(3 points) Determine whether the given set S is a subspace of the vector space V
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Question
(3 points) Determine whether the given set S is a subspace of the vector space V A, V = Pn , and s is the subset of Pn consisting of those polynomials satisfying p(0) = 0. B. V = Ps. and s is the subset of P5 consisting of those polynomials satisfying p(1) > p(0). c. V C(R) (continuously differentiable functions), and S is the subset of V consisting of those functions satisfyingf (0) 20. D. V = Mn×n(R) , and S is the subset of all diagonal matrices E. V = Mnxn(R), and s is the subset of all n × n matrices with det(A) 0. F V CR) (twice continuously differentiable functions), and S is the subset of V consisting of those functions satisfying the differential equation G. V = Rn, and s is the set of solutions to the homogeneous linear system Ax = 0 where A is a fixed m x n matrix.Explanation / Answer
A. Let p1 = a1x+a2x2 +…+anxn and p2 = b1x+b2x2 +…+bnxn be 2 arbitrary elements of S and let k be an arbitrary scalar. Then p1+p2 = a1x+a2x2 +…+anxn + b1x+b2x2 +…+bnxn= (a1+b1)x+(a+b2)x2+…(an+bn)xn. Since (p1+p2)(0) = 0, hence p1+p2 S so that S is closed under vector addition. Further, kp1 = k(a1x+a2x2 +…+anxn) = ka1x+ka2x2 +…+kanxn. Since (kp1)(0) = 0,hence kp1 S so that S is closed under scalar multiplication. Also, the 0 vector apparently belongs to S so that S is a vector space. Hence S is a subspace of V = Rn.
B. Let p be an arbitrary vector in S and let k be an arbitrary scalar. Now, if p(1)> p(0), and if k is negative, then kp(1)< kp(0) so that kp S. Hence S is not closed under scalar multiplication. Therefore, S is not a vector space. Hence S is not a subspace of V =P5.
D. Since the sum of any 2 diagonal matrices is a diagonal matrix, the scalar multiple of a diagonal matrix is a diagonal matrix, and since the zero matrix is also a diagonal matrix, hence S is a vector space, and ,therefore, S is a subspace of V = Mnxn (R ).
E. Since det (A+B) det(A) +det(B), hence, when det(A) = 0,and det(B) = 0, then det(A+B) need not be 0. Thus, S is not closed under vector addition. Therefore, S is a subspace of V = M.nxn (R ).
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