Condition for equality between subspaces.


by peripatein
Tags: condition, equality, subspaces
peripatein
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#1
Dec3-12, 12:12 PM
P: 812
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!
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micromass
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#2
Dec3-12, 12:17 PM
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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
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#3
Dec3-12, 02:04 PM
P: 812
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?

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#4
Dec3-12, 02:10 PM
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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
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#5
Dec3-12, 04:03 PM
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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
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#6
Dec3-12, 05:05 PM
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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
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#7
Dec3-12, 06:42 PM
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You get alpha*0=0. How does that help?
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#8
Dec3-12, 07:04 PM
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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
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#9
Dec3-12, 07:30 PM
P: 812
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?


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