Matrix representation of linear map $T: \mathbb{C}^2 \to \mathbb{C}^2$. Where's the error? 
Let $B=\{(1, i), (-i, 2)\}$ be a basis of $\mathbb{C}^2$ and $T:\mathbb{C}^2 \to\mathbb{C}^2$ the linear map
$$T(x, y) = (x, 0)$$
Find a matrix representation of $T$ with respect to basis $B$.

Observe that $T(1, i) = (1, 0), T(-i, 2) = (-i, 0)$. Solving the systems
$$\begin{pmatrix} 1 & -i &|& 1 \\ i & 2 &|& 0 \end{pmatrix}$$
$$\begin{pmatrix} 1 & -i &|& -i \\ i & 2 &|& 0 \end{pmatrix}$$
we find
$$T(1, i) = 2b_1 -ib_2 \\ T(-i, 2) = -2ib_1-b_2$$

Calculations for the result above :
Let $\lambda_1, \lambda_2$ be the coefficients in the linear combination of $B$. We want to find the $\lambda_1, \lambda_2$ coefficients that map to $(1, 0)$ and $(-i, 0)$ respectively in the basis $B$.
From the first system we have $\lambda_1 = 1 + i\lambda2$ and $i(1+i\lambda_2)+2\lambda_2=0$. From this readily follows
$$\lambda_2 = -i$$
$$\lambda_1 = 1+i(-i)=1+1=2$$
From the second system we deduce $\lambda_1= -i+i\lambda_2$ and $i\lambda_1 + 2\lambda_2 = 0$. It follows
$$i(-i+i\lambda_2)+2\lambda_2=0 \iff 1+\lambda2=0 \iff \lambda_2=-1$$

Then it follows
$$\begin{pmatrix}2 & -2i \\ -i & -1 \end{pmatrix}$$
should be the required matrix. However, the matrix does not satisfy the desired property. Simply observe, for example, that multiplying such matrix by $(2, 3)$ returns $(4 -6i, -2i-3)$ and
$$(4-6i)b_1 + (-2i-3)b_2 = (2-3i, 0) \neq (2, 0)$$
The fact that the resulting value is $(2 -3i, 0)$ instead of $(2, 0)$ tells me I'm not erring by much $-$surely a miscalculation somewhere. However, after several attempts, I can't find where I'm going wrong. My guess is I'm confusing something when calculating with complex numbers, since I have never studied complex analysis in this is the second or third time in my life I deal with imaginary units.
Any help is appreciated.
 A: Your calculation of the matrix is correct, as is your algebra with complex numbers, but your verification is not correct. To check your work, you need to encode the vector $(2, 3)$ in the new basis before multiplying it by the matrix.
What your calculation winds up verifying is that the vector encoded by $(2, 3)$, which is the vector that is called $(2 - 3i, 2i+6)$ in the standard basis, maps to $(2i-3, 0)$. That is indeed true.
A: Your representation is correct, but you're not using it correctly. In the ordered basis $\langle(1,i),(-i,2)\rangle$ for $\Bbb C^2$, to say $T$ is represented by: $$\begin{pmatrix}2&-2i\\-i&-1\end{pmatrix}$$Is to say that if you give this matrix an input in terms of this basis, you get the image of $T$ again expressed in this basis. When you multiply by $(2,3)$, you are asking: "What is $T(2b_1+3b_2)=T(2-3i,2i+6)$?" When you get the result of $2-3i$, then you know things are working. $(2,3)$ did not represent $2+3i$ here.
A: Your matrix is correct. When you tried to check that it acts on $(2,3)$ correctly, you assumed that it should send $(2,3)\mapsto (2,0)$ but the rule $(x,y)\mapsto (x,0)$ was true only in the first basis.
