Proving a Vector Space Cannot be the Union of Two Proper Subspaces

In summary: I think the issue is that the union is not the same as the span. The union of two subspaces will not be a subspace unless one of the subspaces is contained within the other (is a subspace of the subspace) in which case the union is the larger subspace.
  • #1
Dustinsfl
2,281
5
Prove that a vector space cannot be the union of two proper
subspaces.

Let V be a vector space over a field F where U and W are proper subspaces.

I am not sure where to start with this proof.
 
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  • #2
There seems to be something missing in this problem. For example, take the case where U,W are proper subspaces of V, and [itex]U \subset W[/itex], then [itex]U \cap W \equiv W[/itex] is also a proper subspace of V.

Are you sure there's not some other restrictions regarding the subspaces?
 
  • #3
Would it matter that it is the union and you have written the intersection?
 
  • #4
What is the definition of "proper subspace"? You should have included that definition when you posted the problem.
 
  • #5
Mark44 said:
What is the definition of "proper subspace"? You should have included that definition when you posted the problem.

A proper subspace can't be equal to V.
 
  • #6
But how is this term defined? What you gave is not the definition.
 
  • #7
Mark44 said:
But how is this term defined? What you gave is not the definition.

If U is a proper subspace, then the dim U < dim V and U isn't the subspace of just the 0 vector, i.e., not the trivial subspaces.
 
  • #8
So your definition of "U is a proper subspace of V" does not include requiring that U be a subset of V?
 
  • #9
Coto said:
There seems to be something missing in this problem. For example, take the case where U,W are proper subspaces of V, and [itex]U \subset W[/itex], then [itex]U \cap W \equiv W[/itex] is also a proper subspace of V.

Are you sure there's not some other restrictions regarding the subspaces?

Dustinsfl said:
Would it matter that it is the union and you have written the intersection?
That was a typo. Coto meant [itex]U \cup W = W[/itex].
 
  • #10
I think the issue is that the union is not the same as the span. The union of two subspaces will not be a subspace unless one of the subspaces is contained within the other (is a subspace of the subspace) in which case the union is the larger subspace.

So apply that to the question at hand...
 
  • #11
jambaugh said:
I think the issue is that the union is not the same as the span. The union of two subspaces will not be a subspace unless one of the subspaces is contained within the other (is a subspace of the subspace) in which case the union is the larger subspace.

So apply that to the question at hand...

Thanks.
 

Related to Proving a Vector Space Cannot be the Union of Two Proper Subspaces

What is the definition of a "Union of two subspaces"?

The union of two subspaces is the set of all vectors that can be obtained by combining the elements of both subspaces. It is denoted as S ∪ T, where S and T are the two subspaces.

How is the union of two subspaces different from the intersection of two subspaces?

The union of two subspaces contains all the vectors that are in either of the two subspaces, while the intersection of two subspaces contains only the common vectors in both subspaces.

Is the union of two subspaces always a subspace?

No, the union of two subspaces is not always a subspace. It is a subspace only if the two subspaces have a non-empty intersection.

How is the dimension of the union of two subspaces related to the dimensions of the individual subspaces?

If the two subspaces have a non-empty intersection, then the dimension of the union will be equal to the sum of the dimensions of the individual subspaces minus the dimension of the intersection. Otherwise, the dimension of the union will be equal to the sum of the dimensions of the individual subspaces.

Can the union of two subspaces be represented as a direct sum?

Yes, the union of two subspaces can be represented as a direct sum if the two subspaces are disjoint (i.e. have an empty intersection).

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