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In my study, I come across the following curious equality, which I do not know a proof yet (so I am asking it here).

Let $k$, $l\in \Bbb Z$ be fixed, $m$ --- the size of the below matrix $M$ --- is also fixed, and let $a_i$ be independent variables. We define $M$ by the formula: the $(i,j)$-th entry of $M$ is $a_{ik+jl}$.

Example: $k=1$, $l=2$, $m=3$, we have

$M= \left( \begin{array} {a} a_3 & a_5 & a_7 \\ a_4 & a_6 & a_8 \\ a_5 & a_7 & a_9\end{array} \right)$

Let, further, a sequence of integer numbers $\lambda=(c_1, \dots, c_m)$ be fixed. We denote by $S$ the set of all permutations of $c_1, \dots, c_m$ (the cardinality of $S$ can be less than $m!$, because some of $c_i$ can be equal).

Let $\sigma=(d_1,\dots, d_m)$ be an element of $S$. We define a matrix $M_{r,\sigma}$ ($r$ means rows) as follows: its $(i,j)$-th entry is $a_{ik+jl+d_i}$, i.e. the indices of all $a_*$ of the first row of $M$ are increased at $d_1$, the indices of all $a_*$ of the second row of $M$ are increased at $d_2$, etc.

Analogically, we define a matrix $M_{c,\sigma}$ ($c$ means columns) as follows: its $(i,j)$-th entry is $a_{ik+jl+d_j}$, i.e. the indices of all $a_*$ of the first column of $M$ are increased at $d_1$, the indices of all $a_*$ of the second column of $M$ are increased at $d_2$, etc.

Example: let $\lambda=(1,0,0)$. We have: $S$ consists of 3 elements $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. For example, for $\sigma=(0,1,0)$ we have $M_{c,\sigma}=\left( \begin{array} {a} a_3 & a_6 & a_7 \\ a_4 & a_7 & a_8 \\ a_5 & a_8 & a_9\end{array} \right)$. (the indices of $a_*$ of the second column increase at 1, the indices of $a_*$ of the first and third columns remain the same).

Proposition. For all $k$, $l$, $m$, $\lambda$ we have $$\sum_{\sigma\in S} |M_{r,\sigma}| = \sum_{\sigma\in S} |M_{c,\sigma}|$$

Example. For the above $k$, $l$, $m$, $\lambda$ we have: matrices $M_{r,\sigma}$ have two equal rows for $\sigma=(1,0,0)$, $(0,1,0)$, hence the corresponding determinants are 0, and for this case the proposition becomes:

$$\left| \begin{array} {a}a_3 & a_5 & a_7 \\ a_4 & a_6 & a_8 \\ a_6 & a_8 & a_{10} \end{array} \right| = \left| \begin{array} {a} a_4 & a_5 & a_7 \\ a_5 & a_6 & a_8 \\ a_6 & a_7 & a_9\end{array} \right| +\left| \begin{array} {a} a_3 & a_6 & a_7 \\ a_4 & a_7 & a_8 \\ a_5 & a_8 & a_9\end{array} \right| + \left| \begin{array} {a} a_3 & a_5 & a_8 \\ a_4 & a_6 & a_9 \\ a_5 & a_7 & a_{10}\end{array} \right| $$

Question. 1. Is the proposition really true for all $k$, $l$, $m$, $\lambda$? Maybe they must satisfy some conditions?

  1. Is it already known, what is a reference?

Origin of the question: Resultantal varieties related to zeroes of L-functions of Carlitz modules http://arxiv.org/pdf/1205.2900.pdf

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  • $\begingroup$ Nice question! The answer to 1 is "yes", and you can even replace the $a_{ik+jl+r}$ by the less awkward $b_{i,j,r}$, where $\left(b_{i,j,r}\right)$ is a family indexed by three integers. I'll write up a proof shortly. No idea about 2, but I remember seeing something remotely similar in Muir's 5-volume history. $\endgroup$ – darij grinberg Jan 7 '16 at 22:44
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Let me prove a slightly more general claim.

Theorem 1. Let $\mathbb{K}$ be a commutative ring. Let $R$ be a set. Let $n\in\mathbb{N}$ (where $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $). Set $\left[ n\right] =\left\{ 1,2,\ldots,n\right\} $. For every $i\in\left[ n\right] $, $j\in\left[ n\right] $ and $r\in R$, let $b_{i,j,r}$ be an element of $\mathbb{K}$.

Let $t\in R^{n}$ be an $n$-tuple of elements of $R$. Let $T$ be the set of all permutations of this $n$-tuple $t$. For every $n$-tuple $d\in R^{n}$ and every $i\in\left[ n\right] $, we shall write $d_{i}$ for the $i$-th entry of $d$ (so that $d=\left( d_{1},d_{2},\ldots,d_{n}\right) $). Then,

$\sum\limits_{d\in T}\det\left( \left( b_{i,j,d_{i}}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) =\sum\limits_{d\in T}\det\left( \left( b_{i,j,d_{j} }\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) $.

