Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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  • 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.

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Transpose is linear and involutive, and (AB)T=BTAT(AB)^{\mathsf T}=B^{\mathsf T}A^{\mathsf T}

Statement

For matrices of the appropriate shapes and a scalar λ\lambda,

(A+B)T=AT+BT,(λA)T=λAT,(AT)T=A,(A+B)^{\mathsf T}=A^{\mathsf T}+B^{\mathsf T},\qquad (\lambda A)^{\mathsf T}=\lambda A^{\mathsf T},\qquad (A^{\mathsf T})^{\mathsf T}=A,

and

(AB)T=BTAT.(AB)^{\mathsf T}=B^{\mathsf T}A^{\mathsf T}.

Thus transpose is linear and involutive and reverses products.

Facts & Assumptions

Given: A field FF, matrices A,BA,B of the same shape, conformable matrices A,CA,C, and a scalar λ∈F\lambda\in F.

[L1]

Transposition swaps the two entry indices (The transpose ATA^{\mathsf T} of a matrix).

Proof

technique · direct
1.1

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

givenL1
2.1

For conformable A∈Mm×n(F)A\in M_{m\times n}(F) and C∈Mn×p(F)C\in M_{n\times p}(F), one has ((AC)T)ki=(AC)ik=∑j<naijcjk((AC)^{\mathsf T})_{ki}=(AC)_{ik}=\sum_{j<n}a_{ij}c_{jk}.

step 1.1L1
3.1

Commutativity in FF rewrites the sum in step 2.1 as ∑j<n(CT)kj(AT)ji=(CTAT)ki\sum_{j<n}(C^{\mathsf T})_{kj}(A^{\mathsf T})_{ji}=(C^{\mathsf T}A^{\mathsf T})_{ki}, proving the product law entrywise.

step 2.1L1∎

Depends on

Used by

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Sources