TheoremStatement: AI-adaptedProof: AI-adaptedprecheck 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.

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

Statement

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

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

Facts & Assumptions

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

[L1]

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

[L2]

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

Proof

technique · direct

1.1givenL1

Regard $T$ from $B^{'}$ to $C^{'}$ as the composite of the identity on $V$ from $B^{'}$ to $B$ , then $T$ from $B$ to $C$ , then the identity on $W$ from $C$ to $C^{'}$ .

2.1step 1.1L1L2

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

3.1step 2.1L1L2∎

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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

[T]B′C′=PC′←C[T]BCPB←B′[T]B′C′​=PC′←C​[T]BC​PB←B′​

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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