Question

Let W be a subspace of R^n and suppose that v1,v2,w1,w2,w3 are vectors in W. Suppose that v1; v2 are linearly independent and that w1;w2;w3 span W.

(a) If dimW = 3 prove that there is a vector in W that is not equal to a linear combination of v1 and v2.

(b) If w3 is a linear combination of w1 and w2 prove that w1 and w2 span W.

(c) If w3 is a linear combination of w1 and w2 prove that dimW = 2 by showing that dimW >= 2 and dimW<= 2.

Answer 1

(a). Let us assume that v₁, v₂ are linearly independent, dim (W) = 3 and that every vector in W is a linear combination of v₁, v₂. Then the set{ v₁, v₂ } is a spanning set for W. Hence, by the definition of a basis the set { v₁, v₂ } is a basis for W as it is a linearly independent set which spans W. This implies that dim (W) = 2 which is a contradiction. Hence there is a vector in W that is not equal to a linear combination of v₁ and v₂.

(b). Let the vectors w₁,w₂ ,w₃ span W and let w₃ be a linear combination of w₁ and w₂. Then there exist non-zero scalars a, b such that w₃ = aw₁+bw₂. In such a case , for any scalars p, q, r, we have pw₁+qw₂+rw₃ = pw₁+qw₂+r(aw₁+bw₂) = (p+ar)w₁+(q+br)w₂. Thus, every linear combination of w₁,w₂,w₃ is a linear combination of w₁,w₂. Therefore, w₁ and w₂ span W.

(c ). Let the vectors w₁,w₂ ,w₃ span W and let w₃ be a linear combination of w₁ and w₂.

Since the vectors v₁, v₂ ∈ W and are linearly independent, hence dim (W ) ≥ 2. Further, since the vectors w₁,w₂ ,w₃ span W and since w₃ is a linear combination of w₁ and w₂, hence, by part(b) above, w₁ and w₂ span W so that dim(W ) ≤ 2. Therefore, dim(W ) = 2.