The rigid-body gauge system is nonsingular

Suppose $(x,\eta )$ lies in the nullspace of the augmented matrix. Then $Kx+R\eta =0$ and $R^{T}x=0$. Multiplying the first equation by $x^T$ gives $x^{T}Kx+x^{T}R\eta =0$, and the second equation makes the second term zero; hence $x^{T}Kx=0$.

For any $y\in {\mathbb{R}}^N$ and $t\in \mathbb{R}$, positive semidefiniteness gives $(x+ty{)}^{T}K(x+ty)=2t{y}^{T}Kx+t^2{y}^{T}Ky\ge 0$. If $y^{T}Kx\mathrm{\ne }0$, choosing $t$ with the opposite sign and sufficiently small magnitude makes the right side negative. Thus $y^{T}Kx=0$ for every $y$, so $Kx=0$.

Since $\ker K=\mathrm{range}R$, write $x=Ra$. Then $0=R^{T}x=R^{T}Ra$; full column rank makes $R^{T}R$ positive definite, so $a=0$ and $x=0$. The first null equation now gives $R\eta =0$, and full column rank yields $\eta =0$. The augmented square matrix is injective and therefore nonsingular by finite-dimensional rank-nullity.

Nonsingularity gives a unique solution for every right-hand side. For the right-hand side $[f;0]$, multiply $Ku+R\gamma =f$ by $R^T$. Symmetry of $K$ and $KR=0$ imply $R^{T}K=0$, so $R^{T}R\gamma =R^{T}f=0$ and positive definiteness gives $\gamma =0$. Hence $Ku=f$ and $R^{T}u=0$. Conversely, any vector satisfying those two equations with multiplier zero solves the augmented system, so its uniqueness follows from step 3.1.