Vector Spaces, is it or not? I don't understand when a set is determined to be a vector space or not, for example of given question is:
The set $\Bbb{R}^2$ with the usual scalar multiplication, but with addition defined by $[x,y] + [a,b] = [x+a+1, y+b]$ and with scalar multiplication defined by $r[x,y] = [xr+r-1, ry]$
I don't understand how to tell if this follows the properties of vector spaces. 
 A: A vector space typically has $10$ properties to check:


*

*Closure under addition. (If you add two vectors, you get another vector)

*Additive commutativity. (The order in which you add two vectors doesn't matter)

*Additive associativity. (Parentheses can be moved around freely)

*Zero vector. (There is vector that you add to any other vector and nothing changes)

*Additive inverses. (You have the negative of a vector.)

*Closure under scalar multiplication (If you multiply a vector by a scalar, you get another vector)

*Distributive law for multiplication ($c(u+v)=cu+cv$)

*Distributive law for multiplication ($(c+d)u=cu+cd$)

*Associative law for multiplication ($(cd)u=c(du)$)

*Identity for multiplication ($1u=u$).


To determine if this is a vector space, you need to check all of these properties.  Let's get started:


*

*$[x,y]+[a,b]=[x+a+1,y+b]$, when you add two vectors, you get a new vector in $\mathbb{R}^2$, so this is OK.  There would only be a possible problem if one were to restrict the part of $\mathbb{R}^2$ under consideration (which doesn't happen here.)

*$[x,y]+[a,b]=[x+a+1,y+b]$ and $[a,b]+[x,y]=[a+x+1,b+y]=[x+a+1,y+b]=[x,y]+[a,b]$ so additive commutativity holds.

*We compute $([x,y]+[a,b])+[w,z]=[x+a+1,y+b]+[w,z]=[x+a+1+w+1,y+b+z]=[x+a+w+2,y+b+z]$ and $[x,y]+([a,b]+[w,z])=[x,y]+[a+w+1,b+z]=[x+a+w+1+1,y+b+z]=[x+a+w+2,y+b+z]=([x,y]+[a,b])+[w,z]$ so associativity holds.

*We want a vector $[a,b]$ so that $[x,y]+[a,b]=[x,y]$.  Since $[x,y]+[a,b]=[x+a+1,y+b]$, we need $x+a+1=x$ (so $a=-1$ and $y+b=y$, so $b=0$).  Observe that $[x,y]+[-1,0]=[x-1+1,y+0]=[x,y]$ so $[-1,0]$ is the zero.

*We want a vector $[a,b]$ so that $[x,y]+[a,b]=[-1,0]$.  Since $[x,y]+[a,b]=[x+a+1,y+b]$, to get this to equal $[-1,0]$, we need $x+a+1=-1$ so $a=-2-x$ and for $y+b=0$, $b=-y$.  Therefore, the negative of $[x,y]$ is $[-2-x,-y]$.

*Since $r[x,y]=[xr+r-1,ry]$, which si a vector in $\mathbb{R}^2$ this is OK.  There could only be a problem if we restricted our attention to a part of $\mathbb{R}^2$.

*Consider $r([x,y]+[a,b])=r[x+a+1,y+b]=[r(x+a+1)+r-1,r(y+b)]=[rx+ra+2r-1,ry+rb]$.  On the other hand, $r[x,y]+r[a,b]=[rx+r-1,ry]+[ra+r-1,rb]=[rx+r-1+ra+r-1+1,ry+rb]=[rx+ra+2r-1,ry+rb]=r([x,y]+[a,b])$.

*Consider $(c+d)[x,y]=[(c+d)x+(c+d)-1,(c+d)y]=[cx+dx+c+d-1,cy+dy]$ and $c[x,y]+d[x,y]=[cx+c-1,cy]+[dx+d-1,dy]=[cx+c-1+dx+d-1+1,cy+dy]=[cx+dx+c+d-1,cy+dy]=(c+d)[x,y]$.

*Consider $(cd)[x,y]=[cdx+cd-1,cdy]$ and $c(d[x,y])=c[dx+d-1,dy]=[c(dx+d-1)+c-1,cdy]=[cdx+cd-1,cdy]=(cd)[x,y]$.

*Consider $1[x,y]=[1x+1-1,1y]=[x,y]$ so the multiplicative identity holds.
Therefore, all $10$ properties hold and this is a vector space (unless I made a mistake somewhere - very possible).  If anyone wants to make this prettier, please feel free to change this up, I got tired somewhere around property 7.
A: There are two mainly used ways to check if a set is a Vector Space.


*

*The first, and most tiresome way, is to check if the $8$ Axioms of Vector Spaces are properties of the given set.

*The second one, which is easier, is to simply  check if the given set is a subspace of a set that we already know is a Vector Space.



