How can I tell if this is a vector space?

1. Oct 28, 2007

Antineutron

1. The problem statement, all variables and given/known data
A set of objects is given, together with operations of addition and scalar multiplication. Determine which sets are vector spaces under the given operations. For those that are not vector spaces, list all axioms that fail to hold.

(x,y,z) + (x',y',z') = (x+x',y+y',z+z') and k(x,y,z) = (kx,y,z)

2. Relevant equations

1. If u and v are objects in V, then u + v is in V.

2. u + v = v + u

3. u + (v + w) = (u + v) + w

4. There is an object 0 in V, called a zero vector for V, such that 0 + u = u+ 0 = for all un in V

5. For each u in V, there is an object -u in V, called a negative of u, such that u + (-u) = (-u) + u = 0.

6. If k is any scalar and u is any object in V, then ku is in V.

7. k(u + v) =ku + kv

8. (k + m)u = ku + mu

9. k(mu) = (km)(u)

10. 1u=u

3. The attempt at a solution

I have no idea what they are asking for, the back of the books say it fails one axiom which is: (k+m)u = ku +mu Which is axiom 8 in my book.

Last edited: Oct 28, 2007
2. Oct 28, 2007

cristo

Staff Emeritus
Well, since you say there are ten axioms for a vector space, then you should check whether the set satisfies the axioms.

3. Oct 28, 2007

Antineutron

thats the part I am having trouble with. How do see that it fails the axiom that I have just stated? Can you show me?

4. Oct 28, 2007

JasonRox

Don't you have any examples in the textbook?

Do you understand what the axioms are?

List the axioms.

Note: By the way, the worse thing you can do is look up the answer before trying or even understanding the question.

5. Oct 28, 2007

Antineutron

I agree, but I'm not the kind of student that looks up answer unless I am desperate. I'm not a precalculus student you know. And this is not a simple calculation question. Its just a simple question that I cannot comprehend. Also, there is no examples that are anything like this question in my textbook. Thank you.

Here are the axioms:

1. If u and v are objects in V, then u + v is in V.

2. u + v = v + u

3. u + (v + w) = (u + v) + w

4. There is an object 0 in V, called a zero vector for V, such that 0 + u = u+ 0 = for all un in V

5. For each u in V, there is an object -u in V, called a negative of u, such that u + (-u) = (-u) + u = 0.

6. If k is any scalar and u is any object in V, then ku is in V.

7. k(u + v) =ku + kv

8. (k + m)u = ku + mu

9. k(mu) = (km)(u)

10. 1u=u

Thats all, did you ever had time when doing math homework, that you could not understand something, but you know its not so difficult. Just remember that time, and thats how I feel right now.

Last edited: Oct 28, 2007
6. Oct 28, 2007

morphism

OK. So you have a set S of triples (x,y,z) (you should be clear about where x,y,z are coming from -- you didn't state this in your post; I'm going to assume that they're real numbers). Then two binary operations are being defined on S:
"addition": (x, y, z) + (x', y', z') = (x+x', y+y', z+z')
and "scalar multiplication": k*(x, y, z) = (kx, y, z)

Now if S is a vector space under these two operations, it has to satisfy all the axioms you listed. For example, let's try to see if it satisfies the first. Let u=(x,y,z) and v=(x',y',z'). We want to prove that u+v is in S. Since u+v=(x+x',y+y',z+z') by definition, and since x+x', y+y' and z+z' are all real numbers, then u+v is in S. So the first axiom holds.

Can you take it from here?

7. Oct 28, 2007

Antineutron

I understand that part, but can you tell me why axiom 8 does not hold? Thats where I am having trouble. How do I appy axiom 8 to test to see if the scalar multiplication that is stated does is not vector space according to that particular axiom?

Last edited: Oct 28, 2007
8. Oct 28, 2007

ZioX

You would want to find a counterexample, try playing around with it.

9. Oct 29, 2007

Antineutron

Look at axiom 8 and tell me in good mathematically rigorous explaination why that one does not apply to the scalar multiplication. Thats all.

10. Oct 29, 2007

D H

Staff Emeritus
Expand each side of (k+m)*u = ku + mu to see whether this holds for any k, m, and u=(x,y,z). Look at the non-trivial case: none of k,m, x, y, or z are zero.

