Determinant of a Hankel matrix with sequence (1, 2, 3,…, n, 1, 2,…, n-1)












1












$begingroup$


I'm looking for a closed form of the determinant of matrices like



$begin{bmatrix}1 & 2 & 3\2 & 3 & 1\ 3& 1 &2end{bmatrix}$



or



$begin{bmatrix}1 & 2 & 3 &4\2 & 3 &4 & 1\ 3&4& 1 &2\ 4& 1&2&3end{bmatrix}$,



that means Henkel matrices of the size n with sequence (1, 2, 3,..., n, 1, 2,..., n-1)






matrices determinant integers







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  • $begingroup$
    Have you tried something?
    $endgroup$
    – Tito Eliatron
    Jan 7 at 20:38










  • $begingroup$
    Yes, some row operations. I tried to create 1's in the first column by substracting row j from row j+1, but that didn't seem to help. I tried swapping the rows, such that the 1's are on the main diagonal, also not helpful I think.
    $endgroup$
    – Emphikongri
    Jan 7 at 20:43






  • 1




    $begingroup$
    Maybe you can add to the first row (or column) the others
    $endgroup$
    – Tito Eliatron
    Jan 7 at 20:44
















1












$begingroup$


I'm looking for a closed form of the determinant of matrices like



$begin{bmatrix}1 & 2 & 3\2 & 3 & 1\ 3& 1 &2end{bmatrix}$



or



$begin{bmatrix}1 & 2 & 3 &4\2 & 3 &4 & 1\ 3&4& 1 &2\ 4& 1&2&3end{bmatrix}$,



that means Henkel matrices of the size n with sequence (1, 2, 3,..., n, 1, 2,..., n-1)






matrices determinant integers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Have you tried something?
    $endgroup$
    – Tito Eliatron
    Jan 7 at 20:38










  • $begingroup$
    Yes, some row operations. I tried to create 1's in the first column by substracting row j from row j+1, but that didn't seem to help. I tried swapping the rows, such that the 1's are on the main diagonal, also not helpful I think.
    $endgroup$
    – Emphikongri
    Jan 7 at 20:43






  • 1




    $begingroup$
    Maybe you can add to the first row (or column) the others
    $endgroup$
    – Tito Eliatron
    Jan 7 at 20:44














1












1








1





$begingroup$


I'm looking for a closed form of the determinant of matrices like



$begin{bmatrix}1 & 2 & 3\2 & 3 & 1\ 3& 1 &2end{bmatrix}$



or



$begin{bmatrix}1 & 2 & 3 &4\2 & 3 &4 & 1\ 3&4& 1 &2\ 4& 1&2&3end{bmatrix}$,



that means Henkel matrices of the size n with sequence (1, 2, 3,..., n, 1, 2,..., n-1)






matrices determinant integers







share|cite|improve this question









$endgroup$




I'm looking for a closed form of the determinant of matrices like



$begin{bmatrix}1 & 2 & 3\2 & 3 & 1\ 3& 1 &2end{bmatrix}$



or



$begin{bmatrix}1 & 2 & 3 &4\2 & 3 &4 & 1\ 3&4& 1 &2\ 4& 1&2&3end{bmatrix}$,



that means Henkel matrices of the size n with sequence (1, 2, 3,..., n, 1, 2,..., n-1)






matrices determinant integers




matrices determinant integers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 20:32









EmphikongriEmphikongri

83




83












  • $begingroup$
    Have you tried something?
    $endgroup$
    – Tito Eliatron
    Jan 7 at 20:38










  • $begingroup$
    Yes, some row operations. I tried to create 1's in the first column by substracting row j from row j+1, but that didn't seem to help. I tried swapping the rows, such that the 1's are on the main diagonal, also not helpful I think.
    $endgroup$
    – Emphikongri
    Jan 7 at 20:43






  • 1




    $begingroup$
    Maybe you can add to the first row (or column) the others
    $endgroup$
    – Tito Eliatron
    Jan 7 at 20:44


















  • $begingroup$
    Have you tried something?
    $endgroup$
    – Tito Eliatron
    Jan 7 at 20:38










  • $begingroup$
    Yes, some row operations. I tried to create 1's in the first column by substracting row j from row j+1, but that didn't seem to help. I tried swapping the rows, such that the 1's are on the main diagonal, also not helpful I think.
    $endgroup$
    – Emphikongri
    Jan 7 at 20:43






  • 1




    $begingroup$
    Maybe you can add to the first row (or column) the others
    $endgroup$
    – Tito Eliatron
    Jan 7 at 20:44
















$begingroup$
Have you tried something?
$endgroup$
– Tito Eliatron
Jan 7 at 20:38




$begingroup$
Have you tried something?
$endgroup$
– Tito Eliatron
Jan 7 at 20:38












$begingroup$
Yes, some row operations. I tried to create 1's in the first column by substracting row j from row j+1, but that didn't seem to help. I tried swapping the rows, such that the 1's are on the main diagonal, also not helpful I think.
$endgroup$
– Emphikongri
Jan 7 at 20:43




