Condition for equality between subspaces.

In summary, the condition on the vectors in K for V to be equal to W is that the set of vectors \{v_1+v_2,v_2+v_3,v_3+v_4,v_4+v_1\} must be linearly independent. If the vectors in K are linearly dependent, then V will not be equal to W. This is because the dimension of V is equal to or less than the number of elements in K, and if the vectors in K are linearly dependent, then V cannot be spanned by them. Therefore, the condition on the vectors in K is crucial for V to be equal to W.
  • #1
peripatein
880
0
Hi,

Homework Statement



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}

Homework Equations





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!
 
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  • #2
peripatein said:
Hi,

Homework Statement



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}

Homework Equations





The Attempt at a Solution



I believe V would be equal to W if W were linearly independent,

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

[tex]\{v_1+v_2,v_2+v_3,v_3+v_4,v_4+v_1\}[/tex]

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

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!

What did you get when you wrote that mathematically??
 
  • #3
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
peripatein said:
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?

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

[itex]\alpha(v_1+v_2)+\beta(v_2+v_3)+\gamma (v_3+v_4)+\delta(v_4+v_1)=0[/itex]

only if [itex]\alpha=\beta=\gamma=\delta=0[/itex].

Now, try to use that [itex]\{v_1,v_2,v_3,v_4\}[/itex] is a basis.
 
  • #5
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
peripatein said:
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?

OK, so what if you take the equation

[tex]\alpha(v_1+v_2)+\beta(v_2+v_3)+\gamma(v_3+v_4)+ \delta(v_4+v_1)=0[/tex]

and if you substitute [itex]\alpha[/itex] for [itex]\gamma[/itex] and [itex]-\alpha[/itex] for [itex]\delta[/itex] and [itex]\beta[/itex]?
 
  • #7
You get alpha*0=0. How does that help?
 
  • #8
peripatein said:
You get alpha*0=0. How does that help?

It shows that there is always a nontrivial linear combination that ends up in zero. Doesn't that show that your set [itex]\{v_1+v_2,v_2+v_3,v_3+v_4,v_4+v_1\}[/itex] is never linearly independent?
 
  • #9
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?
 

1. What is the condition for equality between subspaces?

The condition for equality between subspaces is that they must have the same dimension and span the same set of vectors. This means that all the vectors in one subspace can be expressed as linear combinations of the vectors in the other subspace.

2. How do you determine if two subspaces are equal?

To determine if two subspaces are equal, you need to compare their dimensions and check if they span the same set of vectors. If both conditions are met, then the subspaces are equal.

3. Can two subspaces with different bases be equal?

Yes, two subspaces with different bases can still be equal as long as they have the same dimension and span the same set of vectors. The basis of a subspace is not unique, so different bases can represent the same subspace.

4. What is the significance of the condition for equality between subspaces?

The condition for equality between subspaces is significant because it allows us to determine when two subspaces are equivalent and can be treated as the same mathematical object. This allows for easier manipulation and analysis of these subspaces in various mathematical applications.

5. Can subspaces of different dimensions be equal?

No, subspaces of different dimensions cannot be equal. The dimension of a subspace is a fundamental property and different dimensions mean that the subspaces are not equivalent. Subspaces with different dimensions can still have some overlap, but they cannot be considered equal.

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