PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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.

If $A$ has full row rank, then $A^{+} = A^{} (A A^{})^{- 1}$

Statement

Let $F \in {R, C}$ and let $A \in M_{m \times n} (F)$ have full row rank $m$ . Then $A A^{*}$ is invertible and

$A^{+} = A^{*} (A A^{*})^{- 1} .$

Facts & Assumptions

Given: A matrix $A \in M_{m \times n} (F)$ of full row rank, where $F \in {R, C}$ .

[L1]

If a matrix has full column rank, then its pseudoinverse is $(A^{*} A)^{- 1} A^{*}$ (If $A$ has full column rank, then $A^{+} = (A^{*} A)^{- 1} A^{*}$ ).

[L2]

Pseudoinversion commutes with adjoints: $(A^{*})^{+} = (A^{+})^{*}$ (Pseudoinversion is involutive, commutes with adjoints, and is equivariant under unitary left and right factors).

Proof

technique · direct

1.1

Because $A$ has full row rank, the adjoint $A^{*}$ has full column rank. Applying [L1] to $A^{*}$ gives $(A^{*})^{+} = ((A^{*})^{*} A^{*})^{- 1} (A^{*})^{*} = (A A^{*})^{- 1} A .$

L1algebra

2.1

Taking adjoints and using [L2], $A^{+} = ((A^{*})^{+})^{*} = ((A A^{*})^{- 1} A)^{*} = A^{*} (A A^{*})^{- 1} .$

L2step 1.1algebra

3.1

In particular $A A^{*}$ is invertible and the displayed formula holds.

step 2.1∎

Depends on

Used by

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

Gene H. Golub, Least squares, singular values and matrix approximations (standard reference, not scraped)

If AA has full row rank, then A+=A∗(AA∗)−1A+=A∗(AA∗)−1