vector space and subspace

To show W is a subspace of V, it needs to be show 0 is in W, and W is closed under addition and scalar multiplication. In the axioms for a vector space, it does not say it needs to be closed under scalar multiplication/addition. Is it a consequence of the axioms?

Re: vector space and subspace

if you list the axioms you use, i will show you it is a consequence of those axioms.

Re: vector space and subspace

ok thanks. Let v,w be in V, t,s in F.

(1) vector addition is associative and commutative.

(2) contains zero vector

(3) additive inverse

(4) t(sv)=(ts)v

(5) t(v+w)=tv+tw

(6) v(t+s)=vt+vs

Re: vector space and subspace

Quote:

Originally Posted by

**Duke** ok thanks. Let v,w be in V, t,s in F.

(1) vector addition is associative and commutative.

(2) contains zero vector

(3) additive inverse

(4) t(sv)=(ts)v

(5) t(v+w)=tv+tw

(6) v(t+s)=vt+vs

Does the definition you use call addition and multiplication "binary operations"?

Re: vector space and subspace

No but I suppose they are. They take 2 objects and map it to one object.

Re: vector space and subspace

The point is that the **definitions** of addition and scalar multiplication must specify that the result is a vector. That is, the addition is a function from VxV to V and scalar multiplication is a function from FxV to V. The fact that the result must be a vector **is** closure for those operations.

Re: vector space and subspace

for any mapping to be defined, you need to specify two things, first: a domain, and co-domain. it is not merely enough to define a formula for f(x,y), you must say what it is allowed to be.

now, in a vector space, the primary operation of interest is the vector sum: f(x,y) = x+y. this is a mapping from VxV--->V, intrinsic to the definition is: you take 2 vectors, do something you call "adding" and you get a 3rd vector. some textbooks aren't very explicit about this, but it is basic to the characterization of any algebraic structure, when defining operations, you have to specify "the type of inputs and the type of outputs". for a vector space, the requirement is that the 2 inputs (since there are 2 this is a 2-ary, or binary operation) are vectors, and the output is also a vector. this is expressed different ways:

1) + is a binary operation on V

2) u+v is the vector that..... (note by calling u+v a vector, we are claiming u+v is in V), or defined as.....

3) +:VxV--->V +(u,v) = u+v (sometimes authors use "s(u,v)" for sum).

often this isn't even called an axiom, but it concealed in some introductory text which many people gloss over, because it isn't all by itself with a big number by it saying: hey! i'm an axiom!

now, in general, not just with vector spaces, but with fields, or polynomials, a sub-thing is a smaller thing you get by using the same operations and rules of the bigger thing, but is a thing in its own right. translated into vector-ese this is:

a subspace U, of a vector space V, is a subset U of the underlying set V, such that U is a vector space in itself, when using the same definition of vector sum and scalar multiplication that we use for V.

this means U has to satisfy all of the axioms for a vector space. so if + on V, takes a pair of V-vectors, and spits out a V-vector, then the first thing we need to do is check that that same operation + on U, spits out a U-vector, when we add two U-vectors. otherwise u1+u2 goes "outside" U, and U is not a self-contained vector space (we have to make it bigger in order for it to work).

the exact same reasoning applies to scalar multiplication, which is a "mixed binary" operation which goes from FxV--->V (often F is the real numbers). if we want a subset U of V to be a self-contained vector space, we need the same scalar multiplication to "stay in U" FxU--->U.

read carefully the paragraaphs just before the place in your text where the axioms are listed, and tell me what you find there.

Re: vector space and subspace

My objection is this. You say 'you take 2 vectors, do something you call "adding" and you get a 3rd vector'. But that statement would still be true if '3rd vector' was from a different vector space W. I think your answer will be that it is implicity understood to be an operation V X V-> V.

Re: vector space and subspace

it shouldn't be implicit. if your textbook does not specifically state that vector addition on V must produce another vector in V, then shame on it.

the term "vector" can have many meanings (a lot of very different sets can be made into vector spaces, some of them are....unusual).

so the set V that gives us our set of "what we are calling vectors" has to be defined beforehand. you can't just assume "people know what vectors are".

Re: vector space and subspace