Determinantal symmetry: proof requested












8












$begingroup$


Consider the determinantal function
$$F(a,b,c):=detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}.$$
I would like to ask:




QUESTION. Can you provide an argument, combinatorial or otherwise, to prove the symmetry
$$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$
without actually computing the its values? I know how to justify this based on direct evaluation.




Example. The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or
$$detbegin{pmatrix} 10&15&21&28\20&35&56&84\35&70&126&210\56&126&252&462
end{pmatrix}=
detbegin{pmatrix} 15&35&70\21&56&126\28&84&210
end{pmatrix}=detbegin{pmatrix} 35&56\70&126
end{pmatrix}.$$






co.combinatorics linear-algebra determinants







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Did you try to express this as the number of tuples of non-intersecting lattice paths?
    $endgroup$
    – Martin Rubey
    9 hours ago








  • 2




    $begingroup$
    The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
    $endgroup$
    – Mark Wildon
    7 hours ago








  • 1




    $begingroup$
    Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
    $endgroup$
    – Gerhard Paseman
    7 hours ago
















8












$begingroup$


Consider the determinantal function
$$F(a,b,c):=detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}.$$
I would like to ask:




QUESTION. Can you provide an argument, combinatorial or otherwise, to prove the symmetry
$$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$
without actually computing the its values? I know how to justify this based on direct evaluation.




Example. The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or
$$detbegin{pmatrix} 10&15&21&28\20&35&56&84\35&70&126&210\56&126&252&462
end{pmatrix}=
detbegin{pmatrix} 15&35&70\21&56&126\28&84&210
end{pmatrix}=detbegin{pmatrix} 35&56\70&126
end{pmatrix}.$$






co.combinatorics linear-algebra determinants







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Did you try to express this as the number of tuples of non-intersecting lattice paths?
    $endgroup$
    – Martin Rubey
    9 hours ago








  • 2




    $begingroup$
    The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
    $endgroup$
    – Mark Wildon
    7 hours ago








  • 1




    $begingroup$
    Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
    $endgroup$
    – Gerhard Paseman
    7 hours ago














8












8








8


2



$begingroup$


Consider the determinantal function
$$F(a,b,c):=detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}.$$
I would like to ask:




QUESTION. Can you provide an argument, combinatorial or otherwise, to prove the symmetry
$$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$
without actually computing the its values? I know how to justify this based on direct evaluation.




Example. The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or
$$detbegin{pmatrix} 10&15&21&28\20&35&56&84\35&70&126&210\56&126&252&462
end{pmatrix}=
detbegin{pmatrix} 15&35&70\21&56&126\28&84&210
end{pmatrix}=detbegin{pmatrix} 35&56\70&126
end{pmatrix}.$$






co.combinatorics linear-algebra determinants







share|cite|improve this question











$endgroup$




Consider the determinantal function
$$F(a,b,c):=detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}.$$
I would like to ask:




QUESTION. Can you provide an argument, combinatorial or otherwise, to prove the symmetry
$$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$
without actually computing the its values? I know how to justify this based on direct evaluation.




Example. The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or
$$detbegin{pmatrix} 10&15&21&28\20&35&56&84\35&70&126&210\56&126&252&462
end{pmatrix}=
detbegin{pmatrix} 15&35&70\21&56&126\28&84&210
end{pmatrix}=detbegin{pmatrix} 35&56\70&126
end{pmatrix}.$$






co.combinatorics linear-algebra determinants




co.combinatorics linear-algebra determinants






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 9 hours ago







T. Amdeberhan

















asked 9 hours ago









T. AmdeberhanT. Amdeberhan

17.4k229127




17.4k229127








  • 4




    $begingroup$
    Did you try to express this as the number of tuples of non-intersecting lattice paths?
    $endgroup$
    – Martin Rubey
    9 hours ago








  • 2




    $begingroup$
    The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
    $endgroup$
    – Mark Wildon
    7 hours ago








  • 1




    $begingroup$
    Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
    $endgroup$
    – Gerhard Paseman
    7 hours ago














  • 4




    $begingroup$
    Did you try to express this as the number of tuples of non-intersecting lattice paths?
    $endgroup$
    – Martin Rubey
    9 hours ago








  • 2




    $begingroup$
    The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
    $endgroup$
    – Mark Wildon
    7 hours ago








  • 1




    $begingroup$
    Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
    $endgroup$
    – Gerhard Paseman
    7 hours ago








4




4




$begingroup$
Did you try to express this as the number of tuples of non-intersecting lattice paths?
$endgroup$
– Martin Rubey
9 hours ago






$begingroup$
Did you try to express this as the number of tuples of non-intersecting lattice paths?
$endgroup$
– Martin Rubey
9 hours ago






2




2




$begingroup$
The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
$endgroup$
– Mark Wildon
7 hours ago






$begingroup$
The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
$endgroup$
– Mark Wildon
7 hours ago






1




1




$begingroup$
Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
$endgroup$
– Gerhard Paseman
7 hours ago




$begingroup$
Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
$endgroup$
– Gerhard Paseman
7 hours ago










2 Answers
2






active

oldest

votes


















11












$begingroup$

Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $atimes b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $atimes b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $atimes btimes c$ box. Thus the symmetry.






share|cite|improve this answer











$endgroup$





















    8












    $begingroup$

    Just for the record:



    In view of Fedor's nice reply, the interpretation gives away
    $$detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}
    =prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}
    frac{i+j+k+2}{i+j+k+1}.$$
    An apparent symmetry!



