A lot of this falls back on my previous question. But I will state the details here again.
Let $W$ be a subspace of $\mathbb{C}^3$. Suppose that $\mathcal{A}=(\alpha_1,\alpha_2)$ and $\mathcal{B}=(\beta_1,\beta_2)$ both form an ordered basis for $W$. Let $\alpha=(0,i-1,i+1)$ be a vector in $W$.
Let $T: W \to W$ be the linear transformation $T(x_1\alpha_1+x_2\alpha_2)=(x_1-x_2)\alpha_1+(x_1+x_2)\alpha_2.$ Find the matrix representaiton ${}_{\mathcal{B}}[T]_{\mathcal{A}}$ and use it to find $T[(\alpha)]_\mathcal{B}$.
I'm not sure how to proceed with this problem. There are a few things bugging me about the notation The ${}_{\mathcal{B}}[T]_\mathcal{A}$ is giving me the impression that we are taking $[\alpha]_\mathcal{B}$ and then feeding it into $[T(\alpha)]_\mathcal{A}$.
Here's a nice approach: note that the first column of the matrix ${}_{\mathcal B}[T]_{\mathcal A}$ is given by $$ {}_{\mathcal B}[T]_{\mathcal A} \pmatrix{1\\0} = [T(1 \cdot \alpha_1 + 0 \cdot \alpha_2)]_{\mathcal B} = [\alpha_1 + \alpha_2]_{\mathcal B} = \\ [(1,0,i) + (1+i,1,-1)]_{\mathcal B} = [(2 + i,1,-1 + i)]_{\mathcal B} \\ [\color{red}{2} \cdot (1,1,0) + \color{red}{i}\cdot (1,i,1+i)]_{\mathcal B} = \\ [\color{red}{2} \cdot \beta_1 + \color{red}{i}\cdot \beta_2]_{\mathcal B} =\pmatrix{2 \\ i} $$ The coefficients $2,i$ either had to be calculated by solving the system $$ c_1 \cdot (1,1,0) + c_2 \cdot (1,i,1+i) = (2 + i,1,-1 + i) $$ for coefficients $c_1,c_2$, or could also be calculated using the transition matrix $U$ which you were meant to compute in the previous question.
All together, we now know that $$ {}_{\mathcal B}[T]_{\mathcal A} = \pmatrix{2 & ?\\i & ?} $$ You can calculate the second column similarly by computing ${}_{\mathcal B}[T]_{\mathcal A} \pmatrix{0\\1}$ in the method outlined above.