A reported singular DIIS Gram system triggers the baseline fallback

The matrices $P_0,P_1,X$ are mutually Frobenius-orthogonal and $\mathrm{\parallel }{P}_0{\mathrm{\parallel }}_F^2=\mathrm{\parallel }{P}_1{\mathrm{\parallel }}_F^2=1$. Put $\mathrm{\Delta }=P_0-P_1$, so the displayed map is $G(A)=⟨A,\mathrm{\Delta }{⟩}_{F}X+⟨A,X{⟩}_F\mathrm{\Delta }$. Since this is a real-linear map into the Hermitian span of $X,\mathrm{\Delta }$, symmetry of the Frobenius inner product gives $⟨G(A),B{⟩}_F=⟨A,\mathrm{\Delta }{⟩}_F⟨X,B{⟩}_F+⟨A,X{⟩}_F⟨\mathrm{\Delta },B{⟩}_F=⟨A,G(B){⟩}_F$ for Hermitian $A,B$. Thus the model data satisfy [F1].

Substitution gives $G(P_0)=X$ and $G(P_1)=-X$. The matrices $X$ and $-X$ have eigenvalues $-1,1$, so both iterates have gap $2$. Direct multiplication gives $[G(P_0),P_0]=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=[G(P_1),P_1]\mathrm{\ne }0.$

Writing the common residual as $J$, the Gram matrix is $B_{ij}=⟨J,J{⟩}_F=2,B=\left(\begin{array}{cc}2& 2\\ 2& 2\end{array}\right),$ which is singular. The equality-constrained bordered system has matrix $\left(\begin{array}{ccc}2& 2& 1\\ 2& 2& 1\\ 1& 1& 0\end{array}\right)$, whose first two rows coincide, so a solver that reports singularity takes the prescribed fallback. With $C=2$, $c=(1\mathrm{/}2,1\mathrm{/}2)$ is feasible and minimizes the objective; the objective is constant on $c_1+c_2=1$, so [F2] selects this unique minimum-norm minimizer. If a rank-revealing solver returns it instead of reporting singularity, then $F_{mix}=\frac{1}{2}X-\frac{1}{2}X=0$ has no strict gap, and the proposal algorithm again falls back by its candidate-gap branch. Thus either outcome of this configured solve uses the baseline.