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Transpose of a product (AB)ᵀ

Short answer

(AB)ᵀ and BᵀAᵀ have same size n×m. Entry (j,i) of (AB)ᵀ is Σ aᵢₖ bₖⱼ; of BᵀAᵀ is Σ bₖⱼ aᵢₖ, equal by commutativity of scalar multiplication, so (AB)ᵀ = BᵀAᵀ.

Prove that (AB)ᵀ = BᵀAᵀ where A is a matrix of size m×p and B is a matrix of size p×n.

Statement. If A is of size m × p and B is of size p × n, then (AB)ᵀ = BᵀAᵀ.

Proof

Orders. AB has size m × n, so (AB)ᵀ has size n × m. Bᵀ is n × p and Aᵀ is p × m, so BᵀAᵀ is n × m. Both sides have the same order.

Entries. Let A = [aᵢₖ] and B = [bₖⱼ], 1 ≤ i ≤ m, 1 ≤ k ≤ p, 1 ≤ j ≤ n. By the definition of a product, (AB)ᵢⱼ = Σₖ aᵢₖ bₖⱼ.

The (j, i) entry of (AB)ᵀ equals the (i, j) entry of AB:

[(AB)ᵀ]ⱼᵢ = Σₖ aᵢₖ bₖⱼ        (k = 1 to p)

The (j, i) entry of BᵀAᵀ is, using (Bᵀ)ⱼₖ = bₖⱼ and (Aᵀ)ₖᵢ = aᵢₖ:

[BᵀAᵀ]ⱼᵢ = Σₖ (Bᵀ)ⱼₖ (Aᵀ)ₖᵢ = Σₖ bₖⱼ aᵢₖ

Multiplication of numbers is commutative, so the two sums are equal for every j and i. Therefore (AB)ᵀ = BᵀAᵀ. ∎

Note. The order reverses: (ABC)ᵀ = CᵀBᵀAᵀ.

✓ Answer checked by Claude Opus 5.5 (high effort)

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