# Spanning sets in R^2

1. Apr 3, 2015

### jdawg

Hi! Can someone please help me with this problem? I need to understand how to do it for my test tomorrow!

I know this a spanning set for R^2, but the way I saw this problem solved was kind of lazy and not very helpful.
S={(-1,4),(4,-1),(1,1)}
I tried testing to see if it had just the trivial solution, but I got stuck...
(0,0)=c1(-1`,4)+c2(4,-1)+c3(1,1)
When I put this into a matrix I couldn't get it to reduce.

Thanks!

2. Apr 3, 2015

### Dick

You are confusing 'spanning' with 'linearly independent'. That equation doesn't have only the trivial solution. It can't. Any three vectors in $R^2$ are linearly dependent. What you want to show is that any vector $(x,y)$ can be written as $(x,y)=c_1 (-1,4)+ c_2 (4,-1)+c_3 (1,1)$. Write down the corresponding linear equations. The solution is far from unique, but any one will do.

3. Apr 3, 2015

### jdawg

I am a little confused about the relationship between linear independence and spanning sets! If you can write the vectors as a linear combination, are they spanning and linearly dependent? Also if you take the determinant and it is non zero, then it is a spanning set and linearly independent, correct? Or do I need to think of spanning sets and linear independence as completely separate things? Sorry my thoughts are so unorganized!

Would I only check for the trivial solution if I'm trying to figure out linear independence?

4. Apr 3, 2015

### Dick

Linear independence is the condition that the equation you wrote in the first post has only the trivial solution. Spanning is the condition I wrote in the second post. The concepts aren't totally separate if you know about dimension. But they are two different things. In your case the set of vectors is linearly dependent. But they do span. Just try and show that. If you have a square matrix, then you can check the determinant and the if it is nonzero then they span AND are linearly independent. You don't have a square matrix here.

Last edited: Apr 3, 2015
5. Apr 3, 2015

### jdawg

Alright, so:

x=-c1+4c2+c3
y=4c1-c2+c3

The way I saw it worked the person just let c3=0 and then they just took the determinate of this:

-1 4
4 -1

And then they concluded that the determinate was nonzero and it was a spanning set.
Why is it ok to just let c3=0?

Is there a more proper way to work this problem?

6. Apr 3, 2015

### Dick

Well, it could probably be explained better. The person explaining this to you probably saw that any two of those vectors are linearly independent. $R^2$ is two dimensional, so ANY two linearly independent vectors will span. That's easy to see because if two vectors are linearly dependent they are multiples of each other and in two dimensions you only need two independent vectors to make a spanning set. No pair of those vectors are linearly dependent. So you can throw any one out. The person that worked this chose to throw the third vector out by setting $c_3=0$. You could also set $c_1=0$ or $c_2=0$ and get the same result. It's really about having experience with solving linear equations.

7. Apr 3, 2015

### jdawg

Oh ok, that makes sense! So what would you do if it was a situation where that method would not work?

8. Apr 3, 2015

### Dick

You go back the the linear equations you have to solve. You can use back substitution or if you want to use a matrix form use row reduction. You can use determinants once you have the same number of vectors as the dimension of the space. Otherwise, it's really about the underlying linear equations. You know how to solve them, right?

9. Apr 3, 2015

### jdawg

Sort of! When I row reduced I got c1+7c3=0
c2+3c3=0

I'm not sure what to do with that information.

10. Apr 3, 2015

### Dick

If you are trying show they are linearly independent, then they obviously aren't right? There are solutions where all of the c's aren't equal to zero. I thought you wanted to show that the spanned. That's different set of equations.

11. Apr 3, 2015

### jdawg

Ohh! Ok, I think I get it now. Thanks so much for all your help!! :)