Let $L\in R^{k\times k}$ a Latin square matrix. Which is the most general form of $A\in R^{k\times k}$ such that $$ A^TLA=L' $$ with $L'$ another Latin square? Thanks!
Fabio
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2$\begingroup$ So you're allowing the entries of your Latin square to be any $k$ different real numbers, possibly different for $L'$ than for $L$? $\endgroup$– Robert IsraelCommented Aug 30, 2016 at 21:14
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I doubt that there's a very nice characterization in general.
Let $L$ be any $k \times k$ Latin square matrix (whose entries are $k$ distinct reals). Let $M$ be a $k \times k$ Latin square with symbolic entries. For $B = A^T L A$ to be a Latin square with the same pattern as $M$, what we need is $B_{i.j} = B_{i'.j'}$ if and only if $M_{i.j} = M_{i',j'}$. We might start by solving a set of $k(k-1)$ quadratic equations (e.g. taking one $(i',j')$ for each symbol) in the $k^2$ unknowns $a_{ij}$. We might hope that for a generic solution the entries of $B$ that are not required to be equal will not be equal. Of course, restricting to real solutions is an additional complication. In some cases, symmetry dictates there is no solution: if one of $L$ and $M$ is symmetric, the other must also be symmetric.
For example, I tried the case $k=3$ with $L = \pmatrix{1 & 2 & 0\cr 2 & 0 & 1\cr 0 & 1 & 2}$ and $M = \pmatrix{a & b & c\cr b & c & a\cr c & a & b\cr}$ (both symmetric). We get a set of $3$ equations in $9$ unknowns, which turns out to have Hilbert dimension $6$. One family of solutions is
$$ A = \pmatrix{a_{1,1} & a_{1,1} & a_{1,1}\cr a_{2,1} & a_{2,1} & a_{2,1}\cr -2 a_{1,1} & -2 a_{1,1} & -2 a_{1,1}\cr} $$ but this doesn't work as $A^T L A$ has all entries equal. One $5$-parameter family that does work is $$ \pmatrix{a_{1,1} & a_{1,2} & a_{1,3}\cr u/d & v/d & w/d\cr -2 a_{1,1} & a_{3,2} & a_{3,3}\cr} $$ where $$ \eqalign{u &= 72\,{a_{{1,1}}}^{2}a_{{1,2}}a_{{1,3}}+36\,{a_{{1,1}}}^{2}a_{{1,2}}a_{{ 3,3}}+36\,{a_{{1,1}}}^{2}a_{{1,3}}a_{{3,2}}+18\,{a_{{1,1}}}^{2}a_{{3,2 }}a_{{3,3}}-4\,a_{{1,1}}{a_{{1,2}}}^{3}+12\,a_{{1,1}}{a_{{1,2}}}^{2}a_ {{3,2}}+15\,a_{{1,1}}a_{{1,2}}{a_{{3,2}}}^{2}-4\,a_{{1,1}}{a_{{1,3}}}^ {3}+12\,a_{{1,1}}{a_{{1,3}}}^{2}a_{{3,3}}+15\,a_{{1,1}}a_{{1,3}}{a_{{3 ,3}}}^{2}+4\,a_{{1,1}}{a_{{3,2}}}^{3}+4\,a_{{1,1}}{a_{{3,3}}}^{3}+9\,{ a_{{3,3}}}^{2}{a_{{1,2}}}^{2}-18\,a_{{3,2}}a_{{1,3}}a_{{3,3}}a_{{1,2}} +9\,{a_{{3,2}}}^{2}{a_{{1,3}}}^{2} \cr