TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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.

$[T]_{\mathcal B'}^{\mathcal C'}=P_{\mathcal C'\leftarrow\mathcal C}[T]_{\mathcal B}^{\mathcal C}P_{\mathcal B\leftarrow\mathcal B'}$

Statement

Let $T:V\to W$ be linear. If $\mathcal B,\mathcal B'$ are ordered bases of $V$ and $\mathcal C,\mathcal C'$ are ordered bases of $W$ , then

$[T]_{\mathcal B'}^{\mathcal C'}=P_{\mathcal C'\leftarrow\mathcal C}[T]_{\mathcal B}^{\mathcal C}P_{\mathcal B\leftarrow\mathcal B'}.$

Facts & Assumptions

Given: The linear map and four ordered bases in the Statement.

[L1]

$P_{\mathcal Y\leftarrow\mathcal X}$ is the matrix of the identity map converting $\mathcal X$ -coordinates to $\mathcal Y$ -coordinates (The change-of-basis matrix $P_{\mathcal C\leftarrow\mathcal B}=[\operatorname{id}_V]_{\mathcal B}^{\mathcal C}$ ).

[L2]

Matrix representation sends a composite of linear maps to the product of their matrices in compatible intermediate bases ( $[S\circ T]_{\mathcal B}^{\mathcal D}=[S]_{\mathcal C}^{\mathcal D}[T]_{\mathcal B}^{\mathcal C}$ ).

Proof

technique · direct

1.1

Regard $T$ from $\mathcal B'$ to $\mathcal C'$ as the composite of the identity on $V$ from $\mathcal B'$ to $\mathcal B$ , then $T$ from $\mathcal B$ to $\mathcal C$ , then the identity on $W$ from $\mathcal C$ to $\mathcal C'$ .

givenL1

2.1

Applying [L2] twice, with the matrix of the last-applied map on the left, gives $[T]_{\mathcal B'}^{\mathcal C'}=P_{\mathcal C'\leftarrow\mathcal C}[T]_{\mathcal B}^{\mathcal C}P_{\mathcal B\leftarrow\mathcal B'}.$

step 1.1L1L2

3.1

The right factor is square of size $\dim V$ , the middle factor has shape $(\dim W)\times(\dim V)$ , and the left factor is square of size $\dim W$ , so the product is defined and has the asserted shape, proving the formula.

step 2.1L1L2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 7 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

S. Axler, Linear Algebra Done Right, 4th ed., §3D, results 3.81 and 3.84 (standard reference, not scraped)
S. Schiavone, MIT 18.700 Day 9, change of basis (standard reference, not scraped)

[T]B′C′=PC′←C[T]BCPB←B′[T]_{\mathcal B'}^{\mathcal C'}=P_{\mathcal C'\leftarrow\mathcal C}[T]_{\mathcal B}^{\mathcal C}P_{\mathcal B\leftarrow\mathcal B'}[T]B′C′​=PC′←C​[T]BC​PB←B′​