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A scalar saddle system has golden-ratio preconditioned eigenvalues

Statement

Take $A = [1]$ , $B = [1]$ , $M_{Q} = [1]$ , and $γ = 1$ in the symmetric saddle-point and block-preconditioning convention of Symmetric saddle-point system, Schur complement, and block preconditioners. Interpret $V = Q = R$ with Euclidean norms, $a (v, w) = v w$ , and $b (v, q) = q v$ ; thus the pressure-Schur hypotheses hold with $α = Λ = C_{b} = β = 1$ . Then $K = (\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array}), P_{1} = I .$ The symmetrically preconditioned eigenvalues are $\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2} .$ For right-hand side $(1, 0)^{T}$ , the unique saddle solution is $(u, p) = (0, 1)$ . This exact example illustrates the spectrum only; it implies no solver-speed or iteration-count claim.

Facts & Assumptions

Given: The scalar SPD blocks $A = M_{Q} = [1]$ , the full-row-rank matrix $B = [1]$ , scaling $γ = 1$ , and right-hand side $(1, 0)^{T}$ .

[F1]

The saddle matrix is $K = [A B^{T}; B 0]$ and the symmetric preconditioner is $P_{γ} = diag (A, γ M_{Q})$ (Symmetric saddle-point system, Schur complement, and block preconditioners).

[F2]

For normalized Schur eigenvalue $μ$ , the saddle eigenvalues are $λ_{\pm} (μ) = (1 \pm \sqrt{1 + 4 μ}) / 2$ (Spectrum of a symmetrically pressure-mass-preconditioned saddle system).

Proof

1.1F1algebra

The scalar matrix $A = [1]$ is SPD, $B^{T} q = 0$ implies $q = 0$ , and the Schur matrix is $S = B A^{- 1} B^{T} = [1]$ . Thus the normalized Schur eigenvalue is $μ = 1$ and $P_{1} = diag (1, 1) = I$ by [F1].

2.1F1F2step 1.1algebra∎

Applying [F2] at $μ = 1$ gives the characteristic polynomial $λ^{2} - λ - 1$ and roots $(1 \pm \sqrt{5}) / 2$ . Direct substitution in $u + p = 1$ and $u = 0$ yields $(u, p) = (0, 1)$ ; the coefficient matrix is invertible because its determinant is $- 1$ , so this solution is unique.

Depends on

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Sources

Michele Benzi, Gene H. Golub, and Jörg Liesen, Numerical solution of saddle point problems (standard reference, not scraped)