# Some help understanding vector space

1. Oct 31, 2014

### MarcL

Hey, so my class just got into vector space and I don't have a clue what is going on. My teacher makes us go through the axiom for every problem but I can't visualize what is happening in our problems ( and our workbook makes me even more confused)

It seems so... useless. The axioms. Why would I need to verify x+y=y+x. Isn't that always true for additions?
Or that c(x,y)=cx,cy. Or telling me that there is a zero vector in V so u+0=u+0=u.

Sorry if this sounds really dumb it just... doesn't make sense to me. It all sounds so redundant because those are basics vector operations and such no? Samething as a vector space, wouldn't it just be a plane?

I'm just all confused as to what this is... just need help clarifying things.

2. Oct 31, 2014

### Fredrik

Staff Emeritus
You wouldn't call the operation "addition" (or denote it by +) if that axiom doesn't hold. But when you define a new operation, it's not obvious that the axiom holds. The verification isn't supposed to tell you anything about "addition". It's supposed to tell you if what you have defined deserves to be called addition.

That's a definition, not an axiom. You use the definition to prove that ordered pairs of real numbers satisfy vector space axioms. One of them says that "for all v in V, we have 1v=v". The proof goes like this (assuming that we have specified that V is the set of ordered pairs of real numbers, and that addition and scalar multiplication are defined the familiar way): Let v be an arbitrary element of V. Let x and y be the unique real numbers such that v=(x,y). We have $1v=1(x,y)=(1x,1y)=(x,y)=v$.

Let's say that V is the set of functions from ℝ into ℝ. Is it obvious that there's a zero vector in V. Is it super trivial to verify that there's a 0 in V such that v+0=0+v=v for all v in V? (It's trivial to those of us with experience, but most students who try this for the first time in our homework forum get it very, very wrong).

Suppose that V is the set of positive real numbers, and we define the following operations instead: $x\oplus y=xy$ for all x,y in V, and $kx=|k|x$ for all k in ℝ and all x in V. Is it still obvious that there's a zero vector in V?

The set of ordered pairs of real numbers, with "addition" and "scalar multiplication" defined the way you have seen, is the simplest non-trivial example of a vector space. That vector space can be interpreted as a plane. The vector space axioms are statements that are easily seen to be satisfied by those two operations. This result is the inspiration for the definition of "vector space", which say that any set with two operations that satisfy those conditions is called a vector space. Every vector space has a lot in common with planes (specifically they satisfy the vector space axioms), but most vector spaces are not planes.

3. Nov 2, 2014

### Stephen Tashi

If you only think of one example of a vector space, you won't appreciate the need for proving things about vector spaces. You need to stretch your mind a little. For example, think of the set of polynomials in the single variable x with real coefficients or think of the set of continuous real valued functions on the unit interval. See if they can be made to fit the definition of a vector space. Think of some vector spaces with an infinite number of dimensions.