I have solved the following question. I don't understand the choice of notation in the question. Can someone explain to me how the author assigns the superscript and lower script ?
For completeness here is my solution:
$L^{\wedge}(\bar{\alpha}^i)(e_j) = \Sigma_{k} b_k^i \alpha^k(e_j) = \Sigma_k b_k^i \delta_j^k = b_j^i$
On the other hand: $L^{\wedge}(\bar{\alpha}^i)(e_j) = \bar{\alpha}^i(L(e_j)) = \bar{\alpha}^i(\Sigma_k a_j^k \bar{e}_k) = \Sigma_k a_j^k \bar{\alpha}^i(e_k) = \Sigma_k a_j^k \delta_k^i(e_k) = a_j^i$.
You are correct, except for the extra $(e_k)$ at the last step.
There are two points with the upper and lower indices. One is that formulas should be compatible with the Einstein summation convention — so that you sum over an index that appears once up and once down. The other is that the position of the index keeps track of whether you're dealing with a contravariant object (vector or multi-vector) or covariant object (dual vector, etc.). Of course, the basis vectors of a vector will have a lower index ($e_j$) and the basis vectors of a dual vector will have an upper index ($\alpha^j$), so the corresponding coordinates are the reverse — the coordinates of a vector have upper indices and the coordinates of a dual vector have lower indices.
Note that a linear map maps vectors to vectors, so it can be written as a tensor of type $(1,1)$. The linear map $T$ with matrix $(a^i_j)$ mapping $V$ to $V$, using basis $\{e_i\}$ on both sides, can be represented as $T = \sum\limits_{i,j} a^i_j e_i\otimes \alpha^j$ (where $\{\alpha^j\}$ is the dual basis for $V^\vee$). Notice that this is equivalent to writing $T(e_j) = \sum\limits_i a^i_j e_i$.