Prove Linear Independence of {u_1,u_2,...,u_k} for k=2,3,...,n-1

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In summary, this conversation is about proving that if a set of vectors is linearly independent, then the vectors themselves are also linearly independent.
  • #1
Phymath
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let me know if u think i did this proof correctly...

use the definition of linear independence to prove that if [tex]{u_1,u_2,...,u_n}[/tex] is a linearly independent set, then [tex]{u_1,u_2,...,u_k}[/tex] is also a linearly independent set for any
[tex]k = 2,3,...,n-1[/tex]

well here's what i did...
if the set of vectors [tex]{u_1,u_2,...,u_n}[/tex] is linearly independent then none of the vectors in that set is linear combination of the others. therefore
[tex]c_1 u_1 + c_2 u_2 + ... + c_n u_n = \vect{0} [/tex] is satisfied only by the set of scalars [tex] {c_1,c_2,...,c_n}[/tex] that is [tex]c_1 = 0, c_2 = 0, ..., c_n = 0[/tex] then.. [tex] c_1 u_1 + c_2 u_2 +...+c_(n-1) u_(n-1) = -c_n u_n[/tex] should only be satisfied by the same set where [tex]c_1 = 0, c_2 = 0, ..., c_n = 0[/tex] because no vector is a linear combination of the others by definition. thus this can be expanded to...[tex]c_1 u_1 + c_1 u_2 + ... + c_k u_k = -c_n u_n -c_(n-1) u_(n-1) - ... -c_(k+1) u_(k+1)[/tex] the only solution is the set where [tex]c_1 = 0, c_2 = 0, ..., c_n = 0[/tex], thus the set [tex] {u_1,u_2, ...,u_k}[/tex] is linearly independent for any [tex]k=2,3,...,n-1.[/tex]

tell me what u think..
 
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  • #2
I think that this question doesn't even need proof.

If there is a linear combination of u_1,..,u_k that is zero, then there is a linear combination of u_1,..,u_n (k<=n) that is also zero. done. You're writing far too much.
 
  • #3
Well, yes, it needs a proof, its just a trivial one!

Phymath: Suppose [tex]\{u_1,u_2,...,u_{n-1}\}[/tex] were not linearly independent. That would mean that there exist constants [tex]{\kc_1,c_2,...,c_{n-1}\}[/tex], not all 0, such that [tex]c_1u_1+ c_2u_2+...+c_{n-1}u_{n-1}= 0[/tex].
What does that tell you about the orginal set [tex]\{u_1,u_2,...,u_n\}[/tex]?
 
  • #4
well idk i guess that's why i asked the question in the first place. I suppose it could mean that u_(n-1) is a linear combination of the procedding u vectors..[tex] u_1, u_2, ...,u_{n-2}[/tex] no? but i thought by defination that can't be...i suppose that i tells excaztly what i said in my proof that there are no scalar constants to be a solution to that set other then all the constants have to be 0. because taking ur equation if they are linearly dependent (aka not linearly independent) then u can do this, which is by defintion only solved if the constants are zero, please help me out here...i don't get ur point
[tex]c_1 u_1 +c_2 u_2 +...+c_(n-2) u_(n-2) = -c_(n-1) u_(n-1) [/tex]
 
  • #5
linear dependence means zero is a non trivial linear combination of some of the vectors.

but if zero is a linear combination of some of the first k vectors, then zero is also (the same) linear combination of some of the full set of n vectors.

how's this?
 
  • #6
another point of view: linear independence means the map sending the standard unit vectors to those vectors is injective. but an injective map on n space restricts to an injective map on a k dimensional subspace.

ok ok i know that makes it worse.
 
  • #7
mathwonk said:
linear dependence means zero is a non trivial linear combination of some of the vectors.

but if zero is a linear combination of some of the first k vectors, then zero is also (the same) linear combination of some of the full set of n vectors.

how's this?
yes but i have to work this backwards...n then to k not k then to n
 
  • #8
you are mistaking the logic. to prove " if all n vectors are independent then so are the first k", is the same as proving: " if the first k vectors are dependent then so are all n".

so i worked from k to n because I was proving the second statement. I.e. I was working with dependency, and proving your statement "by contradiction". proofs by contradiction work backwards, instead of assuming the hypothesis is true and then deducing the conclusion is also true, you assume instead that the conclsuion is false and then deduce that the hypothesis would also be false.

its the same thing.
 
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FAQ: Prove Linear Independence of {u_1,u_2,...,u_k} for k=2,3,...,n-1

What is linear independence?

Linear independence refers to the property of a set of vectors in a vector space where no vector can be written as a linear combination of the other vectors in the set. In other words, the vectors are not redundant and each vector brings something new to the span of the set.

How do you prove linear independence?

To prove linear independence of a set of vectors, you can use the definition of linear independence and show that the only solution to the equation c1u1 + c2u2 + ... + ckuk = 0 is c1 = c2 = ... = ck = 0. This can be done by setting up a system of equations and solving for the coefficients c1, c2, ..., ck. If the only solution is the trivial solution, then the set of vectors is linearly independent.

What is the importance of proving linear independence?

Proving linear independence is important because it allows us to determine whether a set of vectors can form a basis for a vector space. If a set of vectors is linearly independent, then they can span the entire vector space and form a basis. Additionally, linear independence is a necessary condition for many mathematical concepts and calculations, such as finding eigenvalues and eigenvectors.

Can a set of two vectors be linearly dependent?

Yes, a set of two vectors can be linearly dependent if one vector is a scalar multiple of the other. In other words, one vector can be written as a linear combination of the other vector. For example, if u1 = (1, 2) and u2 = (2, 4), then u2 = 2u1, making the set linearly dependent.

Is it possible for a set of n vectors to be linearly independent for all values of k?

No, it is not possible for a set of n vectors to be linearly independent for all values of k. This is because if n > k, then there will always be at least one redundant vector in the set, making it impossible for the set to be linearly independent.

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