v &= 36\,{a_{{1,1}}}^{2}{a_{{1,3}}}^{2}+36\,{a_{{1,1}}}^{2}a_{{1,3}}a_{{3,3 }}+9\,{a_{{1,1}}}^{2}{a_{{3,3}}}^{2}-9\,a_{{1,1}}{a_{{1,2}}}^{2}a_{{3, 3}}+9\,a_{{1,1}}a_{{1,2}}a_{{1,3}}a_{{3,2}}-9/2\,a_{{1,1}}a_{{1,2}}a_{ {3,2}}a_{{3,3}}+9/2\,a_{{1,1}}a_{{1,3}}{a_{{3,2}}}^{2}-2\,{a_{{1,2}}}^ {4}-2\,{a_{{1,2}}}^{3}a_{{3,2}}-9/2\,{a_{{1,2}}}^{2}{a_{{3,2}}}^{2}-2 \,a_{{1,2}}{a_{{1,3}}}^{3}-3\,a_{{1,2}}{a_{{1,3}}}^{2}a_{{3,3}}+3\,a_{ {1,2}}a_{{1,3}}{a_{{3,3}}}^{2}-4\,a_{{1,2}}{a_{{3,2}}}^{3}+2\,a_{{1,2} }{a_{{3,3}}}^{3}+{a_{{1,3}}}^{3}a_{{3,2}}-15/2\,{a_{{1,3}}}^{2}a_{{3,2 }}a_{{3,3}}-6\,a_{{1,3}}a_{{3,2}}{a_{{3,3}}}^{2}-{a_{{3,2}}}^{4}-a_{{3 ,2}}{a_{{3,3}}}^{3} \cr w &= 36\,{a_{{1,1}}}^{2}{a_{{1,2}}}^{2}+36\,{a_{{1,1}}}^{2}a_{{1,2}}a_{{3,2 }}+9\,{a_{{1,1}}}^{2}{a_{{3,2}}}^{2}+9\,a_{{1,1}}a_{{1,2}}a_{{1,3}}a_{ {3,3}}+9/2\,a_{{1,1}}a_{{1,2}}{a_{{3,3}}}^{2}-9\,a_{{1,1}}{a_{{1,3}}}^ {2}a_{{3,2}}-9/2\,a_{{1,1}}a_{{1,3}}a_{{3,2}}a_{{3,3}}-2\,{a_{{1,2}}}^ {3}a_{{1,3}}+{a_{{1,2}}}^{3}a_{{3,3}}-3\,{a_{{1,2}}}^{2}a_{{1,3}}a_{{3 ,2}}-15/2\,{a_{{1,2}}}^{2}a_{{3,2}}a_{{3,3}}+3\,a_{{1,2}}a_{{1,3}}{a_{ {3,2}}}^{2}-6\,a_{{1,2}}{a_{{3,2}}}^{2}a_{{3,3}}-2\,{a_{{1,3}}}^{4}-2 \,{a_{{1,3}}}^{3}a_{{3,3}}-9/2\,{a_{{1,3}}}^{2}{a_{{3,3}}}^{2}+2\,a_{{ 1,3}}{a_{{3,2}}}^{3}-4\,a_{{1,3}}{a_{{3,3}}}^{3}-{a_{{3,2}}}^{3}a_{{3, 3}}-{a_{{3,3}}}^{4} \cr d &= 8\,{a_{{1,2}}}^{3}+12\,{a_{{1,2}}}^{2}a_{{3,2}}+6\,a_{{1,2}}{a_{{3,2}} }^{2}+8\,{a_{{1,3}}}^{3}+12\,{a_{{1,3}}}^{2}a_{{3,3}}+6\,a_{{1,3}}{a_{ {3,3}}}^{2}+{a_{{3,2}}}^{3}+{a_{{3,3}}}^{3} \cr}$$ where some polynomials in the $a_{ij}$ must be nonzero for $B$ to have three distinct entries.
answered Aug 31, 2016 at 0:50
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$\begingroup$ Thanks for your answer! I was looking indeed for some simple characterization but could not find it. Here was my (not conclusive) reasoning: if L is a latin square I can write it as: $\sum_i^k P_i v_i$ for some set of permutations $P_i$ such that $\sum _i P_i =J$ the all-ones matrix and $v\in R^k$. Thus using a similar reasoning for $L'$ I can write $\sum_i^k A^TP_iA v_i=\sum_i \tilde{P}_ir_i$ for some set of permutations $\tilde{P}_i$ and $r\in R^k$. From this i wanted to deduce that $A^TP_iA $ must be a permutation and thus $A$ must be a permutation and $v$ proportional $r$... $\endgroup$– FabioCommented Aug 31, 2016 at 7:32