Since this set has its own definition of a addition and scalar multiplication I would suggest we check using the 8 Axioms. You can do this using the definition included in this link https://en.wikipedia.org/wiki/Vector_space#Definition
I hope I helped.
A: An easier (but slicker) approach than my other approach to this problem is to consider the map 
$$
\phi([x,y])=[x+1,y]
$$
into the usual copy of $\mathbb{R}^2$.  
Now, this map commuted with addition and scalar multiplication.  In fact,
\begin{align*}
\phi([x,y]+[a,b])&=\phi([x+a+1,y+b])=[x+a+2,y+b]\\
\phi([x,y])+\phi([a,b])&=[x+1,y]+[a+1,b]=[x+a+2,y+b]
\end{align*}
and
\begin{align*}
\phi(r[x,y])&=\phi([rx+r-1,ry])=[rx+r,ry]\\
r\phi([x,y])&=r[x+1,y]=[rx+r,ry]
\end{align*}
Therefore, this is really copy of $\mathbb{R}^2$, just in a disguised manner (a pullback of the standard operations).
A: The way to do this is to check whether $\Bbb R_\oplus,\otimes$ (swhere $\oplus$ is your redefined addition, and $\otimes$ is your redefined scalar multiplication), satisfies the definition of a vector space, which consists of eight axioms (plus closure under addition and multiplication, which are both obvious).
Commutativity of addition is OK: 
$$
\mathbf{a} \oplus \mathbf{b}=(a_1+b_1+1, a_2+b_2)=(b_1+a_1+1, b_2+a_2)=\mathbf{b} \oplus \mathbf{a}
$$
Associativity of addition is OK; this proof uses commutativity:
$$
(\mathbf{a} \oplus \mathbf{b}) \oplus \mathbf{c} = (a_1+b_1+1, a_2+b_2) \oplus (c_1,c_2) =
(a_1+b_1+c_1+2, (b_2+c_2)+a_2) = ((b_1+c_1+1)+a_1+1, (b_2+c_2)+a_2)=(\mathbf{b} \oplus \mathbf{c}) \oplus \mathbf{a}=\mathbf{a} \oplus (\mathbf{b} \oplus \mathbf{c})
$$
If $(i_1,i_2)$ is the identity element of $\oplus$, then for all $\mathbf{a}=(a_1,a_2)$, 
$$ (i_1+a_1+1,i_2+a_2)=(a_1,a_2) \implies \mathbf{I_\oplus} = (-1,0) 
$$
and this works for all  $\mathbf{a}$ so additive identity is OK.
Additive inverse:  Given 
$\mathbf{a}=(a_1,a_2)$,
$$
= (a_1,a_2)\oplus (-a_1-2 , -a_2) = (-1,0)=\mathbf{I_\oplus} =  (-a_1-2 , -a_2)\oplus (a_1,a_2)
$$
so additive inverse is satisfied, with $\ominus(a_1,a_2)=(-a_1-2 , -a_2)$.
Multiplicative consistency requires $r\otimes(s\otimes \mathbf{a}) = (rs)\otimes \mathbf{a}$ and that is OK:
$$
r\otimes(s\otimes \mathbf{a})=r\otimes(sa_1+s-1,sa_2)=(rsa_1+rs-r+r-1,rsa_2)\\
=(rsa_1+rs-1, rsa_2)
= (rs)\otimes \mathbf{a} 
$$
The identity element for $\otimes$ has to be the real number $1$ atnd that works:
$$
1\otimes(a_1,a_2) = (1\cdot a_1 - 1 + 1,a_2) = (a_1,a_2)
$$
Distributivity of scalar multiplication over vector addition is OK:
$$
r\otimes (\mathbf a \oplus \mathbf b) =  
r\otimes (a_1+b_1+1,a_2+b_2) 
\\ = (r(a_1+b_1+1)+r-1,r(a_2+b_2)) \\
(r\otimes \mathbf a) \oplus (r\otimes \mathbf b) 
= (ra_1+r-1,ra_2) \oplus (rb_1+r-1,rb_2)\\
= (ra_1+r-1+rb_1+r-1+1,ra_2+rb_2
\\= (r(a_1+b_1+1)+r-1,r(a_2+b_2)) = r\otimes (\mathbf a \oplus \mathbf b)
$$
Finally, distributivity of scalar-vector multiplication over scalar addition is OK:
$$
(r+s)\mathbf{a} = ( (r+s)a_1 + (r+s) -1, (r+s)a_2 )
\\ r\mathbf{a} \oplus s\mathbf{a} = 
(ra_1+r-1,ra_2) \oplus (sa_1+s-1,sa_2)
\\ = ((ra_1+r-1)+(sa_1+s-1)+1,ra_2+sa_2) =
 ( (r+s)a_1 + (r+s) -1, (r+s)a_2 )=(r+s)\mathbf{a} 
$$
Since this combination o operators meets all the defining axioms, the set under these operators forms a vector space.
