PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31

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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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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

Statement

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

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

Facts & Assumptions

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

[L1]

Full column rank means that all $n$ singular values of $A$ are nonzero (The rank of a linear map is the number of its nonzero singular values).

[L2]

Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse (Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse).

[L3]

An invertible matrix has a two-sided inverse (Invertible matrices and the general linear group ${GL}_{n} (F)$ ).

[L4]

Every finite real or complex matrix has a singular value decomposition (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

Proof

technique · direct

1.1

By [L4], choose an SVD $A = U Σ V^{*}$ , and by [L1] the diagonal matrix $Σ$ has the form $diag (σ_{1}, \dots, σ_{n})$ with every $σ_{i} > 0$ .

L1L4

2.1

Then $A^{*} A = V Σ^{*} Σ V^{*} = V diag (σ_{1}^{2}, \dots, σ_{n}^{2}) V^{*},$ so [L3] gives $(A^{*} A)^{- 1} = V diag (σ_{1}^{- 2}, \dots, σ_{n}^{- 2}) V^{*} .$

L3step 1.1algebra

3.1

Multiplying by $A^{*} = V Σ^{*} U^{*}$ yields $B : = (A^{*} A)^{- 1} A^{*} = V diag (σ_{1}^{- 1}, \dots, σ_{n}^{- 1}) U^{*} .$ Direct diagonal multiplication shows that $B$ satisfies all four Penrose equations for $A$ , so uniqueness in [L2] gives $B = A^{+}$ .

L2step 2.1algebra

4.1

Therefore $A^{*} A$ is invertible and $A^{+} = (A^{*} A)^{- 1} A^{*}$ .

step 3.1∎

Depends on

Used by

Dependency tree · two levels

12 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)
Stanford CS205L, Unit 11: Zero Singular Values (standard reference, not scraped)

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