University of Waterloo MATH 136 Assignment 7 Solutions

University of Waterloo MATH 136 Assignment 7 Solutions 1. Determine, with proof, which of the following are subspaces of the given vector space: (a) S1 = {p(x) ∈ P4(R) | p(1) = 0} of P4(R). Solution: By definition, S1 is a subset of P4(R) and z(x) = 0 ∈ S1 since z(1) = 0. Thus, S1 is a non-empty subset of P4(R). Let p(x), q(x) ∈ S1. Then p(1) = q(1) = 0. Hence, (p + q)(1) = p(1) + q(1) = 0 + 0 = 0, and so (p + q)(x) ∈ S1. Now, let t ∈ R, then (tp)(1) = t(p(1)) = t(0) = 0, and so (tp)(x) ∈ S1. Therefore, S1 is a subspace of P4(R). (b) S2 = a b c d     a + b = c − d  of M2×2(R). Solution: By definition, S2 is a subset of M2×2(R) and  0 0 0 0 ∈ S2 since 0+0 = 0 − 0. Thus, S2 is a non-empty subset of M2×2(R). Let  a1 b1 c1 d1  ,  a2 b2 c2 d2  ∈ S2. Then a1 +b1 = c1 −d1 and a2 +b2 = c2 −d2. Hence,  a1 b1 c1 d1  +  a2 b2 c2 d2  =  a1 + a2 b1 + b2 c1 + c2 d1 + d2  . Furthermore, (a1 + a2) + (b1 + b2) = (a1 + b1) + (a2 + b2) = (c1 − d1) + (c2 − d2) = (c1 + c2) − (d1 + d2), and so S2 is closed under addition. Now, let t ∈ R, then t  a1 b1 c1 d1  =  ta1 tb1 tc1 td1  . Furthermore, ta1 + tb1 = t(a1 + b1) = t(c1 − d1) = tc1 − td1, and so S2 is closed under scalar multiplication. Therefore, S2 is a subspace of M2×2(R).