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Linearly Independent Proof

02/13/2011 6:08 PM

Need help with finishing this proof.

Question

Let {x1 , . . . , xk} be a linearly independent finite set in a normed linear space. Show that there exists delta > 0 such that the condition

max||xi-yi||<delta ,1<=i<=k

implies that {y1 , . . . , yk} is also linearly independent

proof:

Assume {y1 , . . . , yk} is also linearly dependent

there exist c1,...,ck not all zero s.t.

c1y1+...+ckyk=0

WLOG suppose c1=1

yi=xi+ei, ,1<=i<=k (where ei is really small)

c1(x1+e1)1+...+ck(xk+ek)=0

c1x1+...+ckxk=-(c1e1+...+ckek)

RHS span{x1,..xk}

therefore,

c1e1+...+ckek=t1x1+...+tkxk

Now, I don't know what to do. I know i need to use that fact that the norms are equivalent to prove that c1 cannot equal 1, but I don't know how to do it.

Thankful for any help

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#1

Re: linearly independet proof

02/13/2011 6:49 PM
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#2

Re: linearly independet proof

02/13/2011 6:58 PM

If your going to uses acronyms please specify them first. It takes a little effort and it means people can understand you! Your post is pure Gobbledegook! You'll find it on Google!

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#3

Re: linearly independet proof

02/13/2011 7:23 PM

Please reformulate this in a browser with a toolbar that allows proper formatting of subscripts and symbols such as ≤.

Engineering and pure math are not the same; I would bet that less than 0.1% of CR4 members would know what a "normed linear space" is; e.g., I've long since forgotten.

Use δ and ε rather than "delta" and "e".

WLOG = "without loss of generality"; RHS = ??

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#4

Re: Linearly Independent Proof

02/14/2011 8:15 AM

Part of the problem states that:

c1y1+...+ckyk=0

and

WLOG suppose c1=1

At the end you say: I know i need to use that fact that the norms are equivalent to prove that c1 cannot equal 1, but I don't know how to do it.

I'm not sure that this statement is true (that c1 ≠ 1). Consider the equivalent equation:

c1'y1+...+ck'yk=0, where c1' ≠ 0, and ≠ 1. Divide both sides of the equation by c1' and you get:

c1'y1/c1' + c2'y2/c1' + ... ck'yk/c1' = 0, which can be rewritten as:

c1y1 + c2y2 + ... ckyk = 0; where c1 clearly equals 1 ( and c2'/c1' = c2, etc).

Thus c2 can equal 1.

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#5
In reply to #4

Re: Linearly Independent Proof

02/14/2011 9:30 AM

The final statement should, of course, have been:

Thus c1 can equal 1.

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