Condition for equality between subspaces.


by peripatein
Tags: condition, equality, subspaces
peripatein
peripatein is offline
#1
Dec3-12, 12:12 PM
P: 816
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!
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
micromass
micromass is offline
#2
Dec3-12, 12:17 PM
Mentor
micromass's Avatar
P: 16,690
Quote Quote by peripatein View Post
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,
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??
peripatein
peripatein is offline
#3
Dec3-12, 02:04 PM
P: 816
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?

micromass
micromass is offline
#4
Dec3-12, 02:10 PM
Mentor
micromass's Avatar
P: 16,690

Condition for equality between subspaces.


Quote Quote by peripatein View Post
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.
peripatein
peripatein is offline
#5
Dec3-12, 04:03 PM
P: 816
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?
micromass
micromass is offline
#6
Dec3-12, 05:05 PM
Mentor
micromass's Avatar
P: 16,690
Quote Quote by peripatein View Post
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]?
peripatein
peripatein is offline
#7
Dec3-12, 06:42 PM
P: 816
You get alpha*0=0. How does that help?
micromass
micromass is offline
#8
Dec3-12, 07:04 PM
Mentor
micromass's Avatar
P: 16,690
Quote Quote by peripatein View Post
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?
peripatein
peripatein is offline
#9
Dec3-12, 07:30 PM
P: 816
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?


Register to reply

Related Discussions
Equality between subspaces Calculus & Beyond Homework 15
equality of integrals => equality of integrands Calculus 4
Product equality and sum of squares equality puzzle Brain Teasers 1
finding the intersection of subspaces, and addition of subspaces Calculus & Beyond Homework 6
Prove Equality given a condition Precalculus Mathematics Homework 7