# Homework Help: Scalar Multiple of Vector - Vector & Scalar = 0?

1. May 6, 2016

### Calaver

Note: I am not in the course where this problem is being offered; it was simply an interesting linear algebra "thought question" that I found online to which I believe I have found a solution. However, there is one step in my solution that I am unsure about, so thank you to anyone who spares the time to assist.

1. The problem statement, all variables and given/known data

Let x and y be vectors in ℝn. Is it possible that x is a scalar multiple of y (i.e., there exists a scalar c such that x = cy), but y is not a scalar multiple of x?

2. Relevant equations
Basically restating the problem in an equation here, from what I see no pure equation other than this is needed:

Let b, c be scalars in ℝ and x, y be vectors in ℝn. Let the scalar c be defined such that x = cy. Is there always a b such that y = bx?

3. The attempt at a solution

There is not always a scalar b for the given vectors x, y and given scalar c to make the equations above true. Take the case c = 0, x = (0,0), y ≠ (0,0). Then (0,0) = 0y. But there does not exist a scalar b such that bx = b⋅(0,0) = y ≠ (0, 0) by the fact that (0,0)⋅b = (0,0) for all b.

I believe I have interpreted the question correctly, and it seems that the first part of my "proof" (may not be completely formal or rigorous - I'll be open to any suggestions to improve it) is valid by the proof here. But I cannot figure out if the underlined statement is simply a postulate of linear algebra, if there is a way that I should prove it, or if it is even correct. Should the last statement of the proof be changed or is it even valid?

Thanks to anyone who takes the time to help.

EDIT: Clarity in last paragraph.

2. May 6, 2016

### andrewkirk

Yes it's valid, but it needs to be proven. The proof that $0\cdot\mathbf x=\mathbf 0$ is only four lines, and 'obvious' once you see it, but the challenge is getting the right way of approaching it, to be able to see that proof.

If you've ever seen the proof for $\mathbb R$ (or for fields more generally) that $0\times x=0$, it's analogous to that.

I'll leave to you the fun of discovering the proof. As a hint, it uses the vector space axiom of 'Distributivity of scalar multiplication with respect to field addition', from the list of vector space axioms here.

Your wondering whether it might be an axiom is well-founded, as it seems natural to think it must be an axiom, and not obvious that it is a theorem. I was amused to see that this UCLA definition of vector spaces includes it as an axiom without realising its redundancy!

3. May 6, 2016

### haruspex

I believe your proof is fine. The underlined statement is true since b(x,y)=(bx,by).

Last edited: May 6, 2016
4. May 6, 2016

### Calaver

Thanks everyone for your responses so far!

I've seen the proof for ℝ before, but didn't recall all of the steps right away. But now I think I've got a proof for 0⋅x=0 (thanks for the hint!).

0⋅x = (0+0)x
because of the property that 0 is the identity element of addition.
= 0⋅x + 0⋅x
because (a+b)v = av + bv .
Transitively,
0⋅x = 0⋅x + 0⋅x.
And because if a=b then a+c=b+c by the substitution property of equality (while I was doing this I originally thought that a+b=b+c was the axiom as Euclid stated, but found out that it's actually more general), then
0⋅x + (-0⋅x) = 0⋅x + 0⋅x + (-0⋅x).
Because -0⋅x is the inverse element of addition for
0⋅x, 0 = 0⋅x + 0.
But
0⋅x + 0 = 0⋅x
because (once again) 0 is the identity element of addition.
So we have:
0 = 0⋅x.
Q.E.D.

I just realized after I went through the proof that I already saw a similar proof in the link in my original post. Oh well, it was worth it because I understand the proof better now after working through each step on my own!

It also occurred to me that my original issue was proving b⋅0 = 0 for a scalar b and vector 0. Then I just saw that your post contains the line 0⋅x=0 for vectors 0 and v. But thinking through the proofs, it seems that the only thing that would change would be which one is the vector and which one is the scalar; the overall structure stays similar throughout.

Once again, thank you both for your help and thank you andrewkirk for your challenge to prove the statement!