$begingroup$
Yes, some row operations. I tried to create 1's in the first column by substracting row j from row j+1, but that didn't seem to help. I tried swapping the rows, such that the 1's are on the main diagonal, also not helpful I think.
$endgroup$
– Emphikongri
Jan 7 at 20:43




1




1




$begingroup$
Maybe you can add to the first row (or column) the others
$endgroup$
– Tito Eliatron
Jan 7 at 20:44




$begingroup$
Maybe you can add to the first row (or column) the others
$endgroup$
– Tito Eliatron
Jan 7 at 20:44










2 Answers
2






active

oldest

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2












$begingroup$

The magnitude of the determinant is $$frac{n^{n-1}(n+1)}{2}.$$ The sign is given by $$f(n) = begin{cases}+ & mathrm{floor}(n/2) text{ is even.} \ - & mathrm{floor}(n/2) text{ is odd} .end{cases}$$A useful formula is $sum_{i=1}^{n} i = frac{n(n+1)}{2}$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    This A052182 Your formula is identical to the last one given there, expect that the signs given there are incorrect. Perhaps you should publish a correction.
    $endgroup$
    – saulspatz
    Jan 7 at 21:46










  • $begingroup$
    It appears that the order of rows in that entry differs from the order of rows in this question (see the 3x3 matrix, for example). Therefore I believe the signs are correct in that list.
    $endgroup$
    – Klint Qinami
    Jan 8 at 1:00










  • $begingroup$
    You're right. Careless of me.
    $endgroup$
    – saulspatz
    Jan 8 at 3:10



















1












$begingroup$

These matrices are actually circulant matrices, whose determinant is well known.



For the matrices at hand, the determinant is
$
Delta_n=prod _{j=0}^{n-1}f(omega _{j})
$,
where $f(z)=1+2z+3z^{2}+dots +nz^{n-1}$ and $w_j$ is the $j$-th root of unity.



We have $f(omega_0)=f(1)=1+2+3+cdots +n = dfrac{n(n+1)}{2}$.



We also have $f(z)=g'(z)$, where $g(z)=1+z+z^2+z^3+cdots+z^n=dfrac{z^{n+1}-1}{z-1}$. Therefore,
$$
g'(z)=dfrac{n z^{n+1} - (n+1) z^n + 1}{(z-1)^2}
$$
and so, for $jne0$,
$$
f(omega_j)=g'(omega_j)=dfrac{n omega_j - n}{(omega_j-1)^2}=dfrac{n}{omega_j-1}
$$
Thus,
$$
Delta_n=prod _{j=0}^{n-1}f(omega _{j})
=dfrac{n(n+1)}{2} prod _{j=1}^{n-1}dfrac{n}{omega_j-1}
=dfrac{n^n(n+1)}{2} dfrac{1}{prod _{j=1}^{n-1}(omega_j-1)}
$$
Now $prod _{j=1}^{n-1}(omega_j-1)=h(1)$ for
$$
h(z)=prod _{j=1}^{n-1}(omega_j-z)
=(-1)^{n-1}prod _{j=1}^{n-1}(z-omega_j)
=(-1)^{n-1}dfrac{z^{n}-1}{z-1}
=(-1)^{n-1}(1+z+z^2+z^3+cdots+z^{n-1})
$$
and so $h(1)=(-1)^{n-1}n$. Thus
$$
Delta_n
=dfrac{n^n(n+1)}{2} dfrac{1}{(-1)^{n-1}n}
=(-1)^{n-1}dfrac{n^{n-1}(n+1)}{2}
$$






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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    The magnitude of the determinant is $$frac{n^{n-1}(n+1)}{2}.$$ The sign is given by $$f(n) = begin{cases}+ & mathrm{floor}(n/2) text{ is even.} \ - & mathrm{floor}(n/2) text{ is odd} .end{cases}$$A useful formula is $sum_{i=1}^{n} i = frac{n(n+1)}{2}$.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      This A052182 Your formula is identical to the last one given there, expect that the signs given there are incorrect. Perhaps you should publish a correction.
      $endgroup$
      – saulspatz
      Jan 7 at 21:46










    • $begingroup$
      It appears that the order of rows in that entry differs from the order of rows in this question (see the 3x3 matrix, for example). Therefore I believe the signs are correct in that list.
      $endgroup$
      – Klint Qinami
      Jan 8 at 1:00










    • $begingroup$
      You're right. Careless of me.
      $endgroup$
      – saulspatz
      Jan 8 at 3:10
















    2












    $begingroup$

    The magnitude of the determinant is $$frac{n^{n-1}(n+1)}{2}.$$ The sign is given by $$f(n) = begin{cases}+ & mathrm{floor}(n/2) text{ is even.} \ - & mathrm{floor}(n/2) text{ is odd} .end{cases}$$A useful formula is $sum_{i=1}^{n} i = frac{n(n+1)}{2}$.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      This A052182 Your formula is identical to the last one given there, expect that the signs given there are incorrect. Perhaps you should publish a correction.
      $endgroup$
      – saulspatz
      Jan 7 at 21:46










