# Equality between subspaces

1. Nov 23, 2012

### peripatein

Hi,
How may I determine whether the subspaces U and W are equal to each other?:

K is linearly independent wrt V, defined thus:
K={v1,v2,v3,v4} subset of V
U and W, subspaces of V, are defined thus:
U=Sp(K); W=Sp{v1-v2,v2-v3,v3-v4,v4-v1}

I am not allowed to use equality between dimensions!

I have tried solving:
a1v1+a2v2+a3v3+a4v4=b1(v1-v2)+b2(v2-v3)+b3(v3-v4)+b4(v4-v1)
But I am not sure it got me anywhere :-/.

I hope one of you could assist. Thanks in advance!

2. Nov 23, 2012

### haruspex

Seems like a reasonable start. Where did that lead?

3. Nov 23, 2012

### peripatein

It led to a1=b1-b4; a2=-b1+b2; a3=-b2+b3; a4=-b3+b4
I tried substituting them into a matrix but am not sure I got anything meaningful.
I also know that when a1v1+a2v2+a3v3+a4v4=0, a1=a2=a3=a4=0.
Any idea how I should proceed?

4. Nov 23, 2012

### HallsofIvy

Staff Emeritus
If you can show that v1-v2, v2-v3, v3-v4, and v4-v1 are independent then you are done. Do you see why?

5. Nov 23, 2012

### peripatein

I am not sure. If b1(v1-v2)+b2(v2-v3)+b3(v3-v4)+b4(v4-v1)=0 necessarily implies b1=b2=b3=b4=0, why does that mean that U=W? Isn't the homogeneous system merely a private case? Were U and W both linearly independent, why would U be equal to W?

6. Nov 23, 2012

### Dick

If you add those equations up you get a1+a2+a3+a4=0. What would that tell you?

7. Nov 23, 2012

### peripatein

Does that tell me that {v1,v2,v3,v4} is linearly dependent as a1, a2, a3, a4 are not forcibly all zero?

8. Nov 23, 2012

### Dick

I would say it means that only SOME vectors in Sp{v1,v2,v3,v4} are also in Sp{v1-v2,v2-v3,v3-v4,v4-v1}. Can you tell me why?

9. Nov 23, 2012

### peripatein

As for the two spans to be equal certain conditions must be met, i.e. a system of equations must be solved which imposes certain limitations on the values of the scalars.

10. Nov 23, 2012

### Dick

That's pretty vague. You wrote a system of equations. One conclusion of that was that if v=a1v1+a2v2+a3v3+a4v4 then v can only be in Sp{v1-v2,v2-v3,v3-v4,v4-v1} if a1+a2+a3+a4=0. Is that condition true for any vectors in Sp{v1,v2,v3,v4}? Is it untrue for any vectors in the span?

11. Nov 23, 2012

### HallsofIvy

Staff Emeritus
My point was that since you are given that v1, v2, v3, b4 are independent, the subspace they span has dimension 4. The other 4 vectors, v1-v2,v2-v3,v3-v4,v4-v1, will span that subset if and only if they are independent.

12. Nov 23, 2012

### peripatein

Okay, so I have therefore tried showing that b1(v1-v2)+b2(v2-v3)+b3(v3-v4)+b4(v4-v1)=0, implies b1=b2=b3=b4=0 (for linear independence). However, based on the fact that a1v1+a2v2+a3v3+a4v4=0 for a1=a2=a3=a4=0, all I got was that b1=b2=b3=b4=any real number (i.e. not necessarily zero)! How shall I proceed?

13. Nov 23, 2012

### Dick

Good job. Wouldn't that show that they are NOT linearly independent?

14. Nov 23, 2012

### peripatein

I was somehow under the impression that U WAS in fact equal to W, so I did expect linear independence in both cases. Is the conclusion then that U is necessarily not equal to W?

15. Nov 23, 2012

### Dick

Yes, U isn't equal to W. And not even 'necessarily'. It just plain isn't. (v1-v2)+(v2-v3)+(v3-v4)+(v4-v1)=0. That doesn't sound linearly independent to me.

16. Nov 23, 2012

Thanks! :-)