Linear Algebra Axioms

1. Jan 5, 2012

TranscendArcu

1. The problem statement, all variables and given/known data
Prove that $\vec0 + \vec{A} = \vec{A}$

3. The attempt at a solution
$$\vec{A} + (-\vec{A}) = \vec0$$
$$\vec0 + \vec{A} = \vec{A} + (-\vec{A}) + \vec{A} = \vec{A} + \vec{A} + (- \vec{A})$$
$$\vec0+ \vec{A} = \vec{A} + \vec0 = \vec{A}$$

I find this last line somewhat unconvincing since it seems it could also be retrieved by the axiom $\vec{A} + \vec{B} = \vec{B} + \vec{A}$. Is this a legitimate proof or do I have to do this another way?

2. Jan 5, 2012

Fredrik

Staff Emeritus
The identity you want to prove is one of the conditions in the definition of "vector space". So if we're dealing with a vector space, no proof is needed.

So I'm guessing that we're dealing with the set $\mathbb R^n$, with addition and scalar multiplication defined in the usual way, and that you haven't yet proved that these definitions turn the set into a vector space. Verifying that this identity holds would be one of the steps involved in doing that. If you want to prove that $\vec 0+\vec A=\vec A$ for all $\vec A$ in $\mathbb R^n$, then you have to use the definition of "+".

3. Jan 5, 2012

TranscendArcu

I have this definition: vector addition assigns to each pair of vectors A,B a vector denoted by A + B, called their sum. That's all I have, but (as of right now) I don't know where to go with this definition.

4. Jan 5, 2012

Fredrik

Staff Emeritus
And how do you define "vector"? I would define it as "a member of a vector space", but if we use that definition, there's nothing left to prove, since the definition includes the identity you want to prove.

5. Jan 5, 2012

TranscendArcu

I was kind of expecting our proof to somehow involve the axioms that we use to define vector spaces.

6. Jan 5, 2012

Fredrik

Staff Emeritus
But the statement you want to prove is one of those axioms.

There is something you haven't told me about this problem. You haven't even told me if we're dealing with an arbitrary vector space or a specific one. If it's some specific vector space, then which one? Is A an arbitrary vector in that space, or a specific one? If it's a specific vector, then which one?

My guess is that this is the correct statement of the problem: Prove that for all $\vec A\in\mathbb R^n$, $\vec 0+\vec A=\vec A$.

If that's it, then you have to use the definition of +, i.e. the definition of the addition operation on $\mathbb R^n$. This is the sort of thing you prove before you even know that $\mathbb R^n$ is a vector space, so you can't use the vector space axioms.

7. Jan 5, 2012

TranscendArcu

I think your assumptions are probably safe. Below is a screenshot of my textbook. The problem above is a very similar proof that makes use of the axioms. The problem below is the proof in question. I think what the text is trying to show is that if we take some of the axioms to be true, then an additional axiom follows as a consequence. But, as you can see, there isn't much in the way of detail to these problems.

http://img689.imageshack.us/img689/9075/screenshot20120105at105.png [Broken]

Edit: The "Proposition 2.1.1" mentioned in the problem doesn't include any additional details, in case you were wondering.

Last edited by a moderator: May 5, 2017
8. Jan 5, 2012

micromass

Can you show me the axioms you are working with??

PS: There is absolutely no need to put an arrow on top of your vectors. It's annoying to do and no text uses that notation.

9. Jan 5, 2012

TranscendArcu

http://img51.imageshack.us/img51/7523/screenshot20120105at110.png [Broken]
http://img254.imageshack.us/img254/7523/screenshot20120105at110.png [Broken]

Last edited by a moderator: May 5, 2017
10. Jan 5, 2012

micromass

Ah yes. Well, the theorem seems to follow directly from axiom 1 and axiom 3, right?

11. Jan 5, 2012

TranscendArcu

I think it does. But I would also like to know if what I wrote in #1 was legitimate or not. Indeed, it may be an indirect way of getting to the answer, but if it works, I'd like to know. (Or it could just be wrong, in which case clearly axioms 1 and 3 are the way to go.)

