Union of subspaces of a linear space

[de_brook]

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

 

[yyat]

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.

 

[de_brook]

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.

 

[yyat]

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?

 

[de_brook]

\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?

 

[de_brook]

\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?

 

[de_brook]

\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?

 

[ThirstyDog]

I think the short answer is No.

 

[de_brook]

what about if you are not working n dimensional space, can you still find such a space

 

[de_brook]

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?

 

[ThirstyDog]

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.

 

[de_brook]

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.