Linear Algebra: Vector Subspaces

AI Thread Summary
The discussion centers on the statement that the union of two vector subspaces is a subspace. It is clarified that this statement is generally false, as the union may not be closed under vector addition. A counter-example is provided using subspaces in R2, specifically the sets {(x, 0)} and {(0, y)}, which do not combine to include all necessary vectors. The participants emphasize that the definition of a subspace requires closure under addition and scalar multiplication, which the union fails to satisfy. Overall, the union of two vector subspaces does not form a subspace.
war485
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Homework Statement



True/false: Union of two vector subspaces is a subspace.

Homework Equations



none

The Attempt at a Solution



I'm unsure if this is true because I'm also unsure if it already assumes that it is closed under scalar multiplication and addition. If it is closed, then I'd like to think it is true, but it might be false if it isn't closed. Help me a lil' to know what's already assumed?
 
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Generally that is false, the union of two vector subspaces is not a subspace. To prove it is false, first assume that it is true and look for a counter-example. Let me see what you can come up with.
 
how about span of e1 and span of e2. Then their union is not closed under vector addition, then the union is not a subspace, woah.
 
war485 said:

Homework Statement



True/false: Union of two vector subspaces is a subspace.

Homework Equations



none

The Attempt at a Solution



I'm unsure if this is true because I'm also unsure if it already assumes that it is closed under scalar multiplication and addition. If it is closed, then I'd like to think it is true, but it might be false if it isn't closed. Help me a lil' to know what's already assumed?
Do you not know what the "union" of two sets is? A question like this doesn't "assume" anything it doesn't say. Since a "subspace" is simply a subset that is closed under addition and scalar multiplication, surely a question that asks whether something is a subspace or not will not "assume" that it is a subspace.

As war485 suggested, in R2, {(x, 0)} is a subspace, containing, say, (2, 0). The set {(0, y)} is a subspace containing (0, 3). Their union does NOT contain (2, 3).
 

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