ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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 worked polar decomposition of an invertible matrix

Example

For

$T = (\begin{array}{cc} 0 & 2 \\ 1 & 0 \end{array}),$

the polar decomposition is

$T = S U, S = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}), U = (\begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array}) .$

Facts & Assumptions

Given: The matrix $T$ above on $R^{2}$ with the standard inner product.

[L1]

Every endomorphism has a polar decomposition $T = S U$ with $U = \sqrt{T^{*} T}$ and $S$ an isometry (Every endomorphism has a polar decomposition T = SU with U non-negative and S an isometry on the orthogonal complement of ker T, and S is unique exactly when T is invertible).

Verification

technique · direct

1.1

Direct multiplication gives $T^{*} T = (\begin{array}{cc} 1 & 0 \\ 0 & 4 \end{array})$ , so its non-negative square root is $U = (\begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array})$ .

L1algebra

2.1

The matrix $S = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ satisfies $S^{T} S = I$ , so it is orthogonal, and $S U = (\begin{array}{cc} 0 & 2 \\ 1 & 0 \end{array}) = T$ . Hence this is the polar decomposition predicted by [L1].

L1step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)