A vector space and linear independent set.

In summary, if K is an independent set, then for every proper subset T of K, Span(T) is a proper subset of Span(K). However, if K is not linearly independent, then finding a proper subset whose span includes all of Span(K) is difficult.
  • #1
MathematicalPhysicist
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let V be a vector space and K a nonempty subset of V prove/disprove :
K is linear independent set iff for every T such that T is a proper subset of K, span(T) is a proper subset of spanK.

im having difficulty finding a counter example, so i think this statement is correct, but how to prove it?

if K is an independent set, then i can show the for every susbset of it its span includes the span of the subset, but if I am not mistaken this is correct also when K isn't an independent set.
so my question is how to show that if for every T a proper subset of K, and span{T} a proper subset of span{K}?
 
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  • #2
If K is linearly independent, then the span of any proper subset doesn't even include all of K. Conversely, assume K is not linearly independent, and see if you can find a proper subset whose span is all of span(K).
 
  • #3
i don't think you understand, K is linear independent set, i need to prove/disprove that K is linear independent iff for every proper subset of K,T SpT is a proper subset of SpK.

perhaps my first post was confusing, sorry for that.
 
  • #4
loop quantum gravity said:
i don't think you understand, K is linear independent set, i need to prove/disprove that K is linear independent iff for every proper subset of K,T SpT is a proper subset of SpK.

perhaps my first post was confusing, sorry for that.
Yes, StatusX did understand! He said "the span of any proper subset doesn't even include all of K." What is the relationship between K and SpK?
 
  • #5
this is a tautology from the definitions, (plus subtraction and scalar division), so be sure you grok the definitions.
 
  • #6
K is obviously included in spanK, but i don't quite follow, does this mean the statement is correct, and if so then how to prove it, i mean,
even if K is included in spanK.
if K is independent and T is a proper subset of K, then T is also independent, and every combination of the vector of T are included in spanT.
now how do i prove the other side of the statement, i.e when I am given a proper subset, T of K such that spanT is a subset of spanK.
i.e if T={v1,...,vk} K={v1,...,vn} n>k then
a1v1+...anvn=0
a1v1+...akvk=-(ak+1vk+1+...+anvn)
so this is in spanT and thus also in spanK, now I am kind of stuck.
any help would be appreciated.
 
  • #7
I meant, if you take some element v out of K, then, since K is linearly independent, there is no way to write v as a linear combination of the remaining elements. Thus the span of the remaining elements doesn't include all of K, and so can't possibly include all of span(K) (which contains K).
 
  • #8
loop quantum gravity said:
K is obviously included in spanK, but i don't quite follow, does this mean the statement is correct, and if so then how to prove it, i mean,
even if K is included in spanK.
if K is independent and T is a proper subset of K, then T is also independent, and every combination of the vector of T are included in spanT.
now how do i prove the other side of the statement, i.e when I am given a proper subset, T of K such that spanT is a subset of spanK.
i.e if T={v1,...,vk} K={v1,...,vn} n>k then
a1v1+...anvn=0
a1v1+...akvk=-(ak+1vk+1+...+anvn)
so this is in spanT and thus also in spanK, now I am kind of stuck.
any help would be appreciated.
Now we're talking about spanKing? This is getting too kinky for me!:rofl:

Yes, K is included in span(k)- that's the whole point. If T is a proper subset of K, since K is independent, K is NOT in span(T) and therefore neither is span(K)!
 
  • #9
I'll give it a try. Consider vector space V.

Given any nonempty subset K of V, given any proper subset T of K
[K linearly independent -> Span(T) is a proper subset of Span(K)].

I'd say this assertion has already pretty well been established.


Claim: The converse is false. So we should be able to prove its negation.


It is false that, given any nonempty subset K of V, given any proper subset T of K [Span(T) is a proper subset of Span(K) -> K is linearly independent].

This is equivalent to:

There exits a nonempty subset K of V, there exists a proper subset T of K such that [Span(T) is a proper subset of Span(K) and K is not a linearly independent].

This is now an existence assertion. We need only find an example.

Let T = {(1,0)}, K = {(1,0),(1,1),(0,1)}. Clearly Span(T) is a proper subset of Span(K), and K is not linearly independent.


Maybe I misread the OP. Anyway, just my (perhaps a bit pedantic) two cents.
 
  • #10
ok, thanks fopc.
those little simple counter examples. (-:
and halls, you kinky geezer!
(-:
 
  • #11
well paddles were always a standard tool in the old school house as i understand. maybe there is a link from the lower chakras to the brain.
 
  • #12
Personally, I prefer the 2 by 4.
 

1. What is a vector space?

A vector space is a mathematical structure that consists of a set of objects, called vectors, and a set of operations that can be applied to those vectors. These operations include addition and scalar multiplication, and they must follow certain rules in order for the set to be considered a vector space.

2. What is a linear independent set?

A linear independent set is a collection of vectors that cannot be written as a linear combination of each other. In other words, none of the vectors in the set can be created by adding or multiplying the other vectors in the set. This means that each vector in the set is unique and necessary for the set to span the entire vector space.

3. How do you determine if a set of vectors is linearly independent?

To determine if a set of vectors is linearly independent, you can use the concept of linear dependence. If a set of vectors is linearly dependent, it means that at least one vector in the set can be written as a linear combination of the other vectors. On the other hand, if a set of vectors is linearly independent, it means that no vector can be written as a linear combination of the others. One way to determine linear independence is by using the determinant of a matrix formed by the vectors, which must be non-zero for the set to be linearly independent.

4. Why is the concept of linear independence important in vector spaces?

The concept of linear independence is important because it allows us to determine which vectors are necessary for spanning a vector space. In other words, a linearly independent set of vectors forms a basis for the vector space, which means that any vector in the space can be written as a unique linear combination of the basis vectors. This is useful in various applications, such as solving systems of linear equations and representing geometric transformations.

5. Can a set of vectors be linearly independent in one vector space and not in another?

Yes, a set of vectors can be linearly independent in one vector space and not in another. This is because the properties and operations of vector spaces can vary, and what might be considered a linearly independent set in one space may not hold true in another. It is important to consider the specific vector space and its properties when determining linear independence.

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