let V be a vector space and let s = {v1, v2 ...vp} be a set containing p vectors
ID: 3085564 • Letter: L
Question
let V be a vector space and let s = {v1, v2 ...vp} be a set containing p vectors from V. If span s = v and if dimV=p, then the elements of s are linearly independent. why?Explanation / Answer
If {v1 , . . . , vm } is linearly independent then it is a basis of V already and we don’t need to discard any vectors. So assume {v1 , . . . , vm } is linearly dependent. By Lemma 1, when we sweep through the list v 1 , v 2 , . . . , vm from left to right, we will ?nd a vector which is a linear combination of the previous ones. Call such a vector “bad”. Let va be the ?rst bad vector encountered and discard it. Note that • a > k since {v1 , . . . , vk } is assumed to be linearly independent. • {v1 , . . . , vk , . . . , va-1 } is linearly independent, since va was the ?rst bad vector enountered. • {v1 , . . . , vk , . . . , va-1 , va+1 , . . . , vm } spans V, since va is in the span of this set, so we can express every element of V without using va . If the new set (with the bad vector va removed) is linearly independent, we have a basis. Otherwise repeat the sweep through the list with va removed and again discard the ?rst bad vector, call it vb , that is a linear combination of the previous ones. As we iterate this procedure, the bad vectors move progressivly to the right on our list, so we will eventually run out of bad vectors. Thus we arrive a linearly independent set {v1 , . . . , vk , . . . , va-1 , . . . , vb-1 , . . . } that still spans V, hence is a basis of V containing {v1 , . . . , vk }, as desired. ¦ As a consequence of Lemma 2 we have that Every spanning set may be shrunk down to a basis.
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