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A stationary split converges from an error missing its unstable eigenspace

Example

Take $A = I_{2}$ , $b = 0$ , $M = diag (- 1, 2)$ , and $N = M - A = diag (- 2, 1)$ . This is a stationary splitting with iteration matrix $G = M^{- 1} N = diag (2, 1 / 2)$ and $ρ (G) = 2$ . Starting from $x^{(0)} = (0, 1)^{T}$ gives $x^{(k)} = (0, 2^{- k})^{T} \to 0$ ; starting from $(1, 0)^{T}$ gives $x^{(k)} = (2^{k}, 0)^{T}$ , which diverges. Thus one start converges because its error has no component in the unstable eigenspace, even though the splitting does not converge for every start.

Facts & Assumptions

Given: The displayed splitting of $A x = b$ with $A = I_{2}$ and $b = 0$ .

[F1]

If $A = M - N$ with invertible $M$ , its stationary iteration is $x^{(k + 1)} = M^{- 1} N x^{(k)} + M^{- 1} b$ and its iteration matrix is $M^{- 1} N$ (Stationary iteration from a matrix splitting $A = M - N$ ).

[L1]

A stationary splitting converges to the unique solution for every $b$ and every start exactly when its iteration matrix has spectral radius below $1$ (A stationary splitting converges for every start if and only if its iteration matrix has spectral radius below $1$ ).

Verification

1.1givenF1algebra

The matrices satisfy $M - N = I_{2} = A$ and $M$ is invertible. By [F1], $G = M^{- 1} N = diag (2, 1 / 2)$ , whose spectral radius is $2$ . The unique solution for $b = 0$ is $0$ .

2.1givenF1L1step 1.1algebra∎

The recurrence is diagonal, so from $x^{(0)} = (0, 1)^{T}$ it gives $x^{(k)} = (0, 2^{- k})^{T} \to 0$ , while from $(1, 0)^{T}$ it gives $x^{(k)} = (2^{k}, 0)^{T}$ and does not converge. This directly exhibits both the missing-unstable-eigenspace case and the failure of the all-start condition [L1] because $ρ (G) = 2$ .

Depends on

Used by

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Sources

Yousef Saad, Iterative Methods for Sparse Linear Systems, 2nd ed., §4.2.1 (standard reference, not scraped)