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For Hermitian $A$ , including the indefinite case, MINRES minimizes the Euclidean residual over $x_{0} + K_{m} (A, r_{0})$

Statement

Let $A$ be Hermitian, let $x_{0}$ be an initial guess for $A x = b$ , let $r_{0} = b - A x_{0}$ , and let $x_{m}$ be an $m$ th MINRES iterate from the Lanczos tridiagonal least-squares problem. Then

$∥ b - A x_{m} ∥_{2} = \min_{x \in x_{0} + K_{m} (A, r_{0})} ∥ b - A x ∥_{2} .$

This remains valid when $A$ is Hermitian indefinite.

Facts & Assumptions

Given: A Hermitian matrix $A$ , an initial guess $x_{0}$ , the residual $r_{0} = b - A x_{0}$ , and an $m$ th MINRES iterate $x_{m}$ .

[F1]

A MINRES iterate has the form $x_{m} = x_{0} + Q_{m} y_{m}$ , where $y_{m}$ minimizes $∥ β e_{1} - {\overset{ˉ}{T}}_{m} y ∥_{2}$ for the Lanczos tridiagonalization (The MINRES iterate from the Lanczos tridiagonal least-squares problem).

[L1]

For Hermitian $A$ , Arnoldi collapses to the Lanczos factorization with tridiagonal ${\overset{ˉ}{T}}_{m}$ (With a Hermitian matrix, Arnoldi collapses to the Lanczos three-term recurrence and a real symmetric tridiagonal projection).

[L2]

Arnoldi reduces GMRES to the small least-squares problem $\min_{y} ∥ β e_{1} - {\overset{ˉ}{H}}_{m} y ∥_{2}$ (Arnoldi reduces GMRES to a least-squares problem for the small Hessenberg matrix).

[L3]

GMRES minimizes the Euclidean residual over $x_{0} + K_{m} (A, r_{0})$ (GMRES minimizes the residual norm over the affine Krylov space $x_{0} + K_{m} (A, r_{0})$ ).

Proof

technique · direct

1.1

By [L1], the Hermitian Arnoldi factorization uses the same orthonormal Krylov basis $Q_{m}$ as Arnoldi but with ${\overset{ˉ}{H}}_{m} = {\overset{ˉ}{T}}_{m}$ tridiagonal. Therefore the least-squares problem in [L2] is exactly the Lanczos least-squares problem recorded in [F1].

F1L1L2

2.1

Step 1.1 shows that the vector $x_{m} = x_{0} + Q_{m} y_{m}$ coming from [F1] is also the GMRES iterate at step $m$ . Applying [L3] gives $∥ b - A x_{m} ∥_{2} = \min_{x \in x_{0} + K_{m} (A, r_{0})} ∥ b - A x ∥_{2} .$ Neither step 1.1 nor [L3] uses positive definiteness, only Hermitian structure, so the same conclusion holds for Hermitian indefinite $A$ .

F1L3step 1.1∎

Depends on

Used by

MINRES still minimizes the residual on a small symmetric indefinite system Example

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard Barrett et al., Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods (standard reference, not scraped)

For Hermitian AA, including the indefinite case, MINRES minimizes the Euclidean residual over x0+Km(A,r0)x0​+Km​(A,r0​)