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.

$[S \circ T]_{B}^{D} = [S]_{C}^{D} [T]_{B}^{C}$

Statement

Let $T : U \to V$ and $S : V \to W$ be linear, with ordered bases $B$ of $U$ , $C$ of $V$ , and $D$ of $W$ . Then

$[S \circ T]_{B}^{D} = [S]_{C}^{D} [T]_{B}^{C} .$

Facts & Assumptions

Given: The composable linear maps and ordered bases in the Statement, and a vector $u \in U$ .

[L1]

A composite of linear maps is linear (Identity maps and composites of linear maps are linear).

[L2]

Coordinate action gives $[R (x)]_{Y} = [R]_{X}^{Y} [x]_{X}$ ( $[T (v)]_{C} = [T]_{B}^{C} [v]_{B}$ ).

[L3]

Matrix multiplication is associative whenever the shapes are compatible (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

Proof

technique · direct

1.1givenL1L2

Apply [L2] to $T$ and then to $S$ : $[S (T (u))]_{D} = [S]_{C}^{D} [T (u)]_{C} = [S]_{C}^{D} [T]_{B}^{C} [u]_{B}$ .

2.1step 1.1L2L3

Associativity from [L3] rewrites step 1.1 as $[S (T (u))]_{D} = ([S]_{C}^{D} [T]_{B}^{C}) [u]_{B}$ for every $u$ .

3.1step 2.1L1L2L3∎

Evaluating at each vector of $B$ makes $[u]_{B}$ a standard coordinate column, so the columns of $[S \circ T]_{B}^{D}$ equal those of the displayed product. Therefore the matrices are equal.

$[S \circ T]_{B}^{D} = [S]_{C}^{D} [T]_{B}^{C}$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

[S∘T]BD=[S]CD[T]BC[S∘T]BD​=[S]CD​[T]BC​

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

$[S \circ T]_{B}^{D} = [S]_{C}^{D} [T]_{B}^{C}$