linear algebra proof


by bonfire09
Tags: algebra, linear, proof
bonfire09
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#1
Dec26-12, 08:49 PM
P: 219
1. The problem statement, all variables and given/known data
Let E1 = (1, 0, ... ,0), E2 = (0, 1, 0, ... ,0), ... , En = (0, ... ,0, 1)
be the standard unit vectors of Rn. Let x1 ... ,xn be numbers. Show that if
x1E1+...+xnEn=0 then xi=0 for all i.


2. Relevant equations



3. The attempt at a solution
Proof By contradiction
Assume to the contrary that x1E1+...+xnEn=0 then xi0 for some i. We also assume that x1...xi-1 and xi+1...xn are zero. Rewriting the equation we get
x1E1+.xpEp+...+xnEn=0 where xpEp is a nonzero scalar. xpEp=-x1E1-...-xpEp-1-xpEp+1-..-xnEn. But this leads to a contradiction since we assumed earlier that x1...xi-1 and xi+1...xn are zero. Thus x1E1+...+xnEn=0 xi=0 for all i.

Let me know where my proof begins to fall apart? And how do I go about it?
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micromass
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#2
Dec26-12, 08:54 PM
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Quote Quote by bonfire09 View Post
We also assume that x1...xi-1 and xi+1...xn are zero.
Why can you assume this??
micromass
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#3
Dec26-12, 08:58 PM
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Hint: making subscripts like you did in your post is very time-intensive. It is much easier for you to learn LaTeX. I will rewrite the first part of your post using LaTeX here, just push on "QUOTE" to see what I did.

Also see http://www.physicsforums.com/showpos...17&postcount=3 for a guide.

1. The problem statement, all variables and given/known data
Let [itex]E_1 = (1, 0, ... ,0), E_2 = (0, 1, 0, ... ,0), ... , E_n = (0, ... ,0, 1)[/itex].
be the standard unit vectors of [itex]R^n[/itex]. Let [itex]x_1,...,x_n[/itex] be numbers. Show that if
[tex]x_1E_1 + ... + x_nE_n=0[/tex]
then [itex]x_i=0[/itex] for all i.

bonfire09
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#4
Dec26-12, 09:32 PM
P: 219

linear algebra proof


Oh thanks. Yeah my proof is bad I just realized I used the conclusion as my assumption.
I don't think I even need to use a proof by contradiction. Isn't just obvious that if one of the scalars is nonzero then the equation is not zero? Wouldn't that suffice as my proof.
micromass
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#5
Dec26-12, 09:36 PM
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Quote Quote by bonfire09 View Post
Oh thanks. Yeah my proof is bad I just realized I used the conclusion as my assumption.
I don't think I even need to use a proof by contradiction. Isn't just obvious that if one of the scalars is nonzero then the equation is not zero? Wouldn't that suffice as my proof.
Well, if your teacher is happy with "it is just obvious", then yes.

If not, try to calculate

[tex]x_1E_1+...+x_nE_n[/tex]

By definition, we know that [itex]E_1=(1,0,....)[/itex]. So what is [itex]x_1E_1[/itex]? What is [itex]x_2E_2[/itex]? What happens if you add them?
bonfire09
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#6
Dec26-12, 09:43 PM
P: 219
Oh x1E1=(x1,...,0) x2E2=(0,x2,...,0) all the way to xnEn=(0,...,xn). So by adding them together you get (x1,x2,...,xn). And they only way to get the zero vector is when x1...xn is zero? Would that be a way to explain it?
micromass
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#7
Dec26-12, 09:46 PM
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Quote Quote by bonfire09 View Post
Oh x1E1=(x1,...,0) x2E2=(0,x2,...,0) all the way to xnEn=(0,...,xn). So by adding them together you get (x1,x2,...,xn). And they only way to get the zero vector is when x1...xn is zero? Would that be a way to explain it?
Yeah, that's what I had in mind.
bonfire09
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#8
Dec26-12, 09:52 PM
P: 219
alright thanks.
algebrat
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#9
Dec27-12, 12:10 AM
P: 428
Think about what relations the basis vectors satisfy, if you notice the right thing, the proof is pretty swift.
algebrat
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#10
Dec27-12, 12:13 AM
P: 428
The problem with the above proof, it doesnt seem to use the fact that the basis is orthonormal. You could potentially "prove something false".
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#11
Dec27-12, 12:20 AM
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Quote Quote by algebrat View Post
The problem with the above proof, it doesnt seem to use the fact that the basis is orthonormal. You could potentially "prove something false".
Why does orthonormal matter?? I really doubt we need it.
algebrat
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#12
Dec28-12, 02:04 AM
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Quote Quote by micromass View Post
Why does orthonormal matter?? I really doubt we need it.
Oops, yeah you're right aren't you. I guess because the slick proof I was thinking of, was take x.e_1=x_1=0. That works right?

So I was just guessing that it relied on some qualities of the basis vector, but maybe the real mistake would be to not refer to the fact that they are linearly independent. We do have to mention that right?

But perhaps the quality of orthogonal was not relavant, as you say.


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