Question

In: Math

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

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.

Solutions

Expert Solution

(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.


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