Union of subspaces of a linear space

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Discussion Overview

The discussion revolves around the properties of unions of subspaces within a linear space, specifically whether the union of any subspaces can itself be a subspace, except in trivial cases. The scope includes theoretical considerations of linear algebra and dimensionality of vector spaces.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • Some participants assert that a vector space V can only have non-trivial subspaces if its dimension is at least 2.
  • It is proposed that for two linearly independent vectors u and w generating 1-dimensional subspaces U and W, the union U ∪ W cannot be a subspace.
  • One participant expresses difficulty in finding examples of spaces where the union of subspaces is a subspace, noting only finding cases for dimensions less than 2.
  • Another participant emphasizes that in n-dimensional space, the dimension is n, questioning if a linear subspace can be found where the union of any subspaces is a subspace.
  • Some participants conclude that the answer is likely "No," regarding the existence of such spaces in higher dimensions.
  • It is noted that if the dimension is less than two, the only subspaces are V and {0}, which addresses the original question in that case.
  • In dimensions greater than or equal to two, it is argued that the sum of linearly independent vectors does not belong to the union of their respective subspaces, implying the union cannot be a subspace.

Areas of Agreement / Disagreement

Participants generally agree that in dimensions less than two, the only subspaces are trivial. However, there is disagreement regarding the existence of a space in higher dimensions where the union of subspaces could be a subspace, with some asserting "No" and others questioning further.

Contextual Notes

The discussion highlights the dependence on the dimensionality of the vector space and the nature of the subspaces involved. There are unresolved assumptions regarding the definitions of subspaces and the implications of linear combinations.

de_brook
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Is there a linear space V in which the union of any subspaces of V is a subspace except the trivial subspaces V and {0}? pls help
 
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A vector space V can only have non-trivial subspaces if \dim V\ge 2.
This means you can choose two linearly independent vectors u, w, which generate 1-dimensional subspaces U, W respectively. Can U\cup W be a subspace? Hint: try to find a linear combination of u,w that is not in U\cup W.
 
yyat said:
A vector space V can only have non-trivial subspaces if \dim V\ge 2.
This means you can choose two linearly independent vectors u, w, which generate 1-dimensional subspaces U, W respectively. Can U\cup W be a subspace? Hint: try to find a linear combination of u,w that is not in U\cup W.

I have tried searching for such spaces but i could only find for spaces whose dimension is less than 2.
 
de_brook said:
I have tried searching for such spaces but i could only find for spaces whose dimension is less than 2.

\dim\mathbb{R}^n=n, surely you knew that?
 
yyat said:
\dim\mathbb{R}^n=n, surely you knew that?

yyat; Do you mean i can obtain a linear subspace V of \mathbb{R}^n such that the union of any subspaces of V is a subspace of V?
 
yyat said:
\dim\mathbb{R}^n=n, surely you knew that?

yyat; Do you mean i can obtain a linear subspace V of \mathbb{R}^n such that the union of any subspaces of V is a subspace of V?
 
yyat said:
\dim\mathbb{R}^n=n, surely you knew that?

yyat; Do you mean i can obtain a linear subspace V of \mathbb{R}^n, such that the union of any subspaces of V is a subspace of V?
 
I think the short answer is No.
 
what about if you are not working n dimensional space, can you still find such a space
 
  • #10
ThirstyDog said:
I think the short answer is No.

what about if you are not working with n-dimensional space, can you still find such a space?
 
  • #11
If the dimension of the space is less than two then the only subspace are V and {0} as yyat pointed out. Hence your question is answered in this case.

If the dimension of the space is greater or equal to two then consider spaces X and Y generated by linearly independent vectors x and y. x+y does not belong to X \Cup Y. Implying you can't pick any subspaces and the union will be a subspace.
 
  • #12
ThirstyDog said:
If the dimension of the space is less than two then the only subspace are V and {0} as yyat pointed out. Hence your question is answered in this case.

If the dimension of the space is greater or equal to two then consider spaces X and Y generated by linearly independent vectors x and y. x+y does not belong to X \Cup Y. Implying you can't pick any subspaces and the union will be a subspace.

thanks so much.
 

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