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Vector space

  • Thread starter Dustinsfl
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  • #1
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If [tex]S\subseteq V[/tex] and V is a vector space, then S is a vector space.

Assume S isn't a vector space. Since S isn't a vector space, then V isn't a vector space; however, V is a vector space. By contradiction, S is a subspace.

Correct?
 

Answers and Replies

  • #2
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Since S isn't a vector space, then V isn't a vector space
Why is this true?
 
  • #3
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Suppose V = R3 and S = {(x, y, z)| x + y - 2z = 3}. S is clearly a subset of V, but is S a subspace of V?
 
  • #4
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Why is this true?
I said assume S isn't a vector space.

We have [tex]P\rightarrow Q\equiv P\and\sim Q[/tex]

I don't know why the latex is looking messed up but that is supposed to say P and not Q is equivalent to P implies Q
 
  • #5
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Which one is A? You started out with S and V.
 
  • #6
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Which one is A? You started out with S and V.
I meant S.
 
  • #7
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P ==> Q <===> ~P ==> ~Q
but you haven't shown that, given that S isn't a subspace, somehow this implies that V is not a subspace. In other words, if S is not a subspace, why does it necessarily follow that V is not a subspace? In fact, you are really concluding the opposite.

I think you are mixing up a proof by the contrapositive with a proof by contradiction.

See post #3.
 
  • #8
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I was using de morgan's laws which says p implies q is equiv to ~p or q which is equiv to p and ~ q.

That is why I assumed q (S isn't a vector space).
 
  • #9
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Based on your original post, here are P and Q.
P: [itex]S\subseteq V[/itex] and V is a vector space
Q: S is a vector space

Forget deMorgan - look at post #3.
 
  • #10
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If [tex]S\subseteq V[/tex] and V is a vector space, then S is a vector space.

Assume S isn't a vector space. Since S isn't a vector space, then V isn't a vector space; however, V is a vector space. By contradiction, S is a subspace.

Correct?
I don't know if this way change anything but it should be worded like so:

If [tex]S\subseteq V[/tex] of a vector space V, then S is a vector space.
 
  • #11
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Take Mark's counterexample (or some other counterexample you can think of), that is the best approach.
 
  • #12
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Take Mark's counterexample (or some other counterexample you can think of), that is the best approach.
The answer is supposed to be true so taking a counterexample seems counter-intuitive.
 
  • #13
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S is defined as an improper subset of V; so if V is a vector space, S must be as well.
 
  • #14
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S is defined as an improper subset of V; so if V is a vector space, S must be as well.
For all we know, it could be a proper subset too. That was just the notation I used.
 
  • #15
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For all we know, it could be a proper subset too. That was just the notation I used.
Then it is false. There is one vector that S can exclude yet still retain it's subset status. Think of the definition of a vector space.
 
  • #16
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Then it is false. There is one vector that S can exclude yet still retain it's subset status. Think of the definition of a vector space.
Look at post #3, which has been said a few times now.
 
  • #17
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The answer is supposed to be true so taking a counterexample seems counter-intuitive.
The answer was true so I am not to sure about finding a counterexample as being the correct method.
 
  • #18
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But a counterexample *does* exist. Therefore, the statement is false (assuming it was meant to be universally true). The statement is true if and only if S also satisfies the closure axioms of a vector space under the operations of addition and multiplication as defined by V (assuming S is a nonempty subset of V).
 
Last edited:
  • #19
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S is defined as an improper subset of V; so if V is a vector space, S must be as well.
Not true. Not all subsets of vector spaces are themselves subspaces. Here's another example, with V = R2. Let S = {(x, y} | y = 1}.
[itex]S~\subseteq V[/itex], and V is a vector space, but S is not a subspace.
 
  • #20
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Not true. Not all subsets of vector spaces are themselves subspaces. Here's another example, with V = R2. Let S = {(x, y} | y = 1}.
[itex]S~\subseteq V[/itex], and V is a vector space, but S is not a subspace.
I said S was an improper subset.
 
  • #21
HallsofIvy
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S is defined as an improper subset of V; so if V is a vector space, S must be as well.
Where was "S defined as an improper subset of V"?
 
  • #22
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S is defined as an improper subset of V; so if V is a vector space, S must be as well.
For all we know, it could be a proper subset too. That was just the notation I used.
Then it is false. There is one vector that S can exclude yet still retain it's subset status. Think of the definition of a vector space.


There you go guys. I said it was improper because of the notation, then I took it back. Sorry to offend you guys by saying something stupid.
 
  • #23
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No offense taken, just trying to clear up incorrect information. In this problem it doesn't matter whether it said [itex]S \subset V[/itex] or [itex]S \subsetex V[/itex]. The latter notation means "S is a subset of V or is equal to V." S can still be a proper subset of V without contradicting this statement.
 

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