Determining a basis of vector space with unusual addition and multiplications: $(a,b)+(c,d)=(ad+bc,bd) $ and $k * (a,b) =(kab^{k-1},b^k)$ Let $\mathbb W=\{(a,b) \in \mathbb R^2 \mid b>0\}$ and define addition by $(a,b)+(c,d)=(ad+bc,bd) $ and define scalar multiplication by $k * (a,b) =(kab^{k-1},b^k)$.

Find a basis for $\mathbb W$ and hence determine the dimension of $\mathbb W$.

Well for finding a basis and the dimension, I'm used to questions that give me an equation or some sort of definition to meet.  But for this question I have no idea where to begin. 
 A: Let's start by finding the zero vector $\mathbf{0} \in \mathbb{W}$. We need $\mathbf{0}+\mathbf{v} = \mathbf{v}$ for all vectors in the space. In components, let $\mathbf{0} = (a,b)$, and then we need $(a,b) + (x,y) = (ay+bx,by) = (x,y)$. We see that $b$ must be one and $a$ zero, so $\mathbf{0} = (0,1)$. Now, we will choose a nonzero vector and see if its scalar multiples span the space. $(0,2)$ will do. Its scalar multiples are $k(0,2) = (k*0*2^{k-1},2^k) = (0,2^k)$, which clearly is not the whole space. Now, pick another nonzero vector that is not a multiple of the first. $(1,1)$ will do. We look at the space spanned by our two vectors. $k(0,2) + l(1,1) = (0,2^k) + (l*1*1^{l-1},1^l) = (0,2^k) + (l,1) = (0*1+2^k*l,2^k) = (l*2^k,2^k)$
This is the whole space. To see this, for any vector in the space $(x,y)$, set $k = \log_2(y)$ and $l = \frac{x}{y}$. The space has dimension $2$.
Because all finite dimensional spaces over the same base field are isomorphic, we should be able to find an isomorphism from $\mathbb{W}$ to $\mathbb{R}^2$. Denote our isomorphism by $\phi$.
We need $\phi(k(0,2)) = \phi(0,2^k) =k\phi(0,2)$, and also that $\phi(l(1,1)) = \phi(l,1) = l\phi(1,1)$. This suggests setting $\phi(0,b) = (0,\log_2(b))$ and $\phi(a,1) = (a,0)$. Then 
$\phi(x,y) = \phi(\log_2(y)(0,2) + \frac{x}{y}(1,1)) =$ 
$\log_2(y)\phi(0,2) + \frac{x}{y}\phi(1,1) = \log_2(y)*(1,1)+\frac{x}{y}*(1,0) = (\log_2(y),\log_2(y))+(\frac{x}{y},0) =$ 
$(\log_2(y)+\frac{x}{y}, \log_2(y)).$ 
What vector maps to $\mathbf{e}_1$? We see it is $(1,1)$. How about $\mathbf{e}_2$? It's $(-2,2)$. 