11. Oct 29, 2007

ice109

(k+m)u=?ku+mu
((k+m)x,y,z)=(kx,y,z)+(mx,y,z)
((k+m)x,y,z)=((k+m)x,2y,2z)

therefore fails

as opposed to vector space with scalar multiplication defined like this: k(x,y,z)=(kx,ky,kz)
(k+m)u=?ku+mu
((k+m)x,(k+m)y,(k+m)z)=(kx,ky,kz)+(mx,my,mz)
((k+m)x,(k+m)y,(k+m)z)=((k+m)x,(k+m)y,(k+m)z) =true

Last edited: Oct 29, 2007
12. Oct 29, 2007

D H

Staff Emeritus
How could it fail axiom 6? V is a vector space over a field F. This means if u=(x,y,z) is are in V, each of x, y, and z are in F. Multiplication by a scalar means a scalar in the underlying field F. Thus k, x, y, and z are each in F. The product kx is in F, so ku=(kx,y,z) is in V.

13. Oct 29, 2007

ice109

14. Oct 29, 2007

cristo

Staff Emeritus
I know ice109 has given you the answer, but you should not be demanding something like this. People are here helping you with your homework; you should not expect people to be doing it for you.

This is a standard type introductory question in vector spaces: one is given a set and required to check whether it satisfies the axioms of a vector space. You have listed the axioms, but then not made any attempt at checking them. I suggest in future that you make some effort before posting, and do not demand that someone do your work for you.

15. Oct 29, 2007

mathwonk

you are in over your head. where did this assignment come from in precalculus? try some more elementary and basic stuff like simple logic.

16. Oct 29, 2007

D H

Staff Emeritus
Antineutron, we have rules against posting the same problem in multiple forums. You violated that rule with this thread and this other thread. Please don't do this again.

17. Oct 29, 2007

Antineutron

I'm sorry, I needed some help really bad, I will not double post again. I did not post to get an answer without trying to find the answer. I just had a hard time formulating a good solid proof to varify all the given axioms, axiom 8 was not apparent as the others when applying it to this problem, it is not that the axioms are difficult. I ask you one thing mathwork, if you think this is simple logic, did you know all the math before going to college? Is something simple logic, so we should all know everything. Are we to know everying because it is logic? What is the point of going to school and learning? What is your point of telling me is simple logic?

18. Oct 29, 2007

matt grime

Perhaps mathwonk's point was that the negation of 'for all' is 'there exists' and you thus need only to find one counter example, and has little to do with the theory of vector spaces at all. Pretty much any thing you try will yield a counter example to 8. Thus the inferrence is that you didn't multiply any notional vectors by any scalars. Not playing with the example to see what is going on is the worst thing you can do: answers do not often magically appear. In fact mathwonk is arguably saying the exact opposite of what you ascribe to him: it isn't about knowing all the answers but about experimenting with things to see what happens.

Multiplication by 2 is adding something to itself. Clearly those two descriptions, in this case, are incompatible.

19. Oct 29, 2007

ice109

he said he's not in precalc? besides what about linear algebra do you think is too abstruse for pre calc students? heck after taking lin alg i think this course could be taught immediately after algebra in middle/high school

20. Oct 29, 2007

mathwonk

after reading post 17, i am more sure than before you are over your head. why are you so indignant that i tell you the truth? get a book on logic and read some of it.

you apparently have no experience at all with axiom systems, or definitions, or counterexamples as matt suggests.

Last edited: Oct 29, 2007
21. Oct 29, 2007

Antineutron

Bingo, I don't, thank you for pointing that out. All the books you read made you who you are mathwonk, I'm sure you are very proud. Should I get a book on logic like you say so I can learn from books like you?

Last edited: Oct 29, 2007
22. Oct 29, 2007

D H

Staff Emeritus
Enough of the trash talk already! I do not want to see fisticuffs at the PF Saloon.

Mathwonk: Telling someone in public that they are in over their heads is an insult. Please desist.

Antineutron: Even though ice109 gave out the answer, please work through how axiom #8 fails in general here. Aside: Being in over your head is not necessarily a bad thing. It forces one to learn and achieve better than almost anything else.

23. Oct 29, 2007

Antineutron

I went over the problems and tried to figure it out after looking at ice's example which was very helpful. I really appreciate the people that helped me figure out this new way of looking at mathematical concepts, that includes you DH. I feel gracious that I can come here for help and get help from people that can understand people's hardships and not judge people because we may not be so experienced. And I'm sure many people have been in that road before succeeding in what ever areas many of you professionals are in now.

Last edited: Oct 29, 2007
24. Oct 30, 2007

ice109

honestly he can't begin to do these types of problems if he has no idea what the solutions even look like.

25. Oct 30, 2007

cristo

Staff Emeritus
But that's what everyone says after they've learnt a course. The fact of the matter is that the majority of students in school firstly, do not like maths, and secondly, are not very good at maths. So attempting to teach linear algebra to them would just be a complete waste of time and effort.