Math Newberger 247 Spring 03
Homework solutions: Section 4.1 #5-8 In Exercises 5-8 determine if the given set is a subspace of Pn for an appropriate value of n. Justify your answer. 5. All polynomials of the form p(t) = at2 . The set of all polynomials of the form p(t) = at2 is all scalar multiples of the polynomial t2 , thus this set is just Span{t2 }. Since all spans are subspaces, this set is a subspace. 6. All polynomials of the form p(t) = a + t2 . Let H denote the set of all polynomials of the form p(t) = a + t2 . Then we can check that the zero polynomial 0 + 0t + 0t2 is not in H . We try to nd a number a such that a + t2 = 0 + 0t + 0t2 . Since the coecient on t2 on the LHS is 1, but the coecient on t2 on the RHS is 0, these can never be equal (since 0 = 1). Thus the zero polynomial is not in H , since there is no value of a that will make a + t2 = 0+0t +0t2 . Since H does not contain the zero polynomial, H is not a subspace. 7. All polynomials of degree at most 3, with integers as coecients. Let H denote the set of all polynomials of degree at most 3, with integers as coecients. Then the zero polynomial is in H since 0 is an integer. Furthermore, if we take p(t) = a0 + a1 t + a2 t2 + a3 t3 and q(t) = b0 + b1 t + b2 t2 + b3 t3 in H , then a0 , ..., a3 and b0 , ..., b3 are integers. So p(t) + q(t) (a0 = + b0 ) + (a1 + b1 )t + (a2 + b2 )t2 + (a3 + b3 )t3 is in H since (a0 + b0 ), (a1 + b1 ), (a2 + b2 ), and (a3 + b3 ) are all integers. So H is closed under addition. However, H is not a subspace of P3 since H is not closed under scalar multiplication. For example, let p(t) = 1 + t. Then p is in H since the coecients are 1, , 0 0, which are all integers. However if 1 and c = 2, then cp(t) = 2 + 2t is not in H since 2 is not an integer. 8. All polynomials in Pn such that p(0) = 0. Let H denote the set of all polynomials in Pn such that p(0) = 0. Suppose p(t) = a0 + a1 t + + an tn is an element of H . Then p(0) = 0. But p(0) = a0 , so we get that H is all polynomials of the form p(t) =
1
2
a0 + + an tn where a0 = 0. In other words, H is all polynomials of the form p(t) = a1 t + + an tn . But this shows that H is just all linear combinations of the vectors {t, t2 , . . . , tn }, i.e. H =Span{t, t2 , . . . , tn }. So H is a subspace, since all spans are subspaces. One additional example: All polynomials of the form p(t) = a + bt + 3at2 . Let H be the set of all polynomials of the form p(t) = a + bt + 3at2 . Then H is all polynomials of the form p(t) = a(1 + 3t2 ) + bt. In other words, H is all linear combinations of 1 + 3t2 and t. So H = Span{1 + 3t2 , t}, and H is a subspace of P2 .