PropositionStatement: Literature-sourcedProof: AI-generatedprecheck 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.

Transpose is linear and involutive, and $(A B)^{T} = B^{T} A^{T}$

Statement

For matrices of the appropriate shapes and a scalar $λ$ ,

$(A + B)^{T} = A^{T} + B^{T}, (λ A)^{T} = λ A^{T}, (A^{T})^{T} = A,$

and

$(A B)^{T} = B^{T} A^{T} .$

Thus transpose is linear and involutive and reverses products.

Facts & Assumptions

Given: A field $F$ , matrices $A, B$ of the same shape, conformable matrices $A, C$ , and a scalar $λ \in F$ .

[L1]

Transposition swaps the two entry indices (The transpose $A^{T}$ of a matrix).

Proof

technique · direct

1.1givenL1

Swapping indices in the entrywise sum and scalar product gives the two linearity identities, and swapping twice gives $(A^{T})^{T} = A$ .

2.1step 1.1L1

For conformable $A \in M_{m \times n} (F)$ and $C \in M_{n \times p} (F)$ , one has $((A C)^{T})_{k i} = (A C)_{i k} = \sum_{j < n} a_{i j} c_{j k}$ .

3.1step 2.1L1∎

Commutativity in $F$ rewrites the sum in step 2.1 as $\sum_{j < n} (C^{T})_{k j} (A^{T})_{j i} = (C^{T} A^{T})_{k i}$ , proving the product law entrywise.

Depends on

Dependency tree · 2 levels

Sources

S. Axler, Linear Algebra Done Right, 4th ed., §3C, Exercises 14–15

Transpose is linear and involutive, and (AB)T=BTAT(AB)T=BTAT