Consider the endomorphism $f: E \to E$ and the basis $B_E$ and $B'_E$ along with the matrices $A = M(f, B_E, B'_E)$ and $P = M(B_E, B'_E)$
How could i get the matrices: $B = M(f, B'_E)$, $C = M(f, B'_E, B_E)$ and $D = M(f, B_E, B'_E)$ ?
For example, if it was the case of a linear map like $f: E \to F$ and knowing $A = M(f, B_E, B_F)$,when considering other basis $B'_E$ and $B'_F$, and given $P = M(B'_E, B_E)$ and $Q = M(B'_F, B_F).$
I know how to get matrices like $B = M(f, B'_E, B'_F)$, $C = M(f, B'_E, B_F)$ and $D = M(f, B_E, B'_F)$ by writing their correspong matrix equations: $A \cdot X = Y$;$\hspace{5pt}$ $P \cdot X' = X$;$\hspace{5pt}$ $Q \cdot Y' = Y$;$\hspace{5pt}$ $B\cdot X'=Y'$;$\hspace{5pt}$ $C\cdot X'=Y$ and$\hspace{5pt}$ $D\cdot X=Y'$
Then it's just a question of isolating the variables to obtain the desired matrices. In the case of $B\cdot X'=Y' \longrightarrow B = Y'\cdot (X')^{-1} \longrightarrow B = Q^{-1}\cdot A\cdot P\cdot X'\cdot (X')^{-1} \longrightarrow B = Q^{-1}\cdot A\cdot P$
$C = A\cdot P$
$D = Q^{-1}\cdot A$
However, for the particular case of the endomorphism I presented I'm not sure how to proceed. Would the matrix equations for that case be like: $A \cdot X = X'$;$\hspace{5pt}$ $P \cdot X = X'$;$\hspace{5pt}$ $B \cdot X' = X'$; $\hspace{5pt}$ $C \cdot X' = X$ and $\hspace{5pt}$ $D \cdot X = X'$
If that were the case, then matrix B would equal the identity matrix, matrix C would equal $A^{-1}$ (or $P^{-1}$?) and matrix D would equal A.
This results seems kind of weird to me as if I were missing something or doing something wrong, what would be the approach to solve this?