I'm currently working on the following problem:
Let $T$ be the linear operator on $\mathbb{C}^2$ defined by $T(x_1,x_2)=(x_1,0)$. Let $B$ be the standard ordered basis for $\mathbb{C}^2$ and let $B'={(1,i), (-i, 2)}$. What is the matrix of $T$ relative to the pair $B, B'$?
Now I had no idea what the writer meant by "relative to the pair $B, B'$", so I figured I'd basically make a guess and go from there. My solution became the following:
Note that $$[T]_B=\begin{bmatrix}1&0\\0&0\end{bmatrix}$$ Let $P$ be the matrix which describes $B'$, then $$P=\begin{bmatrix}1&-i\\i&2\end{bmatrix}$$ We're now interested in finding $P^{-1}$, to that end we'll use Gauss-Jordan elimination to get: $$\begin{bmatrix}1&-i&&1&0\\i&2&&0&1\end{bmatrix} \rightarrow \begin{bmatrix}1&-i&&1&0\\0&1&&-i&1\end{bmatrix} \rightarrow \begin{bmatrix}1&0&&2&i\\0&1&&-i&1\end{bmatrix}$$ Which gives us $$P^{-1}=\begin{bmatrix}2&i\\-i&1\end{bmatrix}$$ Then $$[T]_{B'}=P^{-1}[T]_BP=P^{-1}\begin{bmatrix}1&-i\\0&0\end{bmatrix}=\begin{bmatrix}2&-2i\\-i&-1\end{bmatrix}$$
My first question of course, is whether this is the solution the exercise was asking about? Or if now, what should I be thinking about instead?
To calculate the matrix of your operator with respect to the bases $B,B'$ (with $B$ used for the domain and $B'$ used for the codomain) you need to calculate $T(e_i)$ for each of the basis elements of $B$ and represent the result in terms of the elements of $B'$. Then, put the resulting coefficients as columns of the matrix. In your case, we have
$$ T(1,0) = (1,0) = 2(1,i) - i(-i,2),\\ T(0,1) = (0,0) = 0(1,i) + 0(-i,2) $$
so
$$ [T]_{B'}^{B} = \begin{pmatrix} 2 & 0 \\ -i & 0 \end{pmatrix}. $$