    As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.



    Let $(q)_k=(1-q)(1-q^2)cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
    $$binom{n}k_q=frac{(q)_n}{(q)_k(q)_{n-k}}.tag1$$
    Then, we may generalize the above identity (1) as
    $$detleft[q^{i^2+ai+bj}binom{i+j+a+b}{i+a}_qright]_{i,j=0}^{c-1}
    =prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      11












      $begingroup$

      Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
      Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $atimes b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $atimes b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $atimes btimes c$ box. Thus the symmetry.






      share|cite|improve this answer











      $endgroup$


















        11












        $begingroup$

        Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
        Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $atimes b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $atimes b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $atimes btimes c$ box. Thus the symmetry.






        share|cite|improve this answer











        $endgroup$
















          11












          11








          11





          $begingroup$

          Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
          Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $atimes b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $atimes b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $atimes btimes c$ box. Thus the symmetry.






          share|cite|improve this answer











          $endgroup$



          Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
          Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $atimes b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $atimes b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $atimes btimes c$ box. Thus the symmetry.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 6 hours ago

























          answered 9 hours ago









          Fedor PetrovFedor Petrov

          49k5112229




          49k5112229























              8












              $begingroup$

              Just for the record:



              In view of Fedor's nice reply, the interpretation gives away
              $$detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}
              =prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}
              frac{i+j+k+2}{i+j+k+1}.$$
              An apparent symmetry!



              As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.



              Let $(q)_k=(1-q)(1-q^2)cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
              $$binom{n}k_q=frac{(q)_n}{(q)_k(q)_{n-k}}.tag1$$
              Then, we may generalize the above identity (1) as
              $$detleft[q^{i^2+ai+bj}binom{i+j+a+b}{i+a}_qright]_{i,j=0}^{c-1}
              =prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$






              share|cite|improve this answer











              $endgroup$


















                8












                $begingroup$

                Just for the record:



                In view of Fedor's nice reply, the interpretation gives away
                $$detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}
                =prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}
                frac{i+j+k+2}{i+j+k+1}.$$
                An apparent symmetry!



                As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.



                Let $(q)_k=(1-q)(1-q^2)cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
                $$binom{n}k_q=frac{(q)_n}{(q)_k(q)_{n-k}}.tag1$$
                Then, we may generalize the above identity (1) as
                $$detleft[q^{i^2+ai+bj}binom{i+j+a+b}{i+a}_qright]_{i,j=0}^{c-1}
                =prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$






                share|cite|improve this answer











                $endgroup$
















                  8












                  8








                  8





                  $begingroup$

                  Just for the record:



                  In view of Fedor's nice reply, the interpretation gives away
                  $$detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}
                  =prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}
                  frac{i+j+k+2}{i+j+k+1}.$$
                  An apparent symmetry!



                  As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.



                  Let $(q)_k=(1-q)(1-q^2)cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
                  $$binom{n}k_q=frac{(q)_n}{(q)_k(q)_{n-k}}.tag1$$
                  Then, we may generalize the above identity (1) as
                  $$detleft[q^{i^2+ai+bj}binom{i+j+a+b}{i+a}_qright]_{i,j=0}^{c-1}
                  =prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$






                  share|cite|improve this answer











                  $endgroup$



                  Just for the record:



                  In view of Fedor's nice reply, the interpretation gives away
                  $$detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}
                  =prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}
                  frac{i+j+k+2}{i+j+k+1}.$$
                  An apparent symmetry!



                  As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.



                  Let $(q)_k=(1-q)(1-q^2)cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
                  $$binom{n}k_q=frac{(q)_n}{(q)_k(q)_{n-k}}.tag1$$
                  Then, we may generalize the above identity (1) as
                  $$detleft[q^{i^2+ai+bj}binom{i+j+a+b}{i+a}_qright]_{i,j=0}^{c-1}
                  =prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 6 hours ago









                  Fedor Petrov

                  49k5112229




                  49k5112229










                  answered 8 hours ago









                  T. AmdeberhanT. Amdeberhan

                  17.4k229127




                  17.4k229127






























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                      Origin of the phrase “under your belt”?

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                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 17 1 Today, I found myself discussing what students should have "under their belts" during a lecture, and I wondered to myself if there was some inappropriate undertone here that I might be unaware of. Question : What is the origin of the phrase "under your belt"? A google search revealed some webpages (e.g. [1]) that assert its origin is related to consumption of food -- once a meal has been eaten, it's under your belt (which is a relief). However, with such websites, I have no way to determine fact from "random guy on the internet making stuff up". etymology idioms ...
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