# Math Help - [SOLVED] Subspace of a vector space

1. ## [SOLVED] Subspace of a vector space

If S is a subspace of a vector space V, then S is a vector space.

De Morgan's Laws:
$P\rightarrow Q\equiv P \wedge \sim Q$

Assume S isn't a vector space and since S is a subspace of V, then $S\subseteq V$.

Since S isn't a vector space, it follows that V isn't a vector space. However, V is a vector space; therefore, by contradiction, S is a vector space.

Correct?

2. Originally Posted by dwsmith
If S is a subspace of a vector space V, then S is a vector space.

De Morgan's Laws:
$P\rightarrow Q\equiv P \wedge \sim Q$

Assume S isn't a vector space and since S is a subspace of V, then $S\subseteq V$.

Since S isn't a vector space, it follows that V isn't a vector space. However, V is a vector space; therefore, by contradiction, S is a vector space.

Correct?
I have no idea. My main problem when trying to follow your line of thinking is to figure out what strange definition of "subspace" you might be using.
If you take the definition of "S is a subspace of V" to be that "S is a subset of V that also happens to be a vector space", then there is really nothing to prove here at all. A subspace (of a vector space) is a vector space by definition.

3. Originally Posted by Failure
I have no idea. My main problem when trying to follow your line of thinking is to figure out what strange definition of "subspace" you might be using.
If you take the definition of "S is a subspace of V" to be that "S is a subset of V that also happens to be a vector space", then there is really nothing to prove here at all. A subspace (of a vector space) is a vector space by definition.
This is the question verbatim.
If S is a subspace of a vector space V, then S is a vector space.

How would you approach it then?

4. Originally Posted by dwsmith
How would you approach it then?
You would tell us your definition of a subspace.

5. Originally Posted by Tikoloshe
You would tell us your definition of a subspace.
The normal def: Closed under scalar multiplication and vector addition.

6. Originally Posted by dwsmith
The normal def: Closed under scalar multiplication and vector addition.
I see. Well then, just prove that that subset S of V (that is closed under scalar multiplicaion and vector addition) has all the properties that we require a vector space to have.

You can't do that by just doing fancy footwork with propositional logic alone. It has got to be an argument based on the definition of being closed under those two operations and of being a vector space.

For example, to show that S contains the null vector, you can argue that this follows from its closure under scalar multiplication, because 0x=0.
and so on... Just follow the definition of vector space...

7. Well, if it’s closed under scalar multiplication, then an additive inverse and additive identity follow. What are other properties in your definition of a vector space do not hold trivially?

Regarding your original proof. The only property of a subspace you seem to use is that of its being a subset. There are of course subsets of a vector space which are not subspaces.

8. Originally Posted by Tikoloshe
Well, if it’s closed under scalar multiplication, then an additive inverse and additive identity follow. What are other properties in your definition of a vector space do not hold trivially?

Regarding your original proof. The only property of a subspace you seem to use is that of its being a subset. There are of course subsets of a vector space which are not subspaces.

We are giving the fact that it is a subspace though.

9. Originally Posted by dwsmith
We are giving the fact that it is a subspace though.
Of course you are. My point was (and I see now that Failure explained it better), that the only property you actually use in your original proof was that of being a subset. Specifically you don’t use closure under multiplication or addition. Under that logic, any subset would satisfy your criteria and you end up with arbitrary subsets as subspaces.

To clarify, what I am talking about is here:
Originally Posted by dwsmith
…, then $S\subseteq V$.

Since S isn't a vector space, it follows that V isn't a vector space…
That is, a subset’s not being a vector space does not imply that the whole space is not a vector space.

E.g., $V=\mathbb{R}$ is a vector space and $A=\{1\}$ can be shown not to be a vector space. It’s not true that $V$ is not a vector space just because $A\subseteq V$ and $A$ is not a vector space.

10. So, once more, the real question is "what is your definition of subspace?"

In many books, S is a subspace of vector space V if and only if S is a subset of V which, with the operations defined for V, is a vector space in its own right. That definition makes your original statement trivial!

So apparently, you are using, "as subset of V that is closed under addition and scalar multiplication" which, strictly speaking, is NOT a good definition! It allows the empty set to be a subspace!

Most textbooks will add either "the subset is non empty" or "the additive identity vector (0) is included". Once we know that the set is non-empty, 0v= 0 is in the set, so those are equivalent. But still, the "normal" definition is the one I gave above- a subset that, with the same operations as V is a vector space itself.

In order to show that S is a vector space, you must also show all of the other "axioms" for a vector space- that the operations are commutative and associative, that there is and additive identity and every vector has an additive inverse. For "additive identity", you need to know that there is at least one vector, v, that is in S so that, since it is closed under scalar multiplication, 0s= 0 is in the set.