Vector space proving/disproving

Question

Determine whether this is a (real) vector space:

The set of $\mathbb R^2$, with addition defined by $([x, y] + [a,b]) = [x + a + 1, y+b]$ and with scalar multiplication defined by $r[x, y] = [rx + r - 1, ry]$

This is a vector-space I just don't get how. By the axiom were $u$ is a vector

$$u + 0 = u$$

$$[x, y] + [0,0] = [x+1, y] \neq [x, y]$$

Am I missing something?

UNLESS $[-1, 0]$ is considered a zero vector which I don't get how, then this would make sense $$[x, y] + [-1, 0] = [x, y]$$

Could someone explain? thanks.

Answer 1

Let the additive unit be $[u, v]$. If $[x, y] + [u, v] = [x, y]$, then expanding the addition, we get $[x + u + 1, y + v] = [x, y]$. Solving the equations $x+u+1 = x$ and $y + v = y$ gives us $[u, v] = [-1, 0]$.

In general, a set can have multiple vector space structures with possibly different units and additions. Just because $[0, 0]$ is the unit of the standard addition on $\mathbb{R}^2$, doesn't mean that it has to be the unit of any other addition, as the above example shows.

Answer 2

The additive identity , $\vec{0}$ is the element that if it is being added to any element, $x$, we obtain $x$ again.

$$[x, y]+ [-1, 0]= [x+ (-1) + 1, y+0] = [x, y]$$

and

$$[-1, 0] + [x, y] = [(-1)+ x + 1, 0+y] = [x, y].$$

Hence the additive identity is indeed $[-1, 0]$.