# Vectors In Space

1. Jul 2, 2008

### danago

Given a set of four vectors in space, prove that at least one is a linear combination of the other three.

I think i am able to visualize this one in my head, but dont really know how to write a solid proof for it.

Lets say i take any three of the vectors; The three will be either coplanar or non-coplanar. If they are coplanar, then any of those three vectors can be represented as a linear combination of the other two. If they are non-coplanar, then any vector in R^3 i.e. the fourth vector, can be be written as a linear combination of the three.

Now i think i have visualised it correctly, but it certainly doesnt feel like i have proven much. Can anyone suggest how i could go about doing so?

Thanks,
Dan.

2. Jul 2, 2008

### Defennder

You can use the concept of dimension and basis. If these three vectors are linearly independent in R3, that makes them a basis. The definitions of basis and dimension will help here.

3. Jul 2, 2008

### danago

Ive just finished my first semester of my first year at uni and thought id get ahead with some of the coursework during the break, starting with my linear algebra class which will be next semester. Ive looked ahead and there is content related to dimension and basis, but its further on. Since this question appears before dimension and basis, id assume that there is some way without them to do this? Thanks for the quick reply though

4. Jul 2, 2008

### rootX

Yep, you are right in your first post.
R^3 doesn't have to defined by <0,0,1>,<0,1,0>,<1,0,0>
It can be defined by any three independent vectors.

There is a theorem that says if you have a set of m vectors in R^n space where m>n
then all vectors in m are not independent.
I am sure you will bump into that theorem very soon

5. Jul 2, 2008

### Defennder

Okay here's what you can do given that you haven't covered dimension and basis at present. Suppose you have four or more vectors in R3, then if all of them are linearly independent the following is true:

There exists c1...c4
$$c_1\vec{u_1} + c_2\vec{u_2} + c_3\vec{u_3} + c_4\vec{u_4} = 0 \ \mbox{where c1...c4 are not all zero}$$

This can also be written as a system of 3 linear equations with all their RHS as 0. The question here is, what type of solutions do you expect c1...c4 to be for these homogenous system of linear equations? Remember that there are two types of solutions for a system of homogenous linear equations. Furthermore, what does this imply about the linear dependence of the system?

6. Jul 2, 2008

### danago

Well there would be the trivial solution, where c1=c2=c3=c4=0. The system could have only this trivial solution, or it may have infinitely many solutions, one of which is the trivial solution.

If the second case occurs i.e. there are non trivial solutions, then that tells us that atleast one of the vectors can be written as a combination of the others.

If the first case occurs i.e. there is only the trivial solution, then what will that mean? I would have thought that it implied that the vectors cant be written as combinations of eachother, but then that doesnt really help this proof.

7. Jul 2, 2008

### Defennder

What is the formal definition for a set of vectors to be linearly independent?

8. Jul 2, 2008

### danago

Oh so when the trivial solution is the only solution, the vectors are linearly independent, so none of the four can be written as a combination of the other three? Wouldnt that be disproving the statement?

9. Jul 2, 2008

### Defennder

Is the trivial solution the only solution?

10. Jul 3, 2008

### danago

Id think not, but how do i know that for sure?

11. Jul 3, 2008

### Defennder

How many unknowns are you solving for? How many equations do you have? What does that imply?

12. Jul 3, 2008

### danago

Ohh so we have 3 equations (since we are dealing with 3 dimensions) in 4 unknowns, which implies that we have an infinite number of solutions i.e. non trivial solutions?

13. Jul 3, 2008

### Defennder

Yeah, basically that's the proof.

14. Jul 3, 2008

### danago

Ahh yea ok. Thanks very much for the help