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Let $M,N$ be two $n\times m$ matrices with $n\leq m$ and coefficients in an algebraically closed field of characteristic zero $K$, both of full rank $n$.

Do there exist two upper triangular matrices $A\in SL(n)$ and $B\in SL(m)$ such that $A\cdot M \cdot B^{T} = \lambda N$ for $\lambda\in K\setminus\{0\}$ ?

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  • $\begingroup$ What is the ground field? $\endgroup$ – YCor Nov 4 '17 at 16:34
  • $\begingroup$ An algebraically closed field of characteristic zero $\endgroup$ – J. Ross Nov 4 '17 at 16:35
  • $\begingroup$ If $m=n$, clearly it's true if and only if they have the same determinant. $\endgroup$ – YCor Nov 4 '17 at 16:37
  • $\begingroup$ Sorry, the question was not precisely stated. I am looking at matrices modulo multiplication by non-zero scalars. $\endgroup$ – J. Ross Nov 4 '17 at 16:46
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    $\begingroup$ Still it is not so clear to me. Have you taken into account that $A$ and $B$ are upper triangular? $\endgroup$ – J. Ross Nov 4 '17 at 17:33
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I think the answer to your question is negative. Consider for instance $$M = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ Assume that there exist $$A = \begin{pmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{pmatrix}, \quad B = \begin{pmatrix} b_{11} & b_{12} \\ 0 & b_{22} \end{pmatrix}$$ with $\det(A) = \det(B) = 1$ and such that $$A\cdot M\cdot B^{T} = N$$ Then $$A\cdot B^{T} = \begin{pmatrix} a_{11}b_{11}+a_{12}b_{12} & a_{12}b_{22}\\ a_{22}b_{12} & a_{22}b_{22} \end{pmatrix} = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$ and hence either $a_{22} = 0$ or $b_{22}=0$ which contradict $\det(A) = \det(B) = 1$.

More generally, your action stabilizes the locus of matrices of the form $\begin{pmatrix} m_{11} & m_{12}\\ m_{21} & 0 \end{pmatrix}$.

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  • $\begingroup$ I've corrected the product display. Btw even with the relaxed conditions (no condition on determinant and up to scalar $\lambda$) there is no solution: upper right and lower left corners nonzero and lower right zero cannot simultaneously happen. $\endgroup$ – მამუკა ჯიბლაძე Nov 4 '17 at 19:44
  • $\begingroup$ Sure, you are right. $\endgroup$ – F_L Nov 4 '17 at 19:50

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