Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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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.

Reduced QR gives the full-column and full-row-rank pseudoinverse formulas without forming normal equations

Statement

  1. If A∈Mm×n(F) has full column rank and A=Q^R^ is a reduced QR factorisation, then A+=R^−1Q^∗.
  2. If A has full row rank and A∗=Q^R^ is a reduced QR factorisation of A∗, then A+=Q^R^−∗.

Facts & Assumptions

Given: A compatible reduced QR factorisation.

[L1]

A reduced QR factorisation is A=Q^R^ with Q^∗Q^=I and R^ square upper triangular (Full, reduced, and column-pivoted computational QR factorisations).

[L2]

In the full-column-rank case, A+=(A∗A)−1A∗ (If A has full column rank, then A+=(A∗A)−1A∗).

[L3]

In the full-row-rank case, A+=A∗(AA∗)−1 (If A has full row rank, then A+=A∗(AA∗)−1).

Proof

technique · direct
1.1L1L2algebra

Suppose A has full column rank and A=Q^R^. By [L1], A∗A=R^∗Q^∗Q^R^=R^∗R^. Using [L2], A+=(A∗A)−1A∗=(R^∗R^)−1R^∗Q^∗=R^−1Q^∗.

2.1L1L3step 1.1algebra

Suppose now that A has full row rank and A∗=Q^R^ is a reduced QR factorisation of A∗. Applying step 1.1 to A∗ gives (A∗)+=R^−1Q^∗. Taking adjoints yields A+=((A∗)+)∗=Q^R^−∗.

3.1step 1.1step 2.1∎

These are exactly the reduced-QR pseudoinverse formulas.

Depends on

Used by

Dependency tree · two levels

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Sources