This Theorem generalizes your statement (namely, it suffices to apply it to $b_{i,j,r}=a_{ik+jl+r}$). Actually, it does not take it very far, as it can in turn be derived from it by setting $k$ and $l/k$ large enough; however, it is the kind of generalization that is worth making even if you do not need the generality, as it pulls things apart that don't belong together.

Proof of Theorem 1. We let $S_{n}$ denote the symmetric group on $n$ elements (i.e., the group of all permutations of $\left[ n\right] $). For every $\tau\in S_{n}$, we let $\left( -1\right) ^{\tau}$ denote the signum of $\tau$. For every $d\in T$, we have

$\det\left( \left( b_{i,j,d_{i}}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) =\sum\limits_{\tau\in S_{n}}\left( -1\right) ^{\tau}\prod\limits_{i=1} ^{n}b_{i,\tau\left( i\right) ,d_{\tau\left( i\right) }}$

(by the definition of a determinant). Summing this equation over all $d\in T$, we obtain

$\sum\limits_{d\in T}\det\left( \left( b_{i,j,d_{i}}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) $

$=\sum\limits_{d\in T}\sum\limits_{\tau\in S_{n}}\left( -1\right) ^{\tau}\prod\limits_{i=1} ^{n}b_{i,\tau\left( i\right) ,d_{i}}$

(1) $=\sum\limits_{\tau\in S_{n}}\left( -1\right) ^{\tau}\sum\limits_{d\in T} \prod\limits_{i=1}^{n}b_{i,\tau\left( i\right) ,d_{i}}$.

On the other hand, for every $d\in T$, we have

$\det\left( \left( b_{i,j,d_{j}}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) =\sum\limits_{\tau\in S_{n}}\left( -1\right) ^{\tau}\prod\limits_{i=1} ^{n}b_{i,\tau\left( i\right) ,d_{\tau\left( i\right) }}$

(by the definition of a determinant). Summing this equation over all $d\in T$, we obtain

$\sum\limits_{d\in T}\det\left( \left( b_{i,j,d_{j}}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) $

$=\sum\limits_{d\in T}\sum\limits_{\tau\in S_{n}}\left( -1\right) ^{\tau}\prod\limits_{i=1} ^{n}b_{i,\tau\left( i\right) ,d_{\tau\left( i\right) }}$

(2) $=\sum\limits_{\tau\in S_{n}}\left( -1\right) ^{\tau}\sum\limits_{d\in T} \prod\limits_{i=1}^{n}b_{i,\tau\left( i\right) ,d_{\tau\left( i\right) }}$.

Now, fix $\tau\in S_{n}$. The set $T$ consists of all permutations of $t$; thus, every permutation of every $n$-tuple in $T$ will again be an $n$-tuple in $T$. Thus, the map

$T\rightarrow T,\ d\mapsto\left( d_{\tau\left( 1\right) },d_{\tau\left( 2\right) },\ldots,d_{\tau\left( n\right) }\right) $

is well-defined. This map is furthermore a bijection (indeed, the inverse map is given by $T\rightarrow T,\ d\mapsto\left( d_{\tau^{-1}\left( 1\right) },d_{\tau^{-1}\left( 2\right) },\ldots,d_{\tau^{-1}\left( n\right) }\right) $). Hence, we can substitute $\left( d_{\tau\left( 1\right) },d_{\tau\left( 2\right) },\ldots,d_{\tau\left( n\right) }\right) $ for $d$ in the sum $\sum\limits_{d\in T}\prod\limits_{i=1}^{n}b_{i,\tau\left( i\right) ,d_{i} }$. As a consequence, we obtain

(3) $\sum\limits_{d\in T}\prod\limits_{i=1}^{n}b_{i,\tau\left( i\right) ,d_{i}} =\sum\limits_{d\in T}\prod\limits_{i=1}^{n}b_{i,\tau\left( i\right) ,d_{\tau\left( i\right) }}$.

Now, let us forget that we fixed $\tau$. We thus have shown that (3) holds for every $\tau\in S_{n}$. Now, (1) becomes

$\sum\limits_{d\in T}\det\left( \left( b_{i,j,d_{i}}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) $

$=\sum\limits_{\tau\in S_{n}}\left( -1\right) ^{\tau}\sum\limits_{d\in T}\prod\limits_{i=1} ^{n}b_{i,\tau\left( i\right) ,d_{i}}$

$=\sum\limits_{\tau\in S_{n}}\left( -1\right) ^{\tau}\sum\limits_{d\in T}\prod\limits_{i=1} ^{n}b_{i,\tau\left( i\right) ,d_{\tau\left( i\right) }}$ (by (3))

$=\sum\limits_{d\in T}\det\left( \left( b_{i,j,d_{j}}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) $ (by (2)).

This proves Theorem 1.

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