PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

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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

1.1

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

givenL1

2.1

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

step 1.1L1

3.1

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.

step 2.1L1∎

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

S. Axler, Linear Algebra Done Right, 4th ed., §3C, Exercises 14–15 (standard reference, not scraped)

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