Zero Vector

[matheinste]

Hello Everyone.

In a thread in this forum relating to a problem on Subspaces I read that as long as a Vector SubSpace is closed under addition and multiplication we always have the zero vector. I can see that we can always get the zero vector but do we not have to define a zero vector first as given in the Vector Space axioms or can we make it by manipulation IE multiplying a vector by minus one and adding this to the original vector. This seems somehow contrived to me so is it OK to get the zero vector in this manner or does it first need to exist by definition.If we can always get it by manipulation in this way it seems unnecessary to require it by definition.

( I know that we are allowed the Zero Vector alone to be a vector space and so cannot get it any other way in this case )

Thanks for any clarification on this query as I am still learning and want to understand fully.

Matheinste

[HallsofIvy]

Remember that you are talking about a subspace of an already given vector space. Since the subspace must first be a subset of that space, and use the same operation, you already know that the zero vector exists as part of the original vector space.

Suppose V is a vector space with operations of scalar multiplication and vector addition already defined, 0 as its zero vector. If U is a subset of V closed under both scalar multiplication and vector addition (as defined in V) then for any vector v, (-1)v is in U (closure under scalar multiplication) and so v+ (-1)v= 0. That is, the 0 vector defined in V must also be in U. Since for any vector v in V, v+ 0= v, for any vector u in U (which is also in V) u+ 0= u and so 0 is also the zero vector in U.

[matheinste]

Thanks HallsofIvy.

I was forgetting that the zero vector was already defined in the original vector space.

Matheinste.

[matt grime]

Why all this adding v to -v? It is a trivial exercise in the early weeks of the course to show that 0v=0 for any v, so any set closed under scalar multiplication contains 0 - you do not even need closed under addition. (Actually, you probably do in the proof that 0v=0, I suppose, but given that one is a *subset* of the other with the same operations, I'm OK.)

[Moo Of Doom]

The reason that we require the 0 vector to be an element of the subset for it to be a subspace is simply to rule out the empty set. Surely the empty set is a subset of any vector space, and the empty set is closed under linear combinations. Thus we need another criteria somewhere in order to rule out the empty set.

[matt grime]

A subspace is defined to be a non-empty set satisfying certain rules. 0 is not a priori specified as an element of that non-empty set.

[matheinste]

Thanks for all help and comments.

I think I have realised where I was going wrong. When testing that a vector space is a subspace of another vector space I was wrongly assuming that because we can make the zero vector by manipulating vectors within the possible subspace that the zero vector was in that possible subspace. Of course it may not be. To prove that a vector space is a subspace I need to show that it contains the zero vector and not just that we can make the zero vector with elements from the possible subspace.

Does this seem OK.

Thanks. Matheinste.

[Moo Of Doom]

To prove that a vector space is a subspace I need to show that it contains the zero vector and not just that we can make the zero vector with elements from the possible subspace.

Does this seem OK.

Not quite. If you can make the zero vector with elements from the subset, then the zero vector is in your subset, and you've already shown it. The key is that there actually has to be an element in your subset in order to do that.

I assume your definition of subspace is something like this:

Let V be a vector space. Then a subset U of V is a subspace if and only if both of the following hold:

1. The zero vector is in U.
2. U is closed under linear combinations.

Maybe you'd be more at ease with the following (equivalent) definition:

Let V be a vector space. Then a subset U of V is a subspace if and only if both of the following hold:

1. U is nonempty.
2. U is closed under linear combinations.

Can you see why they are equivalent?

[matheinste]

Thanks Moo Of Doom.

I have thoughtr about it. If U is nonempty it must contain at least one ( for our purposes one vector ) element. If this element is the zero vector there is no more to be said. If it contains a non zero vector then by the vector space axioms we can make the zero vector ( among others ). ?

Matheinste.