MHB What Is Required to Prove a Subset is a Vector Space?

[Yankel]

Hello all,

I have a theoretical question regarding subspaces.

If V is a subset of a vector space, and we wish to show that V is a vector space itself, we need to show 3 things.

Some references say we need to show: a) V is not empty b) V is closed under + c) V is closed under scalar multiplication. Other references are similar, with (a) being that V contains the 0 vector in it.

What I don't understand is:

1) Why being non empty and having the 0 vector is the same thing ?
2) I can't think of any example in which a set is closed under + and scalar multiplication, but does not contain the 0 vector. If this case exist, it ain't a subspace, but if the set isn't empty, it is ??

I am confused...

 

[S.G. Janssens]

Hello all,

I have a theoretical question regarding subspaces.

If V is a subset of a vector space, and we wish to show that V is a vector space itself, we need to show 3 things.

Some references say we need to show: a) V is not empty b) V is closed under + c) V is closed under scalar multiplication. Other references are similar, with (a) being that V contains the 0 vector in it.

What I don't understand is:

1) Why being non empty and having the 0 vector is the same thing ?
2) I can't think of any example in which a set is closed under + and scalar multiplication, but does not contain the 0 vector. If this case exist, it ain't a subspace, but if the set isn't empty, it is ??

I am confused...

ad 1) I understand that you assume b) and c) and then want to show that the two formulations of condition a) are equivalent.

If $V$ is non-empty, then it contains a vector $x$. Now $V$ is closed under scalar multiplication, so $0\cdot x = \mathbf{0} \in V$, where $0$ is a scalar and $\mathbf{0}$ is the zero vector.

Conversely, if $V$ contains $\mathbf{0}$ then it is of course non-empty.

ad 2) You are right, the only such example would be the empty set: It is trivially closed under vector addition and scalar multiplication, but it does not contain the zero vector.

 

[HallsofIvy]

"Being non-empty" and "containing the 0 vector" are not the same thing, alone! However, with the other condirion, that the set is closed under scalar multiplication, they are.

Clearly, if a set "contains the 0 vector" then it "is non-empty" so that way is trivial. If a set "is non-empty" then it contains some, possibly non-zero, vector v. If the set is also "closed under scalar multiplication", then the set also contains 0(v)= 0, the 0 vector.

 

[Yankel]

Thank you both.

Great explanations, all clear now ! :D