TheoremStatement: Literature-sourcedProof: 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.

$[v]_{C} = P_{C \leftarrow B} [v]_{B}$ and $P_{B \leftarrow C} = P_{C \leftarrow B}^{- 1}$

Statement

For ordered bases $B, C$ of a finite-dimensional vector space $V$ and every $v \in V$ ,

$[v]_{C} = P_{C \leftarrow B} [v]_{B} .$

Moreover,

$P_{B \leftarrow C} = P_{C \leftarrow B}^{- 1} .$

Facts & Assumptions

Given: Ordered bases $B, C$ of $V$ and a vector $v \in V$ .

[L1]

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

[L2]

Coordinate action converts a vector's coordinate column by the matrix of the map, and matrix representation sends composites to products ( $[T (v)]_{C} = [T]_{B}^{C} [v]_{B}$ , $[S \circ T]_{B}^{D} = [S]_{C}^{D} [T]_{B}^{C}$ ).

Proof

technique · direct

1.1givenL1L2

Applying coordinate action from [L2] to ${id}_{V}$ and using [L1] gives $[v]_{C} = P_{C \leftarrow B} [v]_{B}$ .

2.1step 1.1L1L2

Represent the identity composition ${id}_{V} \circ {id}_{V}$ first from $B$ through $C$ back to $B$ , and then from $C$ through $B$ back to $C$ .

3.1step 2.1L1L2∎

The composite-matrix formula in [L2] gives $P_{B \leftarrow C} P_{C \leftarrow B} = I$ and $P_{C \leftarrow B} P_{B \leftarrow C} = I$ , so the two matrices are inverses. Empty ordered bases give the same two equations in $M_{0} (F)$ .

$[v]_{C} = P_{C \leftarrow B} [v]_{B}$ and $P_{B \leftarrow C} = P_{C \leftarrow B}^{- 1}$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

[v]C=PC←B[v]B[v]C​=PC←B​[v]B​ and PB←C=PC←B−1PB←C​=PC←B−1​

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

$[v]_{C} = P_{C \leftarrow B} [v]_{B}$ and $P_{B \leftarrow C} = P_{C \leftarrow B}^{- 1}$