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 Post subject: Re: Linear dependencyPosted: Thu, 26 Jul 2012 21:09:03 UTC
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My original solution is "airtight". The only assumptions are those that are given, the v's are dependent, and the w's have the given definitions. I do not need the assumption that the v's are all non-zero, although the case when they are all 0 is vacuous, but the proof is still valid.

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 Post subject: Re: Linear dependencyPosted: Thu, 26 Jul 2012 21:58:14 UTC
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mathematic wrote:
My original solution is "airtight". The only assumptions are those that are given, the v's are dependent, and the w's have the given definitions. I do not need the assumption that the v's are all non-zero, although the case when they are all 0 is vacuous, but the proof is still valid.

Not quite. You wrote

mathematic wrote:
Since the v's are linearly dependent, there is a set of non-zero c's so that c1v1+ c2v2+c3v3=0
.

That's not the definition of linear dependency, so your proof uses an unverified assumption, so it is invalid.

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 Post subject: Re: Linear dependencyPosted: Sat, 28 Jul 2012 00:06:26 UTC
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mathematic wrote:
My original solution is "airtight". The only assumptions are those that are given, the v's are dependent, and the w's have the given definitions. I do not need the assumption that the v's are all non-zero, although the case when they are all 0 is vacuous, but the proof is still valid.

Not quite. You wrote

mathematic wrote:
Since the v's are linearly dependent, there is a set of non-zero c's so that c1v1+ c2v2+c3v3=0
.

That's not the definition of linear dependency, so your proof uses an unverified assumption, so it is invalid.

What's wrong with the definition?

http://en.wikipedia.org/wiki/Linear_independence

(Quotation from above)
Definition

A finite subset of n vectors, v1, v2, ..., vn, from the vector space V, is linearly dependent if and only if there exists a set of n scalars, a1, a2, ..., an, not all zero, such that

a1v1 + a2v2 + ... + anvn = 0.

Note that the zero on the right is the zero vector, not the number zero.

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 Post subject: Re: Linear dependencyPosted: Sat, 28 Jul 2012 00:43:06 UTC
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Yours says "nonzero c's" which means all of them are nonzero, the definition requires only one to be nonzero.

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 Post subject: Re: Linear dependencyPosted: Sat, 28 Jul 2012 03:04:02 UTC
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mathematic wrote:
What's wrong with the definition?

You are reading it wrong --- "not all zero" is not "all not zero"

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 Post subject: Re: Linear dependencyPosted: Sat, 28 Jul 2012 21:56:14 UTC
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outermeasure wrote:
mathematic wrote:
What's wrong with the definition?

You are reading it wrong --- "not all zero" is not "all not zero"

I stand corrected. However the proof itself is valid with the proper wording - in my mind I was always thinking not all zero, even though I reversed "not all" in the text.

Note the final comment: The a's cannot all be 0, since that would lead to all the c's being 0.

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