1. ## linear algebra questions

ok i have two questions that i'm stuck on:

S={x_1,...,x_m} is a linearly independent set in V, and y is also a nonzero vector in V that does not lie in span(S). NTS {x_1,...,x_m,y} is linearly independent.

i'm convinced that this works, but why wouldnt this work if y is not a vector that is in its span? shouldnt it work for all cases?

question 2:
S is the same vector that is linearly indep in V, i need to show that dim(span(S))=m. can someone explain the importance of dimension of span as opposed to dimension of the vector space itself?

thanks!

2. If there is a set of vectors in which one member is a linear combination of others in the set then the set is not linearly independent. So in this case adding a vector not in the span allows the set to remain independent as long as the original set is independent.

Does what you wrote make sense?

3. shouldnt that be true? because if u take the set S and take an arbitrary linear combination of the vectors in S, shouldn't it have the same dimension as set S?

4. Originally Posted by squarerootof2
shouldnt that be true? because if u take the set S and take an arbitrary linear combination of the vectors in S, shouldn't it have the same dimension as set S?
You aren't making much sense here. The dimension of a vector space is the cardinality of its basis, i.e., the number of elements of a linearly independent set that spans the vector space.

5. Originally Posted by squarerootof2
shouldnt that be true? because if u take the set S and take an arbitrary linear combination of the vectors in S, shouldn't it have the same dimension as set S?
You are mixing terms here. A mere set does not have the luxury of a dimension. A set thats a Vector Space, has dimensions.

So there is no meaning attached to the term, dimension of set S. However if S is a set of vectors, then the term dimension of span of S is perfectly meaningful.

If you define the dimension of a vector space to be the number of elements in the basis of that vector space and define the basis as the minimal spanning set for that vector space, you just have to prove that span(S) is a minimal spanning set. But thats obvious, because if you drop any vector from S, then the remaining vectors cant combine to give the dropped vector(why?). Thus the set is a minimal spanning set.

Hence S is a basis for span(S). But S has m vectors and thus dim(span(S)) = m.

6. i am aware of the definition of basis. but consider the example of R^m and the standard basis. then if u take the span of R^m by considering the vectors generated by linear combinations of the vectors in R^m, then we OBVIOUSLY get vectors that are still in R^m. i'm just worried about cases where the vector space is not R^m. what would be the reason for this theorem not being true?

7. Originally Posted by Isomorphism
You are mixing terms here. A mere set does not have the luxury of a dimension. A set thats a Vector Space, has dimensions.

So there is no meaning attached to the term, dimension of set S. However if S is a set of vectors, then the term dimension of span of S is perfectly meaningful.

If you define the dimension of a vector space to be the number of elements in the basis of that vector space and define the basis as the minimal spanning set for that vector space, you just have to prove that span(S) is a minimal spanning set. But thats obvious, because if you drop any vector from S, then the remaining vectors cant combine to give the dropped vector(why?). Thus the set is a minimal spanning set.

Hence S is a basis for span(S). But S has m vectors and thus dim(span(S)) = m.
i should have been more clear. what i meant was the vectors generated by taking all possible linear combination of the vectors in S.