Linear Algebra: Determine Zero Vector

Solution 1:

I don't really understand the revised scalar multiplication operation. And the "addition operation" isn't commutative:

$$\mathbf{u} + \mathbf{v} = (u_1v_1,u_2/v_2) \neq \mathbf{v} + \mathbf{u} = (v_1u_1,v_2/u_2)$$

This doesn't sound like a consistent definition.

Additional details below.

The zero vector $\mathbf{0}$ is a unique member of the vector space $V$ such that:

1. Additive identity: $\forall \mathbf{a}\in V\;\;\;\mathbf{0}+\mathbf{a}= \mathbf{a}$
2. Scalar multiplication by zero: $\forall \mathbf{a}\in V\;\;0\mathbf{a}= \mathbf{0}$

So we need to find a vector $\mathbf{0}=(a_1,a_2) \in V$ that gives us:

(1) $(u_1a_1,u_2/a_2) = (u_1,u_2) = (a_1u_1,a_2/u_2)$ and

(2) $(u_1^0,0)=(a_1,a_2)$

$(2)$ uniquely identifies $\mathbf{0}=(1,0)$.

However, $(1,0)$ cannot serve as an additive inverse:

$$\mathbf{0} + \mathbf{u} = (u_1,0)\neq \mathbf{u} \neq \mathbf{u} + \mathbf{0} = (1,\infty)$$

If we fix scalar multiplication to be $k\mathbf{u} = (u_1^k,u_2^k)$ then scalar multiplication will be consistent with the additive identity.

Was there a typo? Also, if $u$ is supposed to be a positive vector