Proving something is a vector space using pre-defined properties

Question

The potential vector space is defined as the set $(x,y)$ of all positive real numbers.

I have been given a question where we define the addition of two elements in the potential vector space as:

$$(x_1, y_1) + (x_2, y_2) = (x_1x_2, y_1y_2)$$

I am having trouble understanding how to prove the axiom of zero addition: my attempt has been: $(x_1, y_1) + (0, 0) = (x_1*0, y_1*0) = (0, 0)$ which can't possibly be correct since the axiom states

$$(x, y) + (0, 0) = (x, y)$$

Can anyone let me know what I am missing?

Answer 1

The axiom of zero addition states that there exists SOME element $e$ in the vector space such that $v+e=v$. Under conventional addition, the zero element is as you stated. When you change how addition is defined, you will more than likely see a new zero element.

Under vector addition as you defined it, the zero vector is $e=(1,1)$. Take any vector $v=(x,y)$

Then $v+e=(x,y)+(1,1)=(x*1,y*1)=(x,y)$

So you still have a zero vector, it just may not be the conventional one.

Answer 2

If we assume for a second that the operation does in fact provide a vector space, then we can determine the zero element (additive identity). $$(x_1, y_1) + (1, 1) = (x_1\cdot 1, y_1 \cdot 1) = (x_1, y_1).$$

Ok so that works. Now let's investigate things a little more closely to determine if it is in fact a vector space. Ok so we have closure under $+$ as the positive reals are closed under multiplication in $\mathbb{R}$. The $+$ operation also inherits the commutativity and associativity properties from the properties of multiplication in $\mathbb{R}.$ We also have the additive inverse property as $$\forall(x_1, x_2) \in \mathbb{R^+} \times \mathbb{R}^+ = V, \quad (x_1, y_1) + (\frac{1}{x_1}, \frac{1}{y_1}) = (x_1\frac{1}{x_1}, y_1 \frac{1}{y_1}) = (1, 1) = Id_+.$$ Now there is still is the question of how to define the scalar multiplication operation, as traditional scaling over $\mathbb{R}$ will not give you closure.

Following the procedure in this link: http://www.math.uconn.edu/~glaz/math2210f16/Section%20Handouts/sec4_1.pdf

We have that if we define $\lambda(x_1, y_1) := (x_{1}^{\lambda}, y_1^{\lambda}) \quad \forall \lambda \in \mathbb{R}$. We have closure under scalar multiplication as the exponentiation preserves positivity in $\mathbb{R}$. We have distributivity as $$\lambda(\alpha + \beta) = \lambda (\alpha_1 \beta_1, \alpha_2 \beta_2)= ((\alpha_1 \beta_1)^\lambda, (\alpha_2, \beta_2)^\lambda ) = (\alpha_1^\lambda \beta_1^\lambda, \alpha_2^\lambda \beta_2^\lambda ) = \lambda \alpha + \lambda \beta \quad \forall \lambda \in \mathbb{R} \quad \alpha, \beta \in V \\ (\xi + \eta)\alpha = (\alpha_1^{\xi + \eta},\alpha_2^{\xi + \eta})= (\alpha_1^\xi \alpha_1^\eta,\alpha_2^\xi \alpha_2^\eta) = \xi \alpha + \eta \alpha \quad \forall \xi, \eta \in \mathbb{R}, \quad \alpha \in V$$ We have associativity over scalar multiplication as $$ \xi(\eta\alpha) = \xi (\alpha_1^\eta, \alpha_2^\eta) = ((\alpha_1^\eta)^\xi, (\alpha_2^\eta)^\xi) =(\alpha_1^{\xi \eta}, \alpha_2^{\xi \eta}) = (\xi\eta)\alpha$$ with the identity property $$1\alpha = 1(\alpha_1, \alpha_2) = (\alpha_1^1, \alpha_2^1) = \alpha$$ Thus finally, we can say that we have a vector space $V = \mathbb{R}^+ \times \mathbb{R}^+.$