Self consistent projector iterations and certified mixing — Examples

If $t>0$, then $-t<t$ and the lowest eigenvector is $e_1$, so ${\mathrm{\Pi }}_1(H(t))=e_1{e}_1^{\ast }$. If $t<0$, then $t<-t$ and the lowest eigenvector is $e_2$, so ${\mathrm{\Pi }}_1(H(t))=e_2{e}_2^{\ast }$.

At $t=0$, both eigenvalues equal $0$, so the gap is $0$ and every rank-one orthogonal projector onto a line is a lowest-eigenspace projector; there is no unique ${\mathrm{\Pi }}_1(H(0))$ under [F1].

The limits as $t\downarrow 0$ and $t\uparrow 0$ are respectively $e_1{e}_1^{\ast }$ and $e_2{e}_2^{\ast }$, which are distinct because their Frobenius distance is $\sqrt{2}$. Thus no continuous extension of the uniquely gapped projector exists at $t=0$.

For Hermitian $X,Y$, the diagonal entries are real and $⟨G(X),Y{⟩}_F=X_{11}{Y}_{11}+X_{22}{Y}_{22}=⟨X,G(Y){⟩}_F\mathrm{.}$ Thus $G$ is real-linear, Hermitian-valued, and self-adjoint for the Frobenius inner product, so the stated data satisfy [F1]. Each $P_i$ is Hermitian, idempotent, and has trace $1$, hence lies in ${\mathcal{P}}_1$.

Direct substitution gives $F(P_0)=\mathrm{diag}(1,0)$ and $F(P_1)=\mathrm{diag}(0,1)$. Each matrix has eigenvalues $0<1$, so the strict occupied gap is $1$ by [F5]. By [F2], the lowest eigenspace of $F(P_0)$ is $\mathrm{span}\{(0,1{)}^T\}$ and that of $F(P_1)$ is $\mathrm{span}\{(1,0{)}^T\}$; [F4] guarantees these give the unique projectors $\mathrm{\Phi }(P_0)=P_1$ and $\mathrm{\Phi }(P_1)=P_0$. Thus $K=\{P_0,P_1\}$ is invariant.

By [F3], starting at $P_0$ yields $P_0,P_1,P_0,P_1,\dots$. Since $\mathrm{\parallel }{P}_0-P_1{\mathrm{\parallel }}_F=\sqrt{2}>0$, this sequence does not converge. Thus a positive uniform gap at the iterates alone does not imply convergence.

For $x,y\in [0,1]$, $\mathrm{\Phi }(x)\in [1\mathrm{/}4,3\mathrm{/}4]\subseteq [0,1]$ and $\mathrm{\mid }\mathrm{\Phi }(x)-\mathrm{\Phi }(y)\mathrm{\mid }=\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}2$. Solving $x=x\mathrm{/}2+1\mathrm{/}4$ gives the unique fixed point $x_{\ast }=1\mathrm{/}2$.

The exact iterates satisfy $x_{k+1}-1\mathrm{/}2=(x_k-1\mathrm{/}2)\mathrm{/}2$, so induction gives $\mathrm{\mid }{x}_k-x_{\ast }\mathrm{\mid }=2^{-k}\mathrm{\mid }{x}_0-x_{\ast }\mathrm{\mid }$.

For any $x\in [0,1]$, $\mathrm{\mid }x-\mathrm{\Phi }(x)\mathrm{\mid }=\mathrm{\mid }x-1\mathrm{/}2\mathrm{\mid }\mathrm{/}2$, hence $\mathrm{\mid }x-x_{\ast }\mathrm{\mid }=\mathrm{\mid }x-\mathrm{\Phi }(x)\mathrm{\mid }\mathrm{/}(1-1\mathrm{/}2)$. This algebraic identity directly realizes the same residual-to-error formula stated conditionally in [F1], without invoking that theorem's hypotheses.

The matrix $P_0$ is a Hermitian idempotent of trace $1$, so $P_0\in {\mathcal{P}}_1$ and $P_0\in {\mathcal{D}}_1$ by [F1]. The constant response matrix has eigenvalues $0<1$, so [F2] gives $\mathrm{\Phi }(P_0)=P_1=\mathrm{diag}(1,0)$.

Direct calculation gives $D=(1-\alpha )P_0+\alpha \mathrm{\Phi }(P_0)=\frac{1}{2}{I}_2$. By [F3], $D\in {\mathcal{D}}_1$; explicitly it is Hermitian, satisfies $0⪯D⪯I_2$, and has trace $1$.

But $D^2=\frac{1}{4}{I}_2\mathrm{\ne }D$, so $D$ is not idempotent and hence $D\mathrm{\notin }{\mathcal{P}}_1$. This verifies that relaxed density iterates must not be silently identified with projectors.

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.

The computed residual is $\mathrm{\mid }0-6\mathrm{/}25\mathrm{\mid }=24\mathrm{/}100$. By the triangle inequality, $\mathrm{\mid }0-\mathrm{\Phi }(0)\mathrm{\mid }\le \mathrm{\mid }0-\tilde{\mathrm{\Phi }}(0)\mathrm{\mid }+\mathrm{\mid }\tilde{\mathrm{\Phi }}(0)-\mathrm{\Phi }(0)\mathrm{\mid }\le 24\mathrm{/}100+1\mathrm{/}100=25\mathrm{/}100.$

Here $q=1\mathrm{/}2$ and $\mathrm{\mid }0-x_{\ast }\mathrm{\mid }=1\mathrm{/}2$. The correct certified error bound from [F1] is $(25\mathrm{/}100)\mathrm{/}(1-1\mathrm{/}2)=1\mathrm{/}2$, exact here. Omitting $\eta$ would give $(24\mathrm{/}100)\mathrm{/}(1\mathrm{/}2)=48\mathrm{/}100<1\mathrm{/}2$, which is false.