Find the negative of the elements $(p,q)$ in $V$ for these addition and scalar rules $V=\{(p,q)\mid p,q \in \mathbb R, p+q=1\} ,$ $F=\mathbb R$ , the addition and scalar multiplication is defined as
$(p,q)+(r,s)=(p+r-1,q+s)$
$λ(p,q)=(λp-λ+1,λq)$
so I've started linear algebra recently and im not sure how to approach this, if i could get a little hint on how to start this would be wonderful.
 A: If we just answer the question, it's not very interesting. All one requires of a real vector space like this is -where I will note $\oplus$ to distinguish from the classical addition in $\Bbb R^2$-, that one can add two vectors, and multiply a vector by a real number, so that all the axioms of a space vector are checked.
We then know -this is the first lesson on vector spaces- that : $$\forall (p,q)\in \color{blue}V, \exists!(r,s), (p,q)\oplus(r,s)=(r,s)\oplus(p,q)=e$$
where $e$ denotes the origin of $(\color{blue}V,\oplus,.)$
Let's start by determining $e=(r_0,s_0)$ :
$$\forall (p,q)\in \color{blue}V, (p,q)=(p,q)+e=(p+r_0-1,q+s_0)$$
So, $e=(1,0)$
Then, $$(p,q)\oplus(p',q')=(1,0)$$
$$\iff$$
$$\begin{cases} p+p'-1=1  \\ q+q'=0  \end{cases}$$
$$\iff $$
$$(p',q')=(2-p,-q)$$
So, $$\fbox{-(p,q)=(2-p,-q)}$$
Let's admit that so far, all this seems like a somewhat free game without any real interest. So let's try to shed some light on this. What do we do ?

*

*We work in the set $\Bbb R^2$, "the plane", where we recognize lines of more familiar equations $y=1-x$ and $y=-x$.



*Even when we have "started linear algebra recently", we know that $\color{blue}W:=\{(x,y)\in\Bbb R^2: x+y=0\}$ is a subspace of $(\Bbb R ^2,+,.)$. What we are doing here is called transporting the vector space structure from $W$ to $V$ by vector translation $t_{\color{red}u}$ of vector $\color{red}u=(-1,0)$. As far as I'm concerned, I was told about it from my first linear algebra lessons so it should be accessible, especially on an example like here.


*Let's take a very concrete example since it must be recognized all the same that it is a bit technical. (you have to follow on the drawing; nothing better to learn linear algebra than to look for fruitful links between linear algebra and elementary geometry, so drawings...)
Let $w=A=(3,-2)$. We will first associate the vector $a=A-O'$, with $O'=e=(1,0)$. So, $a=(3,-2)-(1,0)=(2,-2)=b$


*In a general way, let $(p,q), (r,s)\in \color{blue}V$ and $\lambda\in \Bbb R$. We first associate $(p-1,q)$ qnd $(r-1,s)$. Then
$$(p,q)\oplus (r,s):=((p-1),q)+(r-1,s))+(1,0)=(p+r-1,q+s)$$
$$\lambda.(p,q):=\lambda(p-1,q)+(1,0)=(\lambda p-\lambda+1,\lambda q)$$


*For example, back to our original exercise : the negative of the element $(p,q)$ of $\color{blue}V$ is $(p',q')$ such that $O'$ is the midpoint, i.e. $\frac12[(p,q)+(p'q')]=(1,0)\iff (p,q)+(p',q')=(2,0) \iff (p',q')=(2-p,-q)$.
So, my hint to start is to start everything upside down, starting with the subspace $\color{blue}W$ of real vector space $(\Bbb R^2,+,.)$. And with hard work it will be wonderful.
