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#1 2007-09-06 15:40:16

mikau
Member
Registered: 2005-08-22
Posts: 1,504

linear algebra help

Let A_1, A_2, ... A_r be vectors in R^n and assume they are mutually perpendicular. (i.e. any two of them are perpendicular), and that none of them is equal to O. Prove they are linearly independant.

btw. in case anyone's forgotten, perpendicular means the dot product is zero.

I've spent countless hours on this problem today and still could not find a solution. I'm a tad discouraged. sad

Last edited by mikau (2007-09-06 15:42:35)


A logarithm is just a misspelled algorithm.

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#2 2007-09-07 05:51:51

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: linear algebra help

It requires a bit of a trick.  Proceed as you normally would in an independence proof, then take the dot product of both sides with one (any) of the vectors you are given.  This must equal 0.  It also happens that every piece of the left side drops out except one.  This allows you to conclude the scalar is 0.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2007-09-07 05:59:46

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: linear algebra help

procede as i normally would in an independance proof? I really don't know of any normal method as it seems to vary per problem.

i usually start with k1 A + k2 B + k2 C ... = 0. where do we go from here?


A logarithm is just a misspelled algorithm.

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#4 2007-09-07 06:04:05

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: linear algebra help

In an independence proof, the line you typed is just about as common as "let epsilon be greater than 0" for real analysis.

Take the dot product of A and (k1 A + k2 B + k3 C + ...).


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2007-09-07 06:14:32

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: linear algebra help

In an independence proof, the line you typed is just about as common as "let epsilon be greater than 0" for real analysis.

naturally, but what then do you mean by 'procede as you normally would'

well distributing the dot product, we get

A*ki*A + A*k2*B + A*k2 *C .... = 0 as far as i know.

wait, are you saying A*A != 0? can't "any pair" also mean, against itself?


A logarithm is just a misspelled algorithm.

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#6 2007-09-07 06:17:06

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: linear algebra help

i guess it would be impossible. if A*A = 0 then (a1)^2 + (a2)^2 + (a3)^2 ... = 0 which, for real numbers, forces them all to be zero, which is not allowed.

doggone it! swear how could i miss that?

well thanks, Ricky. I can't believe you figured that out so quick.

Last edited by mikau (2007-09-07 06:18:48)


A logarithm is just a misspelled algorithm.

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#7 2007-09-07 06:34:04

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: linear algebra help

naturally, but what then do you mean by 'procede as you normally would'

In an independence proof, the first line almost always starts out as "Let k1 A + k2 B + k3 C + ... = 0".  That is what I meant.

Just remember that properties are given to you for a reason.  If you aren't sure what to do and you are given a property about a dot product, take a dot product of something.  But honestly, this is a type of problem where you would normally get, "(Hint: Take a dot product of the linear combination)".


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#8 2007-09-07 06:59:17

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: linear algebra help

yes, except my mathbooks were designed to destroy me!

i wish i had spotted that ealier. See when they said 'any pair' i thought that meant you could use two of the same one as well (sinc ethey never said we couldn't), since i took their word for it i never bothered to try showing it was impossible.


A logarithm is just a misspelled algorithm.

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