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 $C=A^{-1}B$.

Thank you for your time.


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.

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  • $\begingroup$ And what happens if m>n? $\endgroup$ – Samantha Smith Apr 20 at 10:59
  • $\begingroup$ 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 $CA = B$. $\endgroup$ – levap Apr 20 at 11:08
  • $\begingroup$ 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? $\endgroup$ – Samantha Smith Apr 20 at 11:19
  • $\begingroup$ @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 $AC = B$? $\endgroup$ – levap Apr 20 at 11:43
  • $\begingroup$ 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. $\endgroup$ – Samantha Smith Apr 20 at 11:44

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