Prove V is not a vector space

Let $\displaystyle V = \left\{x,y}\right\}$ be a set with exactly two vectors, x and y. Define vector addition and scalar multiplication in V by the following rules:

Vector addition: $\displaystyle x+x=x$, $\displaystyle y+y=x$, $\displaystyle x+y=y$, and $\displaystyle y+x=y$.

Vector multiplication: $\displaystyle cx=x$, and $\displaystyle cy=y$ for all $\displaystyle c \in \mathbb{R}$.

Prove that V is not a vector space by finding one axiom in the definition of a vector space that fails to hold. You must state the axiom clearly and show it does not hold.

I'm at a loss on this one.

Re: Prove V is not a vector space

Hey jeremy5561.

You need to evaluate the axioms one by one. The first step is to write all the axioms down. Once you've done that then check them one by one.

We will wait for you to write all the axioms down and then if needed, we can look at specific axioms if you need specific help.

Re: Prove V is not a vector space

Closure under addition:

Passes because no matter what combination of x and y are added it results in either x and y in V.

Closure under scalar multiplication.

Passes because cx=x and cy=y for all c in real so no matter what scalar multiple you use the result will be in V.

There exists a zero vector such that 0+v = v

The zero vector could be x because x+x=x and x+y=y.

There exists a additive inverse in V such that v + (-v) = 0 for all v in V.

If x and y are both zero vectors then this would work out. x and y would be the additive inverse for both x and y.

Re: Prove V is not a vector space

Commutativity: passes based on last 2 addition statements.

Distributivity of vectors: True since cx = x and cy = y. That means c(x+y)=cx+cy or any other distributive equation you can come up with all evaluates to x+y.

Associativity: x+x=y

Re: Prove V is not a vector space

Multiplicative identity: any c can be used here as the multiplicative identity.

Multiplicative associativity (c1c2)v1=c1(c2v1): am pretty convinced this is true because both of these evaluate right to v1.

Re: Prove V is not a vector space

Oh I think I got it

a=(x+y)+y=y+y

b=x+(y+y)=x+x

a not= b

WAIT NEVER MIND x+x=y and y+y =x so x+x=y+y

It turns out that it is

(a+b)y = ay+by

a,b are real numbers are thus so is a+b

y = y + y

y = x (contradicts distributivity)

Re: Prove V is not a vector space

Just a bit curious but how do I show if x and y have additive inverses or not. I guess both x and y could be their own additive inverse?

Oh I got it since x is the zero vector then for x since x+x = x(the zero vector) x must be its own additive inverse, which does exist because *there is a item in V for which x + x(additive inverse) = x(zero vector).* Likewise for y.

Re: Prove V is not a vector space

I'm a bit confused. Why does showing x=y prove that V isn't a vector space. Why doesn't it just prove that x and y are the same?

Re: Prove V is not a vector space

Because by hypothesis, V has exactly two vectors (according to the first line in your first post). If this is true, then it can't be a vector space.

If you decide to ignore that statement, then you arrive at a vector space which is isomorphic to the vector space containing only a zero element, or $\displaystyle V = \{0\}$.