# What is a linear space?

1. Oct 22, 2011

### Kuma

I have a question that asks me to show that something is a linear space. but what is a linear space exactly and how does it differ from vector space or euclidian space?

2. Oct 22, 2011

### Fredrik

Staff Emeritus
It's the same thing as a vector space. Some people even call it a linear vector space.

3. Oct 22, 2011

### Kuma

ok. so to simply prove that something is a linear space I just have to show that all the axioms of a linear (vector) space hold?

I am given the definition of L which is the set of all vectors z that has inner product of 0 with vectors w for all vectors w in the linear span of x.

in this case I just have to show that all the axioms hold for L?
ie
z1+ z2 = z2 + z1 etc..?

4. Oct 22, 2011

### Fredrik

Staff Emeritus
Yes, that would work. An alternative (that means more work now, but less work in the future) is to prove the following theorem first, and then use it every time you have to solve a problem like this.

Suppose that V is a vector space and that U is a subset of V. Then U is a (linear/vector) subspace of V if and only if the 0 vector of V is a member of U, and U is closed under linear combinations. (The latter condition means that for all numbers a,b and all x,y in V, ax+by is a member of V).

This result simplifies your task a lot, since it tells you that you only need to check two things.

Last edited: Oct 22, 2011
5. Oct 22, 2011

### Kuma

I have learned that theorem, but how would it apply here? I'm asked to prove that L is a vector space itself, so shouldn't I use the definition of a vector/Linear space itself to show that it indeed is a vector space? ie proving the axioms for all elements in L?

6. Oct 22, 2011

### Fredrik

Staff Emeritus
A subspace is by definition a vector space, so if you just show that 0 is in L and that L is closed under linear combinations, then the theorem ensures that L is a vector space. And in this case, you said that your L was defined as a set of "vectors". To me that can only mean that L is defined as a subset of a vector space, but perhaps you didn't mean to imply that by using the word "vectors".

After some thought, I prefer this version of the theorem over the one I mentioned in my previous post:

Suppose that V is a vector space and that U is a subset of V. Then the following statements are equivalent:

(a) U is a vector space.
(b) For all a,b in ℝ and x,y in U, ax+by is in U.
(c) For all a in ℝ and x,y in U, ax and x+y are in U.

This theorem gives us a nice way to define the term "subspace". Suppose that V is a vector space. A subset U of V is said to be a subspace of V if the equivalent conditions of the theorem are satisfied.

Edit: The requirement that 0 is in U is unnecessary, because the requirement that ax+by is in U for all a,b in ℝ and x,y in U implies that 0=0x+0y is in U.

Last edited: Oct 22, 2011
7. Oct 23, 2011

### HallsofIvy

However, you still need to prove that subset is non-empty and typically the simplest way to do that is to prove that 0 is in the set.

8. Oct 23, 2011

### Fredrik

Staff Emeritus
Ah, good point. Thanks. All vector spaces have a member denoted by 0, so no vector space is empty. That makes my (a) not equivalent to (b) and (c), since U=∅ satisfies (b) and (c) but not (a). So I need to add something like U≠∅ to (b) and (c)...but it looks nicer to require that 0 is in U.