Maximal minors of two matrix (II) Let us suppose that we have two $n\times m$ matrices $A$ and a $B$ with coefficients in an algebraically closed field of characteristic 0. Let us assume that $m>n$ and both of them have rank n. Furthermore, let us suppose that the $n\times n$ minors of both matrices are the same. Does there exist an m×m matrix C with determinant 1 such that $A\cdot C=B$.
If m=n, then the statement is true, because in this case A and B have the same determinant and are both invertible, so $C$ can be determined as $C=A^{−1}$.
Thank you for your time.
 A: Let's assume that the minors corresponding to the first $n$ rows and columns of $A$ and $B$ are non-zero. Then we can write $A,B$ in block form as $A = (A_1, A_2)$ and $B = (B_1,B_2)$ where $A_1,B_2 \in M_n(\mathbb{F})$ are square and invertible. Since $A$ has full rank and the first $n$ columns $A_1$ of $A$ span $\mathbb{F}^n$, one can express the rest of the columns $A_2$ uniquely in terms of linear combinations of the columns of $A_1$. Hence, there exists a (unique) $P$ such that $A_2 = A_1 P$. Similarly, there exists $Q$ such that $B_2 = B_1 Q$. Then
$$ (A_1, A_2) \begin{pmatrix} A_1^{-1} B_1 & A_1^{-1}B_1Q - P \\ 0 & I_{m-n} \end{pmatrix} = (A_1 A_1^{-1}B_1 + A_2 \cdot 0, A_1(A_1^{-1}B_1Q - P) + A_2I) \\
= (B_1, B_1Q - A_1P + A_2) = (B_1, B_2 - A_2 + A_2) = (B_1, B_2) = B $$
and the matrix 
$$  \begin{pmatrix} A_1^{-1} B_1 & A_1^{-1}B_1Q - P \\ 0 & I_{m-n} \end{pmatrix} $$
has determinant one since $\det(A_1) = \det(B_1)$.
In the general case, since $A$ and $B$ has full rank, you know that that there is some minor of $A$ and $B$ which has the same non-zero value (it might not be the leftmost) but you can replace $A$ and $B$ with $AP$ and $BP$ where $P$ is a permutation matrix which permutes the required columns so that they become the first $n$ columns and apply the argument to $AP$ and $BP$.
