^2\mathrm{/}(a(\varphi )V_i)$ is finite and strictly positive: its numerator is positive and its denominator is a finite positive real number. These operations refer to real arithmetic, with no floating-point overflow assertion.

By [F3] each $w_i$ has one finite positive square root. Consequently the diagonal matrix $D=\mathrm{diag}(\sqrt{w_i})$, the weighted design $A=DX$, and the target $r=Dz$ are unique finite real matrices or vectors with the specified dimensions. By [F4] the $p\times n$ matrix $A^{+}$ exists uniquely for every rank of $A$, so the returned $A^{+}r$ exists uniquely in ${\mathbb{R}}^p$. No inverse of $X^{\mathsf{T}}WX$ was required at this stage.

For every $\gamma \in {\mathbb{R}}^p$, the diagonal construction gives $\mathrm{\parallel }A\gamma -r{\mathrm{\parallel }}_2^2={\sum }_i{w}_i(x_i^{\mathsf{T}}\gamma -z_i{)}^2$. Since a nonnegative real number and its square have the same minimizing arguments, [F5] makes $A^{+}r$ a member of the displayed weighted least-squares minimizer set and the unique member of minimum Euclidean norm. In particular that set is nonempty even when $X$ is rank deficient. The minimum-norm property is the conclusion of the supplied theorem, not an extra condition imposed on the algorithm's inputs.

The weighted normal matrix is symmetric, and for every $u\in {\mathbb{R}}^p$ one has $u^{\mathsf{T}}Hu={\sum }_i{w}_i(x_i^{\mathsf{T}}u{)}^2=\mathrm{\parallel }Au{\mathrm{\parallel }}_2^2\ge 0$, so $H$ is positive semidefinite. If $Xu=0$, then $Hu=X^{\mathsf{T}}WXu=0$. Conversely, if $Hu=0$, the displayed sum is zero, and its finitely many nonnegative terms with strictly positive $w_i$ force every $x_i^{\mathsf{T}}u=0$; hence $Xu=0$. This proves $\ker H=\ker X$. Steps 2.1–3.1 apply equally when this kernel is nontrivial.

If $X=0$, then $A=0$. The zero matrix satisfies the Penrose equations, so uniqueness in [F4] gives $A^{+}=0$ and the algorithm returns $0$. The weighted objective is then constant in $\gamma$, so every vector is a minimizer and $0$ is its unique minimum-norm member, consistent with step 3.1. A single response, zero observed responses, and other ranks use the same proof whenever the original admissibility conditions hold. Thus all stated well-definedness claims hold without an invertibility condition or a condition that the returned vector remain in the model domain.