Linear algebra proof (surprise)

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Homework Help Overview

The discussion revolves around a proof in linear algebra concerning the linear independence of eigenvectors associated with a matrix A. The original poster seeks to establish that if subsets of eigenvectors corresponding to each eigenvalue are linearly independent, then the entire set of eigenvectors must also be linearly independent.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of a subset of eigenvectors being linearly independent and question how this leads to the linear independence of the entire set. There are discussions about the structure of proofs and the necessity of clarity in arguments.

Discussion Status

There is ongoing dialogue about the validity of the original proof attempt, with some participants providing critical feedback and others suggesting ways to refine the argument. The conversation indicates a mix of interpretations and approaches, with some participants making progress in defining their terms and clarifying their reasoning.

Contextual Notes

Participants note the importance of clearly defining the relationships between eigenvectors and eigenvalues, as well as the implications of linear combinations of these vectors. There is an emphasis on the need for explicit reasoning in proofs, particularly in the context of linear independence.

evilpostingmong
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Homework Statement


Prove that if eigenvectors v1, v2...vn are such that for any eigenvalue c
of A, the subset of all these eigenvectors belonging to c is linearly independent,
then the vectors v1,v2..vn are linearly independent.


Homework Equations





The Attempt at a Solution


One question: How does the fact that a subset is linearly independent
prove that every vector is linearly independent? For example:
if {v1...vi} is linearly independent, i<n but {v1...vn} is not,
then obviously the entire set is not linearly independent, though the
subset is.
 
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As an example to start you off:

If vn was in the linear span of v1 through vn-1 and c was an eigenvalue for each vi i<n, then you can prove c is an eigenvalue for vn too fairly easily
 
Office_Shredder said:
As an example to start you off:

If vn was in the linear span of v1 through vn-1 and c was an eigenvalue for each vi i<n, then you can prove c is an eigenvalue for vn too fairly easily


vn=-c*(a1v1+...+an-1vn-1) so taking some linear combination from <v1...vn-1>
we have c*(a1v1+...+an-1vn-1) but since c(a1v1+...+an-1vn-1)-c(a1v1+...+an-1vn-1)
(or c(a1v1+...+an-1vn-1)-vn)=0
and we can't have c((a1v1+..+an-1vn-1)-(a1v1+...+an-1vn-1))=0 since 0 cannot
be an eigenvector, (c-c)(a1v1+...+an-1vn-1)=0*v1+...+0*vn-1=0 so v1..vn is independent
 
Why are you even bothering to write this junk? You know that's not a 'proof', right? There is nothing in there that even resembles being a proof. If you can give me a reason why that is NOT a proof, I might give you a hint. You really have to start learning how to recognize when you are writing something that is void of content. Sorry to be harsh, but I'm getting tired of this.
 
Dick said:
Why are you even bothering to write this junk? You know that's not a 'proof', right? There is nothing in there that even resembles being a proof. If you can give me a reason why that is NOT a proof, I might give you a hint. You really have to start learning how to recognize when you are writing something that is void of content. Sorry to be harsh, but I'm getting tired of this.

it isn't a proof because its not clear? It still doesn't make sense how a subset being
linearly independent guarantees that the set it is in is linearly independent, if my proof
isn't right.
 
evilpostingmong said:
it isn't a proof because its not clear

That's not good enough. Your first line doesn't even have a "since" before it or a "then" after it. A proof runs like, if A then B, if B then C etc etc. And at the end if Y then Z Where Z is what you want to prove. Your 'proof' shares none of that structure. It's a random fusillade of symbols. It is not a proof. It's not even close. You talk about eigenvectors and you never even mentioned the operator A or the eigenvalues. How is that possible?
 
evilpostingmong said:
vn=-c*(a1v1+...+an-1vn-1) so taking some linear combination from <v1...vn-1>
we have c*(a1v1+...+an-1vn-1) but since c(a1v1+...+an-1vn-1)-c(a1v1+...+an-1vn-1)
(or c(a1v1+...+an-1vn-1)-vn)=0
and we can't have c((a1v1+..+an-1vn-1)-(a1v1+...+an-1vn-1))=0 since 0 cannot
be an eigenvector, (c-c)(a1v1+...+an-1vn-1)=0*v1+...+0*vn-1=0 so v1..vn is independent

actually I didn't really say how I got vn=-c*(a1v1+..+an-1vn-1).
What I should've done was say how vn is in the span but adding it to the
basis to get a span{v1...vn} makes this set not linearly independent, so
setting c(a1v1+...+an-1vn-1)+vn=0 gives vn=-c(a1v1+...+an-1vn-1)
I guess that's one mistake-didn't be explicit enough.

edit: You posted 6 before I posted 7 so I'll consider your explanation and I'll
rewrite the proof.
 
