# Vector space

1. Apr 30, 2010

### Dustinsfl

If $$S\subseteq V$$ 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?

2. Apr 30, 2010

### VeeEight

Why is this true?

3. Apr 30, 2010

### Staff: Mentor

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. Apr 30, 2010

### Dustinsfl

I said assume S isn't a vector space.

We have $$P\rightarrow Q\equiv P\and\sim Q$$

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. Apr 30, 2010

### Staff: Mentor

Which one is A? You started out with S and V.

6. Apr 30, 2010

### Dustinsfl

I meant S.

7. Apr 30, 2010

### Staff: Mentor

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. Apr 30, 2010

### Dustinsfl

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. Apr 30, 2010

### Staff: Mentor

Based on your original post, here are P and Q.
P: $S\subseteq V$ and V is a vector space
Q: S is a vector space

Forget deMorgan - look at post #3.

10. Apr 30, 2010

### Dustinsfl

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

If $$S\subseteq V$$ of a vector space V, then S is a vector space.

11. Apr 30, 2010

### VeeEight

Take Mark's counterexample (or some other counterexample you can think of), that is the best approach.

12. Apr 30, 2010

### Dustinsfl

The answer is supposed to be true so taking a counterexample seems counter-intuitive.

13. Apr 30, 2010

### Squeezebox

S is defined as an improper subset of V; so if V is a vector space, S must be as well.

14. Apr 30, 2010

### Dustinsfl

For all we know, it could be a proper subset too. That was just the notation I used.

15. Apr 30, 2010

### Squeezebox

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. Apr 30, 2010

### Squeezebox

Look at post #3, which has been said a few times now.

17. Apr 30, 2010

### Dustinsfl

The answer was true so I am not to sure about finding a counterexample as being the correct method.

18. Apr 30, 2010

### Subdot

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: Apr 30, 2010
19. Apr 30, 2010

### Staff: Mentor

Not true. Not all subsets of vector spaces are themselves subspaces. Here's another example, with V = R2. Let S = {(x, y} | y = 1}.
$S~\subseteq V$, and V is a vector space, but S is not a subspace.

20. May 1, 2010

### Squeezebox

I said S was an improper subset.