linear algebra proof

bonfire09

1. The problem statement, all variables and given/known data
Let E₁ = (1, 0, ... ,0), E₂ = (0, 1, 0, ... ,0), ... , E_n = (0, ... ,0, 1)
be the standard unit vectors of Rⁿ. Let x₁ ... ,x_n be numbers. Show that if
x₁E₁+...+x_nE_n=0 then x_i=0 for all i.


2. Relevant equations



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

Let me know where my proof begins to fall apart? And how do I go about it?

micromass

Why can you assume this??

micromass

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

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

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

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

Yeah, that's what I had in mind.

bonfire09

alright thanks.

algebrat

Think about what relations the basis vectors satisfy, if you notice the right thing, the proof is pretty swift.

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".

micromass

Why does orthonormal matter?? I really doubt we need it.

algebrat

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.