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A shift near the wanted eigenvalue suppresses its component under repeated implicit restarts

Example

Let A=diag⁡(1,2,3), start from v=(e1+e2+e3)/3, and seek the largest eigenvalue 3. Use an implicit restart from dimension m=2 to r=1 with the fixed shift rule μ=29/10 on each cycle, residual tolerance τ=0, and restart budget at least t when considering t cycles. This shift rule is allowed by the procedure; it is not claimed to be a good selection rule.

Supply the following finite-return projected solver for this example. For a 2×2 Hermitian matrix H=(αbb‾δ), if b=0, return success with the ordered pairs (α,e1) and (δ,e2). If b≠0, set D=(α−δ)2+4∣b∣2,λ±=(α+δ±D)/2, and return success with the ordered pairs (λ−,(b,λ−−α)T) and (λ+,(b,λ+−α)T). Return unavailable on other shapes or non-Hermitian inputs. Each displayed vector is nonzero, direct multiplication verifies its eigenpair equation, and the two factors reproduce the quadratic characteristic polynomial. This supplied finite-dimensional example provider is not a general eigensolver or a new restart method.

Verification

Given: The exact diagonal matrix, starting vector, target, and shift above.

[F1] Exact implicit Arnoldi restart applies its shift polynomial and preserves an Arnoldi residual identity: for r=1, the exact implicit restart retains the one-dimensional filtered Krylov space generated by q(A)v.

1.1givenF1F2algebra

Every projected H2=V2∗AV2 in this example is Hermitian because A∗=A, so the supplied two-dimensional solver returns valid complete projected data in finite exact work. By [F1], q(A)v=(−19e1−9e2+e3)/(103). The squared component weights are proportional to 361,81,1, so the Rayleigh quotient of the restarted vector is (1⋅361+2⋅81+3⋅1)/(361+81+1)=526/443<2, whereas the initial quotient is (1+2+3)/3=2. The wanted component-to-e1 component ratio falls from 1 to 1/19.

2.1givenF1F2step 1.1algebra

Repeating the same exact restart multiplies the e3 component by q(3)=1/10 and the e1 component by q(1)=−19/10 at every cycle. Hence after t cycles their magnitude ratio is (1/19)t, which tends to zero. This shift choice suppresses the wanted eigendirection and gives no general convergence claim.

3.1givenF1F2step 2.1algebra

For each finite t, the restart vector is proportional to q(A)tv, with all coordinates nonzero because none of q(1),q(2),q(3) vanishes. If a Ritz residual were zero, its lifted vector would satisfy Au=θu; the diagonal equations (λi−θ)ui=0 and distinct λi force u to be proportional to some ei. No ei lies in K2(A,q(A)tv): representing one would require c+dλj=0 for the other two distinct eigenvalues, hence c=d=0. Also q(A)tv,Aq(A)tv,A2q(A)tv are independent because their coordinate determinant is the product of the three nonzero coordinates and the Vandermonde determinant for 1,2,3. Thus no pair is accepted at tolerance zero and the space is not invariant before each scheduled restart.

4.1givenF1F2algebra∎

In a separate boundary run with shift μ=3, the filter has q(3)=0 and q(A)v=(−2e1−e2)/3, so the wanted e3 component is removed exactly in one filtering step. The filtered vector is still nonzero, so this is not filtered breakdown; it is a shift-selection failure.

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