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

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

Statement

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

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

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 $[R(x)]_{\mathcal Y}=[R]_{\mathcal X}^{\mathcal Y}[x]_{\mathcal X}$ ( $[T(v)]_{\mathcal C}=[T]_{\mathcal B}^{\mathcal C}[v]_{\mathcal B}$ ).

[L3]

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

Proof

technique · direct

1.1

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

givenL1L2

2.1

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

step 1.1L2L3

3.1

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

step 2.1L1L2L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 33 results over 11 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., §3C, Definition 3.43 and Theorem 3.81 (standard reference, not scraped)
S. Schiavone, MIT 18.700 Day 9, Theorem 35 (standard reference, not scraped)

[S∘T]BD=[S]CD[T]BC[S\circ T]_{\mathcal B}^{\mathcal D}=[S]_{\mathcal C}^{\mathcal D}[T]_{\mathcal B}^{\mathcal C}[S∘T]BD​=[S]CD​[T]BC​