Prove Linear Algebra Statement: x1E1+...+xnEn=0 then xi=0

In summary: Orthogonal means that the vectors are perpendicular to each other, but maybe that's not what we need. Orthogonal means that the vectors are perpendicular to each other, but maybe that's not what we need.
  • #1
bonfire09
249
0

Homework Statement


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.


Homework Equations





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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  • #2
bonfire09 said:
We also assume that x1...xi-1 and xi+1...xn are zero.

Why can you assume this??
 
  • #3
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 https://www.physicsforums.com/showpost.php?p=3977517&postcount=3 for a guide.

Homework Statement


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.
 
  • #4
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.
 
  • #5
bonfire09 said:
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?
 
  • #6
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?
 
  • #7
bonfire09 said:
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.
 
  • #8
alright thanks.
 
  • #9
Think about what relations the basis vectors satisfy, if you notice the right thing, the proof is pretty swift.
 
  • #10
The problem with the above proof, it doesn't seem to use the fact that the basis is orthonormal. You could potentially "prove something false".
 
  • #11
algebrat said:
The problem with the above proof, it doesn't 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.
 
  • #12
micromass said:
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.
 

What does the statement "x1E1+...+xnEn=0" mean?

The statement means that the linear combination of the vectors E1, E2, ..., En, with corresponding coefficients x1, x2, ..., xn, equals zero.

What does it mean to "prove" a linear algebra statement?

To prove a linear algebra statement means to show that it is true by using logical reasoning and mathematical techniques.

How do you prove the statement "x1E1+...+xnEn=0 then xi=0"?

To prove this statement, we can use the following steps:

  • Assume that x1E1+...+xnEn=0 is true.
  • Apply linear algebra properties and manipulations to simplify the equation.
  • Show that the only solution to the equation is when xi=0 for all i.
  • Therefore, the statement is proven to be true.

Can you give an example to illustrate the statement "x1E1+...+xnEn=0 then xi=0"?

Yes, for example, if we have the vectors E1=(1,0,0), E2=(0,1,0), and E3=(0,0,1), and the coefficients x1=2, x2=3, and x3=5, then the statement x1E1+x2E2+x3E3=0 is true. However, since the only solution to this equation is when x1=x2=x3=0, the statement "x1E1+x2E2+x3E3=0 then x1=x2=x3=0" is also true.

Why is this statement important in linear algebra?

This statement is important because it shows that if a linear combination of vectors equals zero, then the only solution is when all the coefficients are equal to zero. This allows us to manipulate and solve linear equations and systems of equations in a systematic and efficient way.

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