Union of subspaces of a linear space

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 [tex]\dim V\ge 2[/tex].
This means you can choose two linearly independent vectors u, w, which generate 1-dimensional subspaces U, W respectively. Can [tex]U\cup W[/tex] be a subspace? Hint: try to find a linear combination of u,w that is not in [tex]U\cup W[/tex].
 
A vector space V can only have non-trivial subspaces if [tex]\dim V\ge 2[/tex].
This means you can choose two linearly independent vectors u, w, which generate 1-dimensional subspaces U, W respectively. Can [tex]U\cup W[/tex] be a subspace? Hint: try to find a linear combination of u,w that is not in [tex]U\cup W[/tex].
I have tried searching for such spaces but i could only find for spaces whose dimension is less than 2.
 
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I have tried searching for such spaces but i could only find for spaces whose dimension is less than 2.
[tex]\dim\mathbb{R}^n=n[/tex], surely you knew that?
 
[tex]\dim\mathbb{R}^n=n[/tex], 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?
 
[tex]\dim\mathbb{R}^n=n[/tex], surely you knew that?
yyat; Do you mean i can obtain a linear subspace V of [tex]\mathbb{R}^n such that the union of any subspaces of V is a subspace of V?
 
[tex]\dim\mathbb{R}^n=n[/tex], surely you knew that?
yyat; Do you mean i can obtain a linear subspace V of [tex]\mathbb{R}^n[/tex], 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
 
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 [tex] X \Cup Y [/tex]. Implying you can't pick any subspaces and the union will be a subspace.
 
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 [tex] X \Cup Y [/tex]. Implying you can't pick any subspaces and the union will be a subspace.
thanks so much.
 

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