I didn't "explain" anything. I haven't even given you a hint yet. I want you to understand that that what you are writing is garbage and give me a good reason why before I give a hint. You have to start learning to introspect. I.e. read you own writing.
 
evilpostingmong said:
actually I didn't really say how I got vn=-c*(a1v1+..+an-1vn-1).
What I should've done was say how vn is in the span but adding it to the
basis to get a span{v1...vn} makes this set not linearly independent, so
setting c(a1v1+...+an-1vn-1)+vn=0 gives vn=-c(a1v1+...+an-1vn-1)
I guess that's one mistake-didn't be explicit enough.

edit: You posted 6 before I posted 7 so I'll consider your explanation and I'll
rewrite the proof.

Ok, I'll bite. So how is vn in the span? Of what? Why?
 
  • #10
Dick said:
Ok, I'll bite. So how is vn in the span? Of what? Why?
Define a basis of the space V as <v1...vn-1> with dimV=n-1
and with Avi=cvi (with 1<=i<=n-1, and with A being the operator).
The set v1...vn-1 is linearly independent, by definition of a basis.
Consider the span{v1...vn} and let vn be a vector
within the span of V, so Avn=cvn as is Avi=cvi. Since
<v1...vn-1> is within {v1...vn}, the basis is a subset of {v1...vn}. (ok I defined the variables here).
 
Last edited:
  • #11
You are making some progress. You did manage to actually define what you are talking about. It is true that if <v1...vn-1> are eigenvectors with eigenvalue c then any linear combination of them also has eigenvalue c. The proof could use a little work in terms of being explicit, but that's still ok. And after reading stuff you've posted before, I'm not going to diminish that accomplishment. Are you supposed to assume all of the basis vectors are eigenvectors? if so write c1*v1+...cn*vn=0 and operate on both sides of the equation with (A-cI) where c is an eigenvalue. What happens?
 
  • #12
Dick said:
You are making some progress. You did manage to actually define what you are talking about. It is true that if <v1...vn-1> are eigenvectors with eigenvalue c then any linear combination of them also has eigenvalue c. The proof could use a little work in terms of being explicit, but that's still ok. And after reading stuff you've posted before, I'm not going to diminish that accomplishment. Are you supposed to assume all of the basis vectors are eigenvectors? if so write c1*v1+...cn*vn=0 and operate on both sides of the equation with (A-cI) where c is an eigenvalue. What happens?

That's good, Rome wasn't built in a day, but at least something is being built :biggrin:

We have a linear combination from the span {v1...vn} as c1v1+...+cnvn.
Applying A-cI to this equation gives A(c1v1+...+cnvn)-c(c1v1+...+cnvn)=0
or A(c1v1+...+cnvn)=c(c1v1+...+cnvn). Is this what I should've done?
 
Last edited:
  • #13
I'm going to give you the idea in words. Start by writing 0=a1*v1+...+an*vn. You want to prove all of the ai's must be zero. If ALL of the v's corresponded to the SAME eigenvalue then you would be done. Because the problem says to assume that the subset of vectors corresponding to a single eigenvector is linearly independent. And if {v1...vn} were linearly independent, you know the ai's must be zero. Following this so far? The problem is that the vi's correspond to several DIFFERENT eigenvalues. Hmm, you should say, is there some way to eliminate all of the eigenvectors corresponding to all but one eigenvalue? Then you would have the much simpler case. Now how can you use (A-cI)v=0 if v is an eigenvector with eigenvalue c? In words, (A-cI) will wipe out any terms in the sum corresponding to the eigenvalue c.
 
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