Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-08-30
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  • 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.
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Givens QR can zero selected entries of a sparse matrix one at a time

Example

For A=(344302), 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=−7/149, s=10/149. The product zeros one selected entry at a time and yields R=(52450149500).

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).

Verification

technique · direct
1.1L1algebra

The first rotation G1=(3/54/50−4/53/50001) 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.

2.1step 1.1L1algebra

The second rotation G2=(1000−7/14910/1490−10/149−7/149) acts only on rows 2,3, so it keeps the first column fixed and sends (−7/5,2)T to (149/5,0)T. Hence R=G2G1A=(524/50149/500), the displayed upper-triangular matrix.

3.1step 2.1L2algebra

Since both rotations are unitary, Q=(G2G1)∗ and A=QR. This verifies the Givens QR factorisation promised by [L2].

4.1step 1.1step 2.1step 3.1∎

Steps 1.1-3.1 verify the example.

Depends on

Used by

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Dependency tree · two levels

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Sources