Matrix representation of linear map - returns a non-square matrix?

What thought about matrix representations is incorrect. Here is how you would go about finding the entries of the $2 \times 2$ matrix of $\psi$ relative to the basis $B$. Let $v_1,v_2$ denote the elements of $B$. We find that $$ \psi(v_1) = \pmatrix{0\\1\\-1} = 0 \cdot v_1 + 1 \cdot v_2. $$ Correspondingly, the first column of $M_B(\psi)$ is given by $(0,1)$. That is, we have $$ M_{B}(\psi) = \pmatrix{0&?\\1&?}. $$ The second column of $M_B(\psi)$ can be found be expressing $\psi(v_2)$ as a linear combination of $v_1$ and $v_2$. We find that $\psi(v_2) = (-1)\cdot v_1 + (-1)\cdot v_2$, so that the second column of $M_B(\psi)$ is $(-1,-1)$. Putting all that together, we find that the matrix is given by $$ M_B(\psi) = \pmatrix{0&-1\\1&-1}. $$