# Union of subspaces of a linear space

## Main Question or Discussion Point

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 $$U\cup W$$.

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.

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?

$$\dim\mathbb{R}^n=n$$, 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?

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

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?

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.

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.