12. Jan 5, 2012

micromass

Yes, what you wrote in post 1 is correct!!

However, it requires more axioms than saying it's true by 1 and 3. So it might be less preferable to do it that way. It's still correct though, and that's what matters.

13. Jan 5, 2012

Fredrik

Staff Emeritus
Hm...since your book only includes A+0=0 in axiom 3 (and not 0+A=0), it does make sense to prove 0+A from axioms 3 and 1. (I'm used to seeing a version of axiom 3 that says that 0+A=A+0=0).

This appears to be a problem about arbitrary vector spaces. So I think my previous guess was wrong. You don't need to use the definition of a specific addition operation. You just need the axioms.

14. Jan 6, 2012

TranscendArcu

I have another, similar sort of problem that I have posted below.
http://img502.imageshack.us/img502/3047/screenshot20120106at124.png [Broken]
http://img819.imageshack.us/img819/3047/screenshot20120106at124.png [Broken]

My understanding is that any set V, along with vector addition and scalar multiplication, is a vector space as long as it satisfies the axioms above. I think that the space described in the problem is not a vector space because I think it fails the 8th axiom (ie. $1A = A$)

For example, let $r=1$ and $d≠0$ in the rule for multiplication. Then it seems that any scalar multiplication of the form $1(c,d)$ will not return $(c,d)$, but rather $(c,0)$. Because the 8th axiom seems to fail, I don't think this is a vector space.

Am I doing this right?

Last edited by a moderator: May 5, 2017
15. Jan 6, 2012

micromass

That is correct.

16. Jan 6, 2012

TranscendArcu

Thanks!

My next problem states: "Show that if a vector space contains two elements, then it contains infinitely many."

So I understand that every vector space must have a 0-vector. This is one element, I believe. Let us call the other element A. Since there are supposed to be two elements, A=0. Yet there must also be a -A. So we have three elements.

At this point I can see how there are at least three elements in this vector space, but I'm having a hard time turning three into infinity.

17. Jan 6, 2012

micromass

So we have a nonzero element A.

Multiplying by 1 gives us A.
Multiplying by 0 gives us 0.
Multiplying by -1 gives us -A.

Maybe we can multiply by some other scalar to obtain more elements.

Please note that you need to prove that all these elements are truly different. For example: why is A different from -A?

18. Jan 7, 2012

TranscendArcu

Okay, that makes sense. I did the proof for A ≠ -A by contradiction, which seemed the easiest way.

What I found kind of less trivial was my last problem: http://img842.imageshack.us/img842/8237/screenshot20120107at104.png [Broken]As it happens, I've already done this problem by checking each of the axioms to ensure that it holds. I concluded that such a V is a vector space.

However, checking all of the axioms was quite a bit of work, occupying almost 3 pages (which is part of the reason I'm not retyping it all here). So I'm wondering if there is an easier way to do this problem? Or, perhaps I've missed something and this is not a vector space...

Last edited by a moderator: May 5, 2017
19. Jan 7, 2012

micromass

I don't see an easier way to do that problem. Your way seems fine.

20. Jan 9, 2012

Deveno

this is not true. there exist finite vector spaces. a piece of information is missing, which is this: information about the underlying field. if this field is of characteristic zero (that is, if it contains all the integers), then this is a true result. for example, we have the following finite vector space:

F = {0,1}, where:

0+0 = 0
0+1 = 1
1+0 = 1
1+1 = 0 (<--thus this field is of characteristic 2)

0*0 = 0
0*1 = 0
1*0 = 0
1*1 = 1

V = {0,1,a,b}

vector addition is defined as follows:

0+0 = 0
0+1 = 1
0+a = a
0+b = b

1+0 = 1
1+1 = 0
1+a = b
1+b = a

a+0 = a
a+1 = b
a+a = 0
a+b = 1

b+0 = b
b+1 = a
b+a = 1
b+b = 0

scalar multiplication is defined as follows:

0x = 0
1x = x, for all x in V.

this is a very odd vector space, but one that does crop up occasionally.