Let $V=\mathbb{R}$. For $u,v\in V$ and $a\in\mathbb{R}$, define vector addition by $$u\boxplus v:=u+v+2$$ and scalar multiplication by $$a\boxdot u:=au+2a−2.$$ It can be shown that $(V,\boxplus,\boxdot)$ is a vector space over the scalar field $\mathbb{R}$. Find the following...
I have found the sum, scalar multiple and the zero vector. I don't know where the x is coming from, let alone how to find the additive inverse. Can someone assist me? Thanks.
This vector space is nice, never encountered such an example.
If you found that the zero vector is $-2$, the additive inverse is not too far away. You actually don't need to care about scalar multiplication for this part.
The problem of finding the additive inverse is to find a function $i:V\to V$ such that for every $u\in V$, you have $u\boxplus i(u)=e$, where $e$ is the zero vector of $(V,\boxplus,\boxdot)$.
So the first step is to find the zero vector, which you claim you had. Just to be sure, I show here how it is found: for all $u\in V$, we want $u\boxplus e=u$. This means for all $u\in \mathbb R$, we have $u+e+2=u$. We immediatly get $e=-2$, by choosing any $u$ (for instance it suffices to look at $u=0$).
Now we want to find the function $i$ which gives us the additive inverse of any vector of $V$. Given $u$, we need to find $i(u)$ such that $u\boxplus i(u)=e$. This last equation translates $u+i(u)+2=-2$. we solve the equation for $i(u)$, and we get $i(u)=-4-u$, for all $u\in V$. So this shows that the additive inverse in $V$ is given by the formula $i(u)=-4-u$. Notice that for instance $i(e)=-4-(-2)=-4+2=-2=e$, which is normal: zero is always its own inverse.
The additive inverse is defined when the identity element for the set is known like here +(u,v) = u + v + 2 so, we find +(u,x) = u where in x must be in Vspace so u+x+2 = u we get x = -2 hence, it is in Vspace. So now identity element is -2 so now finding additive inverse goes this way +(u,y) = -2 where y is additive inverse for u so by by performing the operation we get y = -4 - u .
