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Verified linear systems and spectral enclosures — Examples

1 · Prerequisites

2 · Summary

Examples

These examples check an exact residual enclosure for a system with condition number $2^{20}$ , show why an interval family containing a singular member cannot receive a uniform inverse certificate, and carry out ordered Hermitian eigenvalue bisection with exact inertia queries. The bisection example records the exact shifts and counts, including the finite query budget.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A certified residual enclosure remains valid for an ill-conditioned 2 by 2 solve

Example

Let $A = (\begin{array}{cc} 1 & 0 \\ 0 & 2^{- 20} \end{array}), b = (\begin{array}{c} 1 \\ 2^{- 20} \end{array}), R = A^{- 1} = (\begin{array}{cc} 1 & 0 \\ 0 & 2^{20} \end{array}), \hat{x} = (\begin{array}{c} 1 \\ 1 + 2^{- 10} \end{array}) .$ These point data have a usable residual certificate with $q = 0$ and $r = 2^{- 10}$ . Its certified error radius equals the actual error even though the spectral condition number of $A$ is $2^{20}$ .

Facts & Assumptions

Given: The exact dyadic matrix, right-hand side, inverse candidate, and approximate solution displayed above.

[F1]

A usable residual certificate gives a unique solution within $r / (1 - q)$ (A uniform approximate inverse and residual enclose every linear solution).

[F2]

The point case of the residual certificate is specified by $∥ I - R A ∥_{2} \leq q$ and $∥ R (b - A \hat{x}) ∥_{2} \leq r$ (Verified point and uniform linear residual certificate data).

Verification

1.1givenF2algebra

Direct multiplication gives $R A = I$ , so $∥ I - R A ∥_{2} = 0$ . Also $A \hat{x} = (1, 2^{- 20} + 2^{- 30})^{T}$ and $R (b - A \hat{x}) = (0, - 2^{- 10})^{T}$ , whose Euclidean norm is $2^{- 10}$ . Thus [F2] holds with $q = 0, r = 2^{- 10}$ .

2.1givenF1step 1.1algebra∎

The exact solution is $x = (1, 1)^{T}$ , so $∥ x - \hat{x} ∥_{2} = 2^{- 10}$ . The singular values of this positive diagonal matrix are $1$ and $2^{- 20}$ , hence its condition number is $2^{20}$ . Applying [F1] gives the same error radius $r / (1 - q) = 2^{- 10}$ , which contains the solution with equality.

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

An interval matrix family can contain a singular member

Example

The real family

A = {A (t) = (\begin{array}{cc} 1 & t \\ t & 1 \end{array}) : - 1 \leq t \leq 1}

is represented by the entrywise interval box with off-diagonal entries $[- 1, 1]$ . It contains a singular matrix, and its natural Leibniz interval evaluation $[0, 2]$ does not certify uniform nonsingularity.

Facts & Assumptions

Given: The interval family $A$ displayed above.

[F1]

A determinant interval excluding zero is sufficient for nonsingularity; an interval containing zero without being exactly ${0}$ yields no such assertion (An interval determinant excluding zero certifies nonsingularity and sign).

[F2]

A usable uniform residual certificate with defect $q < 1$ would make every family member invertible (A uniform approximate inverse and residual enclose every linear solution).

Verification

1.1givenF1algebra

For each $t \in [- 1, 1]$ , $\det A (t) = 1 - t^{2} \in [0, 1]$ . The direct interval product $[- 1, 1] \cdot [- 1, 1] = [- 1, 1]$ encloses $t^{2}$ , so the Leibniz interval expression $1 - [- 1, 1] = [0, 2]$ contains every determinant. At $t = 1$ , $A (1) (1, - 1)^{T} = 0$ with a nonzero vector, so $A (1)$ is singular. The interval $[0, 2]$ includes zero and is not ${0}$ ; [F1] therefore makes no uniform nonsingularity claim from this enclosure.

2.1givenF2step 1.1contradiction∎

If uniform residual data with $q < 1$ existed for this whole family and any nonempty right-hand-side set, [F2] would make $A (1)$ invertible. This contradicts the nonzero null vector just exhibited. Hence such a usable uniform residual certificate cannot hold for the family.

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Ordered eigenvalue bisection by exact inertia queries

Example

For the singleton Hermitian family $A = diag (- 1, 2)$ , a valid outward Frobenius upper bound $M = 3$ , tolerance $ε = 3 / 8$ , query budget $B = 8$ , and the scalar pivot plan with threshold $1$ , exact inertia queries return the ordered eigenvalue brackets

λ_{1} (A) \in [- 9 / 8, - 3 / 4], λ_{2} (A) \in [15 / 8, 9 / 4] .

Facts & Assumptions

Given: The singleton matrix family, initial bracket $[- 3, 3]$ , tolerance, budget and fixed pivot plan above.

[F1]

A successful bisection loop preserves each ordered eigenvalue bracket after every exact count below a nonsingular shift (Certified inertia counts maintain ordered eigenvalue brackets).

[F2]

The interval inertia certifier reports the exact inertia on success (The interval block LDL certifier returns the exact inertia on success).

[F3]

The finite bisection procedure uses the initial uniform Frobenius bracket and updates the least unfinished index at each query (Budgeted ordered Hermitian eigenvalue bisection by verified inertia).

[F4]

A successful run's returned intervals enclose the corresponding eigenvalues (The ordered Hermitian bisection certifier returns valid intervals or an inconclusive status).

Verification

1.1givenF2algebra

The Frobenius norm of $A$ is $\sqrt{5} < 3$ , so the stated initial bracket is valid. On each queried shift, the shifted matrix remains diagonal with nonzero scalar pivots; the exact midpoint factorization succeeds, and its interval pivots are degenerate intervals with the same nonzero signs. Thus [F2] certifies the counts shown below.

2.1F1F3step 1.1algebra

For $k = 1$ , the successive shifts $0, - 3 / 2, - 3 / 4, - 9 / 8$ have respectively $1, 0, 1, 0$ negative eigenvalues of $A - τ I$ . The update rule gives brackets $[- 3, 0]$ , $[- 3 / 2, 0]$ , $[- 3 / 2, - 3 / 4]$ , and $[- 9 / 8, - 3 / 4]$ . For $k = 2$ , shifts $0, 3 / 2, 9 / 4, 15 / 8$ have respectively $1, 1, 2, 1$ negative eigenvalues, giving brackets $[0, 3]$ , $[3 / 2, 3]$ , $[3 / 2, 9 / 4]$ , and $[15 / 8, 9 / 4]$ . These are eight successful queries, and both final widths are $3 / 8$ .

3.1F1F2F4step 2.1algebra∎

The exact increasing eigenvalues are $- 1$ and $2$ , which lie in the displayed final intervals. More generally, the interval conclusions follow from [F4] applied to the certified run; no failed-pivot case occurs here.

Verified linear systems and spectral enclosures — Examples

1 · Prerequisites

2 · Summary

Examples

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A certified residual enclosure remains valid for an ill-conditioned 2 by 2 solve

Example

Facts & Assumptions

Verification

An interval matrix family can contain a singular member

Example

Facts & Assumptions

Verification

Ordered eigenvalue bisection by exact inertia queries

Example

Facts & Assumptions

Verification

Sources