matrices that represent the same linear transf. If we have two  square matrices of the same order n over the field $F$  and the same rank, is there a linear transformation from $F^n$  to $F^n$ such that the two matrices are the representation of it in some basis (not necesary the same basis of domain and codomain) ?
 A: $\def\rank{\operatorname{rank}}\def\Mat{\operatorname{Mat}}\def\GL{\operatorname{GL}}\def\im{\operatorname{rng}}$Let $A \in \Mat_n(F)$ with $\rank A = k$. Let $a_1, \ldots, a_k$ be a basis of $\im A$, complete these vectors to a basis $a_1, \ldots, a_n$ of $F^n$ and let $S = (a_1\mid \cdots \mid a_n) \in\GL_n(F)$ (columnswise). Let $\alpha_{k+1}, \ldots, \alpha_{n}$ a basis of $\ker A$, choose $\alpha_i$ for $1 \le i \le k$ such that $A\alpha_i = a_i$. Then $\alpha_1, \ldots, \alpha_n$ is a basis of $F^n$ and set $T = (\alpha_1\mid\ldots\mid \alpha_n) \in \GL_n(F)$. Now define $C_k := S^{-1}AT$, for $1 \le i \le k$ we have
\[ C_ke_i = S^{-1}ATe_i = S^{-1}A\alpha_i = S^{-1}a_i = e_i \]
and for $k+1 \le i \le n$
\[ C_ke_i = S^{-1}ATe_i = S^{-1}A\alpha_i = 0 \]
So
\[ C_k = \begin{pmatrix} 1 & 0 & \cdots &  0\\
                        0 & 1 & \cdots &  0 \\
                        0 & 0 & \ddots &  \vdots \\
                        0 & 0 &        &  0\end{pmatrix} 
\]
with $k$ ones on the diagonal.
Now let $A, B \in \Mat_n(F)$ with $\rank A = \rank B = k$ be given. Find $S_1, T_1, S_2, T_2 \in \GL_n(F)$ with $S_1AT_1 = S_2BT_2 = C_k$. Hence in the bases given by the $T_i$ and $S_i$ both $A$ and $B$ represent a projection onto the first $k$ coordinates.
A: The answer is no.  For example if $F$ is a field with more than two elements, then the matrix $\begin{pmatrix} 1&0\\0&1\end{pmatrix}$ can represent only the trivial transformation, while $\begin{pmatrix} 2&0\\0&2\end{pmatrix}$ can represent only th transformation the double each vector. 
In general, two matrices $A,B$ represent the same linear transformation in different bases if and only if $A$ is similar to $B$, i.e. there exists an invertible matrix $M$ such that $A=M^{-1} B M$. Indeed, $M$ will be the transition matrix from one basis to another. 
