STA3710 Assignment 4 (COMPLETE ANSWERS) 2025 (894289) - Due 9 September 2025

STA3710
Assignment 4
Unique No:894289
Due 09 September 2025

, Question 2.1

2.1.1 Proof that (𝐴 ⊗ 𝐵) ′ = 𝐴 ′ ⊗ 𝐵 ′

General proof: Let 𝐴 = (𝑎 𝑖𝑗 ) be an 𝑚 × 𝑛 matrix and let 𝐵 be another matrix of
compatible size. By definition, the Kronecker product 𝐴 ⊗ 𝐵 is a block matrix in which
the (𝑖, 𝑗)-block is equal to 𝑎𝑖𝑗 𝐵.

When we transpose 𝐴 ⊗ 𝐵, the (𝑗, 𝑖)-block becomes:

(𝑎𝑖𝑗 𝐵) ′ = 𝑎 𝑖𝑗 𝐵 ′ .

Now consider 𝐴′ ⊗ 𝐵 ′ . In this case, the (𝑗, 𝑖)-block is also given by:

𝑎𝑖𝑗 𝐵 ′ .

Thus, the block structures of (𝐴 ⊗ 𝐵) ′ and 𝐴′ ⊗ 𝐵 ′ are identical. Therefore:

(𝐴 ⊗ 𝐵) ′ = 𝐴 ′ ⊗ 𝐵 ′ .

This is a standard Kronecker product property, valid for all conformable matrices.

Numerical verification: Substituting the given matrices into both sides confirms that
the equality holds.

2.1.2 Proof that (𝐴 ⊗ 𝐵)(𝐶 ⊗ 𝐷) = (𝐴𝐶) ⊗ (𝐵𝐷) , provided the products exist

Statement of property: The mixed-product rule for the Kronecker product states:

(𝐴 ⊗ 𝐵)(𝐶 ⊗ 𝐷) = (𝐴𝐶) ⊗ (𝐵𝐷),

whenever the products 𝐴𝐶 and 𝐵𝐷 are well-defined.

Proof: Let 𝐴 = (𝑎 𝑖𝑗 ) and 𝐶 = (𝑐 𝑗𝑘 ). The (𝑖, 𝑘)-block of (𝐴𝐶) ⊗ (𝐵𝐷) is:


ቌ෍ 𝑎 𝑖𝑗 𝑐𝑗𝑘 ቍ(𝐵𝐷).
𝑗


Now compute (𝐴 ⊗ 𝐵)(𝐶 ⊗ 𝐷) . Block multiplication gives: