Find basis of $\ R^2 $ given transformation matrix $\ T: \mathbf R^2 \rightarrow \mathbf R^2 $ linear transformation $\ T(x,y) = (x+2y,y) $ and I need to find a basis $\ B$ so the matrice $\ [T]_B$ 
$$\ [T]_B = \begin{bmatrix} 1 & 0 \\ 2 & 1\end{bmatrix}$$
So i'm trying to find $\ B = \{ v_1, v_2 \} $ so $\ T(v_1) = 1 \cdot v_1 + 2 \cdot v_2, \ T(v_2) = 0 \cdot v_1 + 1 \cdot v_2 $
maybe define $\ v_1 =( x,y) , \ v_2 =  (s,t) $ and then define them using regular basis so $\ v_1 = x \cdot e_1 + y \cdot e_2, \ v_2 = s \cdot e_1 + t \cdot e_2 $
and then I tried $\ T(v_1) = T(x \cdot e_1 + y \cdot e_2) = x\cdot T(e_1) + y \cdot T(e_2)$ but it didn't get anywhere.
 A: HINT
We have that
$$\ [T] = \begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}$$
and we are looking for $M=[v_1\, v_2]$ such that
$$Tx=y \iff TMu=Mv\iff M^{-1}TMu=v$$
then
$$[T]_B= M^{-1}[T]M \iff M[T]_B= [T]M$$
A: It’s a good idea look for special features of the problem at hand before grinding away with a general method.  
Observe that the required matrix can be obtained from the matrix of $T$ relative to the standard basis by swapping its rows and columns. That is, $$\begin{bmatrix}1&0\\2&1\end{bmatrix} = \begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}1&2\\0&1\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}.$$ The permutation matrix is its own inverse, so this product can be interpreted as a change of basis operation that swaps the standard basis vectors.  
Alternatively, compare the equations that you’ve derived, $$T(\mathbf v_1)=\mathbf v_1+2\mathbf v_2 \\ T(\mathbf v_2)=\mathbf v_2$$ to the effect of $T$ on the standard basis vectors: $$T(\mathbf e_1) = \mathbf e_1 \\ T(\mathbf e_2) = 2\mathbf e_1+\mathbf e_2.$$ It should be fairly obvious that taking $\mathbf v_2=\mathbf e_1$ and $\mathbf v_1=\mathbf e_2$ will satisfy your equations. In fact, you can multiply both of these vectors by the same nonzero scalar to get another basis that works.
