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? 


2. 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>



On a determinantal equality
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.


