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Download AB− BA = A12B21 − A21B12 A11B12 + A12B22 − A12B11
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Selected Solutions Math 420 Homework 2 1/20/12 1.5.8 Let C11 C12 C= C21 C22 be a 2 × 2 matrix. We inquire when it is possible to find 2 × 2 matrices A and B such that C = AB − BA. Prove that such matrices can be found if and only if C11 + C22 = 0. It is easy to see that if A11 A12 A= A21 A22 B11 B12 and B = , B21 B22 then A12 B21 − A21 B12 A11 B12 + A12 B22 − A12 B11 − A22 B12 AB − BA = . A21 B11 + A22 B21 − A11 B21 − A21 B22 A21 B12 − A12 B21 If there exist 2 × 2 matrices A and B such that C = AB − BA, then we have C11 + C22 = ( A12 B21 − A21 B12 ) + ( A21 B12 − A12 B21 ) = 0. Conversely, suppose C11 + C22 = 0, so we have C11 C12 C= . C21 −C11 If C21 6= 0, let " A= 1 1 − C11C+21C12 # " and B = 0 C11 +C12 C21 1 C12 − 1 − C21 0 # . If C21 = 0 but C12 6= 0, let " A= If C21 = C12 = 0, let 1 − CC11 12 1 0 # " and B = 0 C11 A= −C11 0 # 1 + C12 . 0 1 11 − CC12 0 1 and B = . 1 0 In each case, it is easy to check that AB − BA = C. 1.6.6 Suppose A is a 2 × 1 matrix and that B is a 1 × 2 matrix. Prove that C = AB is not invertible. 1 Let Then, we have a A = 1 and B = b1 b2 . a2 a1 b1 a1 b2 C = AB = . a2 b1 a2 b2 If a1 = 0, the first row of C is all zeroes, so C is not invertible. Otherwise, subtracting aa12 times the first row from the second row gives a second row of all zeroes, and again this implies C is not invertible. 1.6.7 Let A be an n × n (square) matrix. Prove the following two statements: (a) If A is invertible and AB = 0 for some n × n matrix B, then B = 0. (b) If A is not invertible, then there exists an n × n matrix B such that AB = 0 but B 6= 0. (a) If A in invertible and AB = 0, then B = A−1 AB = A−1 0 = 0. (b) If A is not invertible, the system AX = 0 has a nontrivial solution X0 . Let B be the n × n matrix whose columns are all equal to X0 . Then B 6= 0 but AB = 0. 1.6.10 Prove the following generalization of Exercise 6. If A is an m × n matrix, B is an n × m matrix and n < m, then AB is not invertible. The homogeneous system of linear equations given by BX = 0 consists of n equations in m unknowns. Since n < m, this system will always have a nontrivial solution. Let X0 be a nontrivial solution. Then ( AB) X = A( BX ) = A(0) = 0, so X0 is a nontrivial solution to ( AB) X = 0. Since a nontrivial solution to this system exists, Theorem 13 implies that AB is not invertible. 2.1.3 If C is the field of complex numbers, which vectors in C3 are linear combinations of (1, 0, −1), (0, 1, 1), and (1, 1, 1)? It is easy to check, by performing elementary row operations, that the matrix 1 0 1 A : = 0 1 1 −1 1 1 is invertible. Thus, the system AX = Y has a solution for each 3 × 1 matrix Y. Since the column Y is a linear combination of the columns of A (Exercise 1.6.6), this means that any vector in C3 is a linear combination of the given vectors. 2 2.1.4 Let V be the set of all pairs ( x, y) of real numbers, and let F be the field of real numbers. Define ( x, y) + ( x1 , y1 ) = ( x + x1 , y + y1 ) c( x, y) = (cx, y). Is V, with these operations, a vector space over the field of real numbers? No, V is not a vector space with these operations, as the distributive proberty 4(d) fails: (1 + 1)(0, 1) = 2(0, 1) = (2(0), 1) = (0, 1), but 1(0, 1) + 1(0, 1) = (0, 1) + (0, 1) = (0 + 0, 1 + 1) = (0, 2). 2.1.5 On Rn , define two operations α⊕β = α−β c · α = −cα. The operations on the right are the usual ones. Which of the axioms for a vector space are satisfied by (Rn , ⊕, ·)? Consider each of the axioms. 3(a) This axiom fails. Addition is not commutative, since α − β 6= β − α whenever α 6= β. 3(b) This axiom fails. Addition is not associative. For example, take α = β = 0 and γ 6= 0. Then (α ⊕ β) ⊕ γ = −γ, but α ⊕ ( β ⊕ γ) = γ. 3(c) This axiom does hold. Taking 0 to be the usual zero vector in Rn , we have α ⊕ 0 = α for all α ∈ Rn , and this vector is unique since α − β = α in Rn implies β = α − α = 0. 3(d) This axiom also holds. We define −α = α for all α ∈ Rn . Then α ⊕ (−α) = α − α = 0. Uniqueness easily follows from uniqueness of additive inverses in Rn (with usual operations). 4(a) This axiom fails, as 1 · α = −α 6= α for nonzero α. 4(b) This axiom fails, as (c1 c2 ) · α = −c1 c2 α, but c1 · (c2 · α) = c1 c2 α, and these are not equal when c1 , c2 , and α are nonzero. 4(c) This axiom holds, since c · (α ⊕ β) = −c(α − β) = −cα − (−cβ) = c · α ⊕ c · β. 4(d) This axiom fails. For example, for α 6= 0, (1 + 1) · α = −2α 6= 0, but 1 · α ⊕ 1 · α = −α − (−α) = 0. 3