rank($A$)=rank($B^{-1}A$) Suppose $A$ is a matrix of dimension $n\times m$ and $B$ is nonsingular of dimension $n\times n$. It is easy to prove that rank($A$)=rank($B^{-1}A$) if $A$ is full rank. But, is the statement true in general, i.e., when rank($A$) $=r\leq m$? 
 A: Hint  Let simply $M=B^{-1}$ which is nonsingular. Suppose that $\operatorname{rank}(A)=r$ and let $c_{1},\cdots,c_{n}$ the $n$ columns of $A$ and $c_{i_1},\cdots,c_{i_r}$ the $r$ linearly independent columns of $A$. Then the $Mc_1,\ldots,Mc_n$ are the $n$ columns of $MA$ and it's easy to see (using definition) that $Mc_{i_1},\ldots,Mc_{i_r}$ are linearly independent.
A: $\operatorname{rank}A$ is the dimension of the image of the linear map $f_A:K^m\longrightarrow K^n$ associated to $A$ and, as $B^{-1}$ is associated to an automorphism of $K^n$ since $B$ and $B^{-1}$ are non singular, it is also the dimension of the image of the composition $f_{B^{-1}}\circ f_A$, which is the linear map associated to the product $B^{-1}A$. So we have the chain of equalities
$$\operatorname{rank}A=\dim\operatorname{Im}f_A=\dim\operatorname{Im}f_{B^{-1}A}=\operatorname{rank}(B^{-1}A).$$
A: Note that the $j$-th column of a matrix $A$ is equal to the product $Ae_j$ with $e_j$ the $j$-th canonical vector. Hence, if the columns $i_1,  \dots, i_k$ of $A$ are linearly dependent, there exist sclars $a_1,\dots,a_k$ such that
$$
a_1Ae_{i_1} + \dots a_kAe_{i_k} = 0. \tag{1}
$$
Multiplying by $B$ to the right, one obtains
$$
a_1B^{-1}Ae_{i_1} + \dots a_kB^{-1}Ae_{i_k} = 0, \tag{2}
$$
and so the same (number of) columns are linearly dependent for $B^{-1}A$. Conversely, if  $B^{-1}A$ has some linearly dependent columns then we are in the situation of $(2)$, and right multiplying by $B$ we obtain $(1)$.
In particular, this shows that the minimum number $d$ of linearly dependent columns of $A$ and $B^{-1}A$ is the same, and this number determines the rank. If $d = 0$ then both have full rank, and if not, they have rank $d-1$ by the very definition of $d$.
