ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-08-30

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Givens QR can zero selected entries of a sparse matrix one at a time

Example

For $A = (\begin{array}{cc} 3 & 4 \\ 4 & 3 \\ 0 & 2 \end{array}),$ first apply a Givens rotation in rows $1, 2$ with $c = 3 / 5$ , $s = 4 / 5$ , then a second Givens rotation in rows $2, 3$ with $c = - 7 / \sqrt{149}$ , $s = 10 / \sqrt{149}$ . The product zeros one selected entry at a time and yields $R = (\begin{array}{cc} 5 & \frac{24}{5} \\ 0 & \frac{\sqrt{149}}{5} \\ 0 & 0 \end{array}) .$

Facts & Assumptions

Given: The displayed matrix $A$ and the two named Givens rotations.

[L1]

Givens transformations are unitary and can annihilate a chosen second coordinate (Householder reflectors and Givens transformations are unitary and can annihilate prescribed entries).

[L2]

Successive Givens transformations produce QR factorizations (Successive Householder or Givens transformations produce full and reduced QR factorisations with the standard dense operation counts).

Verification

technique · direct

1.1

The first rotation $G_{1} = (\begin{array}{ccc} 3 / 5 & 4 / 5 & 0 \\ - 4 / 5 & 3 / 5 & 0 \\ 0 & 0 & 1 \end{array})$ sends the first column $(3, 4, 0)^{T}$ to $(5, 0, 0)^{T}$ , so it zeros the $(2, 1)$ entry while preserving the zero in position $(3, 1)$ . Applied to the second column, it gives $(24 / 5, - 7 / 5, 2)^{T}$ .

L1algebra

2.1

The second rotation $G_{2} = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & - 7 / \sqrt{149} & 10 / \sqrt{149} \\ 0 & - 10 / \sqrt{149} & - 7 / \sqrt{149} \end{array})$ acts only on rows $2, 3$ , so it keeps the first column fixed and sends $(- 7 / 5, 2)^{T}$ to $(\sqrt{149} / 5, 0)^{T}$ . Hence $R = G_{2} G_{1} A = (\begin{array}{cc} 5 & 24 / 5 \\ 0 & \sqrt{149} / 5 \\ 0 & 0 \end{array}),$ the displayed upper-triangular matrix.

step 1.1L1algebra

3.1

Since both rotations are unitary, $Q = (G_{2} G_{1})^{*}$ and $A = Q R$ . This verifies the Givens QR factorisation promised by [L2].

step 2.1L2algebra

4.1

Steps 1.1-3.1 verify the example.

step 1.1step 2.1step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, Givens rotations section (standard reference, not scraped)