ExampleConstruction: AI-generatedVerification: AI-generatedprecheck 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.

A full-column-rank pseudoinverse can be read off directly from a reduced QR factorisation

Example

Let

$A = [\begin{array}{cc} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{array}] = \hat{Q} \hat{R},$

where

$\hat{Q} = [\begin{array}{cc} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{2}{\sqrt{6}} \end{array}], \hat{R} = [\begin{array}{cc} \sqrt{2} & \frac{1}{\sqrt{2}} \\ 0 & \frac{\sqrt{6}}{2} \end{array}] .$

Then

$A^{+} = {\hat{R}}^{- 1} {\hat{Q}}^{*} = [\begin{array}{ccc} \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \end{array}] .$

Facts & Assumptions

Given: The displayed reduced QR factorisation of $A$ .

[L1]

For full column rank, reduced QR gives $A^{+} = {\hat{R}}^{- 1} {\hat{Q}}^{*}$ (Reduced QR gives the full-column and full-row-rank pseudoinverse formulas without forming normal equations).

Verification

technique · computation

1.1

Direct multiplication shows ${\hat{Q}}^{*} \hat{Q} = I$ and $\hat{Q} \hat{R} = A$ , so the displayed matrices are indeed a reduced QR factorisation.

algebra

2.1

The inverse of $\hat{R}$ is ${\hat{R}}^{- 1} = [\begin{array}{cc} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} \\ 0 & \frac{2}{\sqrt{6}} \end{array}] .$ Therefore [L1] gives the displayed matrix for $A^{+}$ .

L1step 1.1algebra

3.1

Multiplying $A^{+} A$ gives the $2 \times 2$ identity, confirming that the matrix from step 2.1 is the full-column-rank pseudoinverse.

step 2.1algebra∎

Depends on

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

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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.