ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-01

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MINRES still minimizes the residual on a small symmetric indefinite system

Example

Take

$A = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}), b = (\binom{1}{1}), x_{0} = (\binom{0}{0}) .$

Then $A$ is symmetric and indefinite. MINRES at step $1$ gives the residual minimizer in the one-dimensional affine Krylov space.

Facts & Assumptions

Given: The displayed Hermitian indefinite system.

[L1]

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

Verification

technique · direct calculation

1.1

Here $r_{0} = b$ , so $q_{1} = r_{0} / ∥ r_{0} ∥_{2} = (1, 1)^{T} / \sqrt{2}$ and $A q_{1} = (\binom{1 / \sqrt{2}}{- 1 / \sqrt{2}}), α_{1} = q_{1}^{T} A q_{1} = 0, β_{1} = 1, q_{2} = (\binom{1 / \sqrt{2}}{- 1 / \sqrt{2}}) .$ Thus the first Lanczos matrix is ${\overset{ˉ}{T}}_{1} = (\binom{0}{1}),$ and the step- $1$ least-squares problem is $\min_{y \in R} {∥ (\binom{\sqrt{2}}{0}) - (\binom{0}{y}) ∥}_{2},$ whose unique minimizer is $y_{1} = 0$ . Therefore $x_{1} = x_{0} + q_{1} y_{1} = 0$ .

algebra

2.1

The affine space $x_{0} + K_{1} (A, r_{0})$ is ${t (1, 1)^{T} : t \in R}$ . For such a vector, $b - A (\binom{t}{t}) = (\binom{1 - t}{1 + t}), {∥ b - A (\binom{t}{t}) ∥}_{2}^{2} = (1 - t)^{2} + (1 + t)^{2} = 2 + 2 t^{2},$ which is minimized exactly at $t = 0$ . So the explicit computation in step 1.1 matches [L1].

L1step 1.1algebra∎

Depends on

For Hermitian $A$, including the indefinite case, MINRES minimizes the Euclidean residual over $x_0+K_m(A,r_0)$

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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)