Question

1. Complete the proof

Let V be a nontrivial vector space which has a spanning set {x_i} ^k_i=1. Then there is a subset of {x_i} ^k_i=1 which is a basis for V.

Proof. We will divide the set {x_i} ^k_i=1 into two sets, which we will call good and bad. If x₁ ≠ 0, then we label x₁ as good and if it is zero, we label it as bad. For each i ≥ 2, if x_i ∉ span{x₁, . . . , x_i−1}, then we label x_i as good, and otherwise we label it as bad. Let B be the set of all the good elements. We claim that B is a basis for V.

B is nonempty because not all the x_i are zero and the first nonzero x_i will be good. We must first verify that B is a linearly independent set. Let B = {b₁, b₂, . . . , b_j} and assume we have scalars c₁, c₂, . . . , c_j such that c₁b₁+c₂b₂+· · ·+c_jb_j = 0. Then c₁b₁+c₂b₂+· · ·+c_j−1b_j−1 = −c_jb_j, and by the construction of the set B, b_j is not a linear combination of the previous elements, which therefore implies that c_j must be zero. But we can repeat this argument. We now know that c₁b₁+c₂b₂ +· · ·+ c_j−1b_j−1 = 0, so that c₁b₁+c₂b₂ +· · ·+c_j−2b_j−2 = −c_j−1b_j−1 and again this implies that c_j−1 = 0. Repeating this will prove that all the coefficients must be zero. Therefore, B is a linearly independent set.

Now show that the span of the elements of B is all of V.

2. Extend to the case when S is an arbitrary subset of V (i.e., do not assume that S is finite).

The linear span of a finite set of vectors in a vector space V is a subspace of V.

Proof. Let S = {x₁, x₂, . . . , x_n} ⊂ V be a given set of vectors, and let W = span(S). Let u and v be vectors in W, and write u = c₁x₁ + c₂x₂ + · · · + c_nx_n and v = d₁x₁ + d₂x₂ + · · · + d_nx_n. Then u + v = Σ ⁿ_i=1(c_i + d_i)x_i , which is a linear combination of the vectors x_i , so u + v is an element W, thus proving that W is closed under addition. If a is a scalar, then au = ac₁x₁ + ac₂x₂ + · · · + ac_nx_n is again a linear combination of {x₁, x₂, . . . , x_n}. Since W is closed under addition and scalar multiplication, it is a subspace of V.

Extend to the case when S is an arbitrary subset of V (i.e., do not assume that S is finite)

Answer 1