Equivalent matrices represent the same linear map in different bases It is clear from the change of basis formula that the matrices $A$ and $B$ representing the same linear map in different bases are equivalent: there exists invertible matrixes $Q$ and $P$ such that $A=Q^{-1}BP$.
My question is about the other way, I mean given two equivalent matrices $A$ and $B$ in $M_{n,p}(\mathbb K)$ why they must represent the same linear map in different bases ? thank you for your help!
 A: Clearly $B$ being an $m \times n$ matrix represents a transformation $\Bbb R^n \rightarrow \Bbb R^m$ where we take the standard basis in each space.  If $A = Q^{-1}BP$, then take a basis for $\Bbb R^n$ made up of the columns of $P$ and a basis for $\Bbb R^m$ made up of the columns of $Q$.  Then $A$ represents the transformation with respect to these new bases.
A: The matrix elements $\{\langle i|A|j\rangle,1\le i\le n,1\le j\le p\}$ of $A$ with respect to a given basis sets $\{\langle i|\}$ and $\{|j\rangle\}$, are the same as $\{\langle i|Q^{-1}BP|j\rangle,1\le i\le n,1\le j\le p\}$, which are the matrix elements of $B$ with respect to new basis sets given by $\{P|j\rangle,1\le j\le p\}$ and $\{\langle i|Q^{-1},1\le i\le n\}$.
A: This is just a simple generalization of similarity (where you consider only $n\times n$ matrices with the same basis in domain and range). Here $P$ is the change-of-basis matrix in the domain ($\Bbb K^p$) and $Q$ is the change-of-basis matrix in the range ($\Bbb K^n$).
