Let $\alpha$ be a linear map from $\mathbb{R}^{m}$ to $\mathbb{R}^{n}$.
Let {$\mathbf{e_1}, . . . , \mathbf{e_m}$} be a basis of $\mathbb{R}^{m}$ and {$\mathbf{f_1}, . . . , \mathbf{f_n}$} a basis of $\mathbb{R}^{n}$. Explain how to represent $\alpha$ by a matrix $A$ relative to the given bases.
A second set of bases {$\mathbf{e_1'}, . . . , \mathbf{e_m'}$} and {$\mathbf{f_1'}, . . . , \mathbf{f_m'}$} is now used to represent $\alpha$ by a matrix $A′$ . Relate the elements of $A′$ to the elements of $A$.
So far: Let $\mathbf{x} \in \mathbb{R}^{m}$. Then $\mathbf{x} = x_{j}\mathbf{e_{j}}$ for some $x_{j}$. So $\alpha(\mathbf{x})_i = [\alpha(\mathbf{e_{j}})]_ix_j$.
We have $\alpha(\mathbf{e}_k)=\sum_{i=0}^n a_{ik}\mathbf e_i$, for some $a$, where the vectors on LHS are in $\mathbb R^m$; the vectors on RHS are in $\mathbb R^n$. $a_{ik}$ are the entries of the matrix $\mathbf A$. $k$ denotes the colomn in which the element lies.
For a second basis, $\alpha(\mathbf{e}_k')=\sum_{i=0}^n a_{ik}'\mathbf e_i'$ . This gives the entries of the matrix $\mathbf A'$ . To relate the two matrix, we need to relate the old and the new basis first in $\mathbb R^m$. Let $$ \mathbf e_i'=\sum_{k=0}^m b_{ki}\mathbf e_k. $$ Where $b_{ki}$ are the entries of another matrix $\mathbf B$. Then we can comfortably write $$ \mathbf A=\mathbf B \mathbf A' \mathbf B^{-1} $$ This can be done more beautifully if you use the tensor notation.