Solutions for Discussion 1, Question B

1. Let f,g be two functions in W. Then by the sum rule for derivatives, (f+g)'' - 3(f+g)' + 2(f+g) = f'' + g'' - 3f' - 3g' + 2f + 2g = (f'' - 3f' + 2f) + (g'' - 3g' + 2g) = 0 + 0 = 0. Hence, f + g is in the set W. Similarly, if t is a real number, then by the scalar multiple rule for derivatives, (tf)'' - 3(tf)' + 2(tf) = tf'' - 3tf' + 2tf = t(f'' - 3f' + 2f) = t.0 = 0. Hence tf is also in the set W. Since W is closed under sums and scalar multiples, it is a vector subspace of Fun(R).
2. The elements of L({e^x,e^2x}) are the functions of the form ae^x+be^2x, where a,b are real scalars. To show every one of these functions is contained in W, we proceed as follows: By the sum and scalar multiple rules for derivatives again, (ae^x+be^2x)'' - 3(ae^x+be^2x)' + 2ae^x+be^2x = (a - 3a + 2a)e^x + (4b - 6b + 2b)e^2x = 0. Hence L({e^x,e^2x}) is contained in W.
3. Let f = a₁ e^x + b₁e^2x and g = a₂ e^x + b₂e^2x be two elements of L({e^x,e^2x}). Then f + g = (a₁ + a₂) e^x + (b₁ + b₂)e^2x, which is also in L({e^x,e^2x}). Similarly, if t is a scalar, tf = (ta₁) e^x + (tb₁)e^2x is an element of L({e^x,e^2x}). Therefore, L({e^x,e^2x}) is a vector subspace of Fun(R).
4. If E is any subset of a vector space V, and A,B are elements of the linear span of E, then possibly by adding terms with zero coefficients, we may assume that the same collections of vectors from E are "used" in making both A,B: A = c₁A₁ + ... + c_nA_n, and B = d₁A₁ + ... + d_nA_n, where c_i,d_i are real scalars, and A_i are vectors in E. Then A + B= (c₁+d₁)A₁ + ... + (c_n+d_n)A_n, which is also in L(E). Similarly, the scalar multiple tA = (tc₁)A₁ + ... + (tc_n)A_n, which is a vector in L(E). Therefore, L(E) is a vector subspace of V. The linear span of the empty set can be defined to be the subset of V consisting of just the zero vector from V (every vector subspace of V must contain the zero vector, and the set containing the zero vector is a subspace.)
5. Done in class on Monday, February 1.