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A finite-basis occupied projector map can two-cycle despite a uniform gap

Example

Take n=2, r=1, overlap S=I2, Cholesky factor L=I2, h=0, and Exc=0 in Finite-basis projector energy model. Define G:H2→H2 by G(X)=diag⁡(X11,X22). Here matrix-entry labels 1,2 mean the first and second rows/columns, corresponding to canonical indices 0,1. The associated response map is F(P)=G(P). Let P0=(1000),P1=(0001). For the undamped occupied-projector iteration, P0 and P1 form a strict two-cycle even though the occupied/unoccupied eigenvalue gap equals 1 at both iterates.

Facts & Assumptions

Given: The finite-basis model data above, the rank-one projectors P0,P1, and Φ(P) equal to the projector onto the lowest eigenvalue of F(P) whenever the gap is positive.

[F1]

The finite-basis model permits Hermitian h, real-linear self-adjoint G, and real C2 Exc, with F(P)=h+G(P)+∇Exc(P) (Finite-basis projector energy model).

[F2]

The occupied projector is selected from the r lowest eigenvalues when a strict gap is present (The occupied-projector response map on its strict-gap domain).

[F3]

The undamped update is Pk+1=Φ(Pk) and checks the gap before updating (Finite-budget undamped self-consistent projector iteration).

[F4]

A strict gap makes the selected occupied projector unique (A strict occupied gap defines a unique rank-r spectral projector).

[F5]

The exact diagonal eigensolver returns the ordered eigenvalues and occupied projector of each displayed diagonal matrix.

Verification

1.1givenF1algebra

For Hermitian X,Y, the diagonal entries are real and ⟨G(X),Y⟩F=X11Y11+X22Y22=⟨X,G(Y)⟩F. Thus G is real-linear, Hermitian-valued, and self-adjoint for the Frobenius inner product, so the stated data satisfy [F1]. Each Pi is Hermitian, idempotent, and has trace 1, hence lies in P1.

1.2givenF2F4F5algebra

Direct substitution gives F(P0)=diag⁡(1,0) and F(P1)=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(P0) is span⁡{(0,1)T} and that of F(P1) is span⁡{(1,0)T}; [F4] guarantees these give the unique projectors Φ(P0)=P1 and Φ(P1)=P0. Thus K={P0,P1} is invariant.

2.1step 1.2F3algebra∎

By [F3], starting at P0 yields P0,P1,P0,P1,…. Since ∥P0−P1∥F=2>0, this sequence does not converge. Thus a positive uniform gap at the iterates alone does not imply convergence.

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