# Condition for equality between subspaces.

1. Dec 3, 2012

### peripatein

Hi,
1. The problem statement, all variables and given/known data

What would be the/a condition on vectors in K so that V=W, where V is a vector space which K={v1,v2,v3,v4} spans, and W is a subspace of V defined thus:
W=Sp{v1+v2,v2+v3,v3+v4,v4+v1}

2. Relevant equations

3. The attempt at a solution

I believe V would be equal to W if W were linearly independent, but by writing that mathematically I get a condition for the scalars, not the vectors in K themselves.

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

2. Dec 3, 2012

### micromass

Staff Emeritus
I know what you mean, but your terminology is wrong. You can't say that W is linearly independent because it is not true. What you mean is that the four vectors

$$\{v_1+v_2,v_2+v_3,v_3+v_4,v_4+v_1\}$$

are linearly independent. That would indeed be the correct condition.

What did you get when you wrote that mathematically??

3. Dec 3, 2012

### peripatein

I have tried to find conditions so that:
a1v1 + a2v2 + a3v3 + a4v4 = v1(b1+b4) +
v2(b2+b1) + v3(b3+b2) + v4(b4+b3).
But that yielded conditions on the scalars, not the vectors. Can conditions on the vectors themselves be found?

4. Dec 3, 2012

### micromass

Staff Emeritus
How did you get that? In order for $\{v_1+v_2,v_2+v_3,v_3+v_4,v_4+v_1\}$ to be a basis, you must prove that any linear combination of the form

$\alpha(v_1+v_2)+\beta(v_2+v_3)+\gamma (v_3+v_4)+\delta(v_4+v_1)=0$

only if $\alpha=\beta=\gamma=\delta=0$.

Now, try to use that $\{v_1,v_2,v_3,v_4\}$ is a basis.

5. Dec 3, 2012

### peripatein

These yielded alpha=-delta=-beta=gamma.
But how does this affect the vectors in K themselves? I mean, what is then the condition on v1,v2,v3,v4?

6. Dec 3, 2012

### micromass

Staff Emeritus
OK, so what if you take the equation

$$\alpha(v_1+v_2)+\beta(v_2+v_3)+\gamma(v_3+v_4)+ \delta(v_4+v_1)=0$$

and if you substitute $\alpha$ for $\gamma$ and $-\alpha$ for $\delta$ and $\beta$?

7. Dec 3, 2012

### peripatein

You get alpha*0=0. How does that help?

8. Dec 3, 2012

### micromass

Staff Emeritus
It shows that there is always a nontrivial linear combination that ends up in zero. Doesn't that show that your set $\{v_1+v_2,v_2+v_3,v_3+v_4,v_4+v_1\}$ is never linearly independent?

9. Dec 3, 2012

### peripatein

Let us go back a bit, momentarily.
I am slightly confused. Why is it that for V to be equal to W, the elements in W must be linearly independent? Is it because dimV is equal to or less than the number of elements in K, i.e. 4?
Furthermore, I know that if the elements in K are linearly independent, then V is not equal to W. Does that mean that for any K whose elements are linearly dependent, V would be equal to W?