I am trying to figure out the relationship between two elements from two dual bases of a vector space, i.e. say ${e_1,...,e_n}$ and ${f_1,...,f_n}$ are 2 bases of $V$ with ${w^1,...,w^n}$ and ${y^1,...,y^n}$ are their dual bases. Write $f_i = \sum_j A_{ij}e_j$. I want to express $y^i$ in terms of the $w^j$.
My attempt. Write $y^m = \sum_k B^{mk} w^k$. Then $$ \delta_i^m = y^m(f_i) = \sum_k B^{mk} w^k (\sum_j A_{ij} e_j) = \sum_k B^{mk} A_{ik} $$ which implies that $B = (A^{-1})^T$. So I guess I've reduced the problem down to expressing the entries of the transpose inverse of a matrix in terms of its entries, but I found this to be quite difficult to do. Can someone give a hint or offer a better way of approaching this? Thanks in advance!
It seems like you're looking for a closed-form expression for the entries of $B = (A^{-1})^T$ in terms of the entries of $A$. The most common way to accomplish this is to use the adjugate formula for the inverse of a matrix, $$ A^{-1} = \frac 1{\det(A)}\cdot \operatorname{adj(A)}. $$ Taking the transpose of both sides (noting that $(A^T)^{-1} = (A^{-1})^T$) yields $$ B = A^{-T} = \frac 1{\det(A)} \operatorname{cof}(A) $$ where $\operatorname{cof}(A)$ denotes the cofactor matrix of $A$ (see the linked article). To be a bit more explicit, we could write $$ B_{ij} = \frac{\det(A^{(ij)})}{\det(A)}, $$ where $A^{(ij)}$ denotes the matrix $A$ with its $i$th row and $j$th columns removed.
Note: the adjugate formula is rarely used as a basis for computation due to the prohibitive computational cost of determinants of large matrices. In practice, one would compute $A^{-1}$ using a standard method such as the Gauss-Jordan algorithm.