Definition D1
The sum of two matrices A, B of the same shape is defined by (A + B)ᵢⱼ = Aᵢⱼ + Bᵢⱼ.
Proof.
The sum is defined entrywise. Unlike the product, composition plays no part in the sum.convention
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Definition D2
A² means AA, and 2A means A + A. A 2×2 matrix with first row a b and second row c d is written (a b ; c d).
Proof.
The square and the double are not new operations but notation.convention
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Lemma L1
A(B + C) = AB + AC and (A + B)C = AC + BC
Proof.
(1)(A(B + C))ᵢⱼ = Σₖ Aᵢₖ(B + C)ₖⱼA1
(2)(B + C)ₖⱼ = Bₖⱼ + CₖⱼD1
(3)Aᵢₖ(Bₖⱼ + Cₖⱼ) = AᵢₖBₖⱼ + AᵢₖCₖⱼA3
(4)Σₖ (AᵢₖBₖⱼ + AᵢₖCₖⱼ) = Σₖ AᵢₖBₖⱼ + Σₖ AᵢₖCₖⱼ(3), A4, A5, I
(5)Σₖ AᵢₖBₖⱼ + Σₖ AᵢₖCₖⱼ = (AB)ᵢⱼ + (AC)ᵢⱼ = (AB + AC)ᵢⱼA1, D1
(6)All entries agree, so A(B + C) = AB + AC(1), (2), (4), (5)
(7)((A + B)C)ᵢⱼ = Σₖ (Aᵢₖ + Bᵢₖ)CₖⱼA1, D1
(8)(Aᵢₖ + Bᵢₖ)Cₖⱼ = AᵢₖCₖⱼ + BᵢₖCₖⱼA3, A6
(9)By the same method, (A + B)C = AC + BC(7), (8), A4, A5, I, A1, D1
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Theorem T1
"For all matrices A, B, AB = BA" is false.
Proof.
(1)Set A = (0 1 ; 0 0), B = (0 0 ; 1 0).convention
(2)(AB)₁₁ = A₁₁B₁₁ + A₁₂B₂₁ = 0·0 + 1·1 = 1A1, N
(3)Computing the remaining entries the same way gives AB = (1 0 ; 0 0)A1, N
(4)By the same method, BA = (0 0 ; 0 1)A1, N
(5)The (1, 1) entries are 1 and 0, which differ, so AB ≠ BA(3), (4)
(6)Therefore this A, B is a counterexample to the universal statement.(5)
(7)One counterexample exists, so the statement is false.(6), A7
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Theorem T2
The matrix product satisfies associativity and distributivity. The only law that fails is commutativity.
Proof.
(1)(AB)C = A(BC) holds, because the product is composition and composition is associative.A2
(2)A(B + C) = AB + AC and (A + B)C = AC + BC also hold.L1
(3)AB = BA does not hold.T1
(4)Therefore, of the three laws, the only one lost is commutativity.(1), (2), (3)
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Corollary C1
(A + B)² = A² + AB + BA + B², and this is not equal to A² + 2AB + B².
Proof.
(1)(A + B)² = (A + B)(A + B)D2
(2)(A + B)(A + B) = A(A + B) + B(A + B)L1
(3)A(A + B) = AA + AB, B(A + B) = BA + BBL1
(4)(A + B)² = A² + AB + BA + B²(1), (2), (3), D2
(5)2AB is AB + AB, so merging AB + BA into 2AB requires AB = BA.D2
(6)For the A, B of T1, A² = (0 0 ; 0 0), B² = (0 0 ; 0 0)T1, D2, A1, N
(7)A + B = (0 1 ; 1 0), so (A + B)² = (1 0 ; 0 1)T1, D1, D2, A1, N
(8)A² + AB + BA + B² = (1 0 ; 0 1), A² + 2AB + B² = (2 0 ; 0 0)(6), T1, D1, D2, N
(9)Therefore (A + B)² ≠ A² + 2AB + B²(7), (8)
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Corollary C2
A³ is determined uniquely regardless of parenthesization.
Proof.
(1)A³ can be written as (AA)A or as A(AA).D2
(2)(AA)A = A(AA)A2
(3)Therefore A³ has a single meaning even written without parentheses.(1), (2)
(4)This holds even without commutativity. All that is needed is associativity.(3), T1
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