linear algebra proof

by bonfire09
Tags: algebra, linear, proof
 P: 202 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 xi≠0 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?
PF Patron
Thanks
Emeritus
P: 15,671
 Quote by bonfire09 We also assume that x1...xi-1 and xi+1...xn are zero.
Why can you assume this??
 PF Patron Sci Advisor Thanks Emeritus P: 15,671 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 $E_1 = (1, 0, ... ,0), E_2 = (0, 1, 0, ... ,0), ... , E_n = (0, ... ,0, 1)$. be the standard unit vectors of $R^n$. Let $x_1,...,x_n$ be numbers. Show that if $$x_1E_1 + ... + x_nE_n=0$$ then $x_i=0$ for all i.
P: 202

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.
PF Patron
Thanks
Emeritus
P: 15,671
 Quote by bonfire09 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

$$x_1E_1+...+x_nE_n$$

By definition, we know that $E_1=(1,0,....)$. So what is $x_1E_1$? What is $x_2E_2$? What happens if you add them?
 P: 202 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?
PF Patron
Thanks
Emeritus
P: 15,671
 Quote by bonfire09 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.
 P: 202 alright thanks.
 P: 428 Think about what relations the basis vectors satisfy, if you notice the right thing, the proof is pretty swift.
 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".
PF Patron
Thanks
Emeritus
P: 15,671
 Quote by algebrat 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.
P: 428
 Quote by micromass 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.

 Related Discussions Calculus & Beyond Homework 4 Calculus & Beyond Homework 4 Calculus & Beyond Homework 3 Calculus & Beyond Homework 2 Calculus & Beyond Homework 15