A contractive implicit Runge-Kutta stage branch has a nonsingular residual Jacobian

Let $Y,W\in \mathcal{B}$. Its convexity puts $Y+t(W-Y)$ in $\mathcal{B}$ for $0\le t\le 1$. For each block $i$, the curve ${\gamma }_i(t)=F_i(Y+t(W-Y))$ is continuous on $[0,1]$ and differentiable inside because $F$ is $C^1$ on an open neighborhood. Its derivative obeys $\mathrm{\parallel }{\gamma }_i^{\mathrm{\prime }}(t){\mathrm{\parallel }}_2\le \mathrm{\parallel }{D}_{Y}F(Y+t(W-Y))(W-Y){\mathrm{\parallel }}_{\mathrm{\infty },2}\le L\mathrm{\parallel }W-Y{\mathrm{\parallel }}_{\mathrm{\infty },2}\mathrm{.}$ Apply [F3] to each block and take the maximum to obtain $\mathrm{\parallel }F(W)-F(Y){\mathrm{\parallel }}_{\mathrm{\infty },2}\le L\mathrm{\parallel }W-Y{\mathrm{\parallel }}_{\mathrm{\infty },2}$. The fixed-point map $\mathrm{\Phi }(Y)=\mathbf{1}\otimes y+h(A\otimes I_d)F(Y)$ satisfies $\mathrm{\parallel }\mathrm{\Phi }(Y)-\mathbf{1}\otimes y{\mathrm{\parallel }}_{\mathrm{\infty },2}\le h\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}M\le R$, so it maps $\mathcal{B}$ into itself. Thus [F2] supplies the unique stage root in $\mathcal{B}$.

Differentiating [F1] with respect to $Y$ gives $D_{Y}G=I-h(A\otimes I_d)D_{Y}F$. If $D_{Y}GZ=0$, the block row-sum norm and the derivative bound imply $\mathrm{\parallel }Z{\mathrm{\parallel }}_{\mathrm{\infty },2}\le h\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}L\mathrm{\parallel }Z{\mathrm{\parallel }}_{\mathrm{\infty },2}=q\mathrm{\parallel }Z{\mathrm{\parallel }}_{\mathrm{\infty },2}\mathrm{.}$ Since $q<1$, this forces $Z=0$; thus the square finite-dimensional map $D_{Y}G$ has trivial kernel.

By [F4], a linear map on the stage space has rank equal to its dimension when its kernel has dimension zero. Therefore $D_{Y}G$ is onto as well as one-to-one, hence invertible. This holds at every point of the prescribed branch ball.