matrix of orthogonal projection with respect to the ordered basis. Orthogonal projection onto the line $y = 2x$ gives a linear transformation $T: R2 → R2$ such that 
$$T(1,2) = (1,2)$$ and $$T(−2,1) = (0,0)$$
Then the matrix of T with respect to the ordered basis 
$$B = \{(1, 2), (−2, 1)\}$$ 
is?
Attempt to solve:
$$\begin{bmatrix}
1 & -2\\ 
2 & 1
\end{bmatrix}*T=\begin{bmatrix}
1 & 0\\ 
2 & 0
\end{bmatrix}$$
solving for T: 
$$T=\begin{bmatrix}
1 & -2\\ 
2 & 1
\end{bmatrix} ^{-1} *\begin{bmatrix}
1 & 0\\ 
2 & 0
\end{bmatrix}$$
$$T=\begin{bmatrix}
1 & 0\\ 
0 & 0
\end{bmatrix}$$
when I check it geometrically (for projection), it always gives me the zeros for the second coordinate.
Can you kindly tell me what's going wrong with it?
Or my solution is completely incorrect? 
And what about getting T in the standard canonical basis?

Your answer is correct...the diagonal form of a projection matrix always has only 1's and $0$'s on the diagonal...if you think about it this it makes sense, since vectors in the projection space is perfectly preserved, and vectors orthogonal to it will vanish.
To get the matrix representation of $T$ with respect to the canonical basis you just use the familiar similarity relation to find a diagonal representation, and since you already have this diagonal representation and the ordered basis that "creates" it, you have \begin{equation} [T]_\epsilon=\begin{bmatrix}
1 & -2\\ 
2 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0\\ 
0 & 0
\end{bmatrix}\begin{bmatrix}
1 & -2\\ 
2 & 1
\end{bmatrix}^{-1}, \end{equation} where $\epsilon$ is the canonical basis
