# Homework Help: Proving a vector space.

1. Oct 19, 2011

### -Dragoon-

1. The problem statement, all variables and given/known data
Let V be the set of ordered pairs (x, y) of real numbers with the operations of vector addition and scalar multiplication given by:
(x, y) + (x', y') = (y + y', x + x')
c(x, y) = (cx, cy)

V is not a vector space. List one of the properties from the definition of vector space that fails to hold for V. Justify your answer.

2. Relevant equations
N/A

3. The attempt at a solution

I really do not have any idea on how to get started on this question. I've tried all 8 properties of a vector space and, at least to me, it holds for all 8 properties and should thus be a vector space, but the question explicitly says it is not. I have the answer to the question, but would like to try a different approach to proving it is not a vector space. Any help?

2. Oct 19, 2011

### lanedance

not the first one interchanges x & y, try and use that to find a counterexample... does a zero element exist such that v+0=0 for all v?

3. Oct 19, 2011

### HallsofIvy

lanedance means "v+ 0= v".

What is (x, y)+ (0, 0)?

You should see that (0, 0) is NOT the "0 vector" but then you have to show that no other vector can be:
(x, y)+ (u, v)= (y+ v, x+ u)= (x, y) gives what?

4. Oct 19, 2011

### -Dragoon-

How exactly did you get (x, y) from (y + v, x + u)?

As to your first question: (x, y) + (0, 0) = (y + 0, x + 0) = (y, x)

But, (x, y) - (y, x) = (y - x, x - y) ≠ (0, 0) assuming, of course, that x ≠ y.

Is this sufficient to prove that V is not a vector space?

5. Oct 19, 2011

### HallsofIvy

No, that only shows that (0, 0) is not the 0 vector. It doesn't mean that some other vector, a, say, does not have the property that v+ a= v for all v.

That was why I wrote (x, y)+ (u, v)= (y+ v, x+ u)= (x, y).
I did NOT get (x,y) from (y+ v, x+ u), I am saying that if (u, v) is the 0 vector then that must be true. Now, show that u and v cannot be independent of x and y.

6. Oct 19, 2011

### -Dragoon-

Ah, yes. I was thinking in terms of subspaces (where the standard zero vector exists in the vector space it is a subset of).

Let's say (a, b) is part of the set V, then using the associativity property of addition:

(a,b) + ((x, y) + (u, v)) = (y + v, x + u) + (a, b) = (x + u + b, y + v + a)

(x, y) + ((a, b) + (u, v)) = (b + v, a + u) + (x, y) = (a + u + y, b + v + x)

Since (a + u + y, b + v + x) ≠ (x + u + b, y + v + a), the associativity property of vector addition does not hold, and hence, V is not a vector space. Correct?

7. Oct 19, 2011

### lanedance

exactly, cheers

8. Oct 19, 2011

### lanedance

good enough for me

9. Oct 19, 2011

### lanedance

thw way I was hinting about the zero vector was as follows:

say the additive identity element 0 exists, lets call it 0=(p,q)

now take an arbitrary element (x,y), it must satisfy
(x,y)+(p,q) = (x,y)

but
(x,y)+(p,q) = ((y+q), (x+p))

for the "zero: property to be true we must have
q=x-y
p=y-x

hence no universal 0=(p,q) element exists