# Maximal minors of two matrix

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 $$n>m$$ and both of them have rank m. Furthermore, let us suppose that the $$m\times m$$ minors of both matrices are the same. Does there exist an $$m\times m$$ 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 $$C=A^{-1}B$$.

Consider $$A,B \in M_3(\mathbb{C})$$ given by:
$$A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}, B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 1 & 1 \end{pmatrix}.$$
Note that for an invertible matrix $$C$$, the column space of $$AC$$ is the same as the column space of $$A$$ so if there exists an invertible matrix $$C$$ such that $$AC = B$$ it means that $$A,B$$ must have the same column space which is clearly not the case for the two matrices above.
• You need to state precisely your question. But regarding to the original question, note that there is an invertible matrix $C$ with determinant one such that $CA = B$. – levap Apr 20 at 11:08
• I mean that if in the statement of the original question we suppose that $m>n$ instead of $n>m$, then is it true? How you compute matrix C? – Samantha Smith Apr 20 at 11:19
• @SamanthaSmith: And what do you assume? That the rank of both $A,B$ is $n$ and you ask whether there an invertible $C$ with determinant one such that $AC = B$? – levap Apr 20 at 11:43
• Okey, the hypotheses are: A and B are n\times m matrices with m>n. Both of them have rank n, The $n\times n$ minors of both of them are the same. – Samantha Smith Apr 20 at 11:44