Let V be a finite dimensional vector space. Also let v 1 , v 2 ,.... , v k be a
ID: 2941254 • Letter: L
Question
Let V be a finite dimensional vector space. Also let v1 , v2 ,.... , vk be a collection of vectors in v.
prove that {v1 , v2 , ...., vk} are independent if and only if vj is not an element of span ( v1 ,v2 ,.. , vj-1 ) for every j = 1, .... , k
you must show that the independence of {v1 , v2 , ...., vk} implies that vj is not the given span for every j. you must also show that vj not in the given span (for all j) implies that all vectors {v1 , v2 , ...., vk} are independent. the second part require inductive argument
Explanation / Answer
WE need to show 2 things:
(i) {v1,v2,...,vk} is independent => vj is not an element of span ( v1 ,v2 ,.. , vj-1 ) for every j = 1, .... , k.
(ii) vj is not an element of span ( v1 ,v2 ,.. , vj-1 ) for every j = 1, .... , k => {v1,v2,...,vk} is independent.
(i) Since the set is independent, no vj is a linear combination of the other vis. In particular, vj is not an element of span ( v1 ,v2 ,.. , vj-1 ) for every j.
(ii) We shall show by induction that each set {v1 ,v2 ,.. , vj-1} is linearly independent. In particular, for j = k we are through. For j = 1 it follows since v1 is not zero, else it is a linear combination of the empty set. Suppose we have proven the statement for j. Now we are given that in addition to the above we also have vj+1 is not a linear combination of v1 ,v2 ,.. , vj. In particular, by induction, v1 ,v2 ,.. , vj are linearly independent. Suppose then that
a1v1 + a2v2 + ... + aj+1vj+1 = 0. If aj+1 = 0 then this gives a linear combination of v1 ,v2 ,.. , vj to the zero vector => all the ai = 0. So, suppose aj+1 is not zero. Then we can write
a1v1 + a2v2 + ... + ajvj = -aj+1 vj+1 and since aj+1 is not zero by assumption, we can divide both sides by this number and obtain that vj+1 is a linear combination of the vectors v1 ,v2 ,.. , vj. Since this is not the case, the induction is complete.
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