    • $begingroup$
      It appears that the order of rows in that entry differs from the order of rows in this question (see the 3x3 matrix, for example). Therefore I believe the signs are correct in that list.
      $endgroup$
      – Klint Qinami
      Jan 8 at 1:00










    • $begingroup$
      You're right. Careless of me.
      $endgroup$
      – saulspatz
      Jan 8 at 3:10














    2












    2








    2





    $begingroup$

    The magnitude of the determinant is $$frac{n^{n-1}(n+1)}{2}.$$ The sign is given by $$f(n) = begin{cases}+ & mathrm{floor}(n/2) text{ is even.} \ - & mathrm{floor}(n/2) text{ is odd} .end{cases}$$A useful formula is $sum_{i=1}^{n} i = frac{n(n+1)}{2}$.






    share|cite|improve this answer









    $endgroup$



    The magnitude of the determinant is $$frac{n^{n-1}(n+1)}{2}.$$ The sign is given by $$f(n) = begin{cases}+ & mathrm{floor}(n/2) text{ is even.} \ - & mathrm{floor}(n/2) text{ is odd} .end{cases}$$A useful formula is $sum_{i=1}^{n} i = frac{n(n+1)}{2}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 7 at 21:27









    Klint QinamiKlint Qinami

    1,137410




    1,137410








    • 2




      $begingroup$
      This A052182 Your formula is identical to the last one given there, expect that the signs given there are incorrect. Perhaps you should publish a correction.
      $endgroup$
      – saulspatz
      Jan 7 at 21:46










    • $begingroup$
      It appears that the order of rows in that entry differs from the order of rows in this question (see the 3x3 matrix, for example). Therefore I believe the signs are correct in that list.
      $endgroup$
      – Klint Qinami
      Jan 8 at 1:00










    • $begingroup$
      You're right. Careless of me.
      $endgroup$
      – saulspatz
      Jan 8 at 3:10














    • 2




      $begingroup$
      This A052182 Your formula is identical to the last one given there, expect that the signs given there are incorrect. Perhaps you should publish a correction.
      $endgroup$
      – saulspatz
      Jan 7 at 21:46










    • $begingroup$
      It appears that the order of rows in that entry differs from the order of rows in this question (see the 3x3 matrix, for example). Therefore I believe the signs are correct in that list.
      $endgroup$
      – Klint Qinami
      Jan 8 at 1:00










    • $begingroup$
      You're right. Careless of me.
      $endgroup$
      – saulspatz
      Jan 8 at 3:10








    2




    2




    $begingroup$
    This A052182 Your formula is identical to the last one given there, expect that the signs given there are incorrect. Perhaps you should publish a correction.
    $endgroup$
    – saulspatz
    Jan 7 at 21:46




    $begingroup$
    This A052182 Your formula is identical to the last one given there, expect that the signs given there are incorrect. Perhaps you should publish a correction.
    $endgroup$
    – saulspatz
    Jan 7 at 21:46












    $begingroup$
    It appears that the order of rows in that entry differs from the order of rows in this question (see the 3x3 matrix, for example). Therefore I believe the signs are correct in that list.
    $endgroup$
    – Klint Qinami
    Jan 8 at 1:00




    $begingroup$
    It appears that the order of rows in that entry differs from the order of rows in this question (see the 3x3 matrix, for example). Therefore I believe the signs are correct in that list.
    $endgroup$
    – Klint Qinami
    Jan 8 at 1:00












    $begingroup$
    You're right. Careless of me.
    $endgroup$
    – saulspatz
    Jan 8 at 3:10




    $begingroup$
    You're right. Careless of me.
    $endgroup$
    – saulspatz
    Jan 8 at 3:10











    1












    $begingroup$

    These matrices are actually circulant matrices, whose determinant is well known.



    For the matrices at hand, the determinant is
    $
    Delta_n=prod _{j=0}^{n-1}f(omega _{j})
    $    ,
    where $f(z)=1+2z+3z^{2}+dots +nz^{n-1}$ and $w_j$ is the $j$-th root of unity.



    We have $f(omega_0)=f(1)=1+2+3+cdots +n = dfrac{n(n+1)}{2}$.



    We also have $f(z)=g'(z)$, where $g(z)=1+z+z^2+z^3+cdots+z^n=dfrac{z^{n+1}-1}{z-1}$. Therefore,
    $$
    g'(z)=dfrac{n z^{n+1} - (n+1) z^n + 1}{(z-1)^2}
    $$
    and so, for $jne0$,
    $$
    f(omega_j)=g'(omega_j)=dfrac{n omega_j - n}{(omega_j-1)^2}=dfrac{n}{omega_j-1}
    $$
    Thus,
    $$
    Delta_n=prod _{j=0}^{n-1}f(omega _{j})
    =dfrac{n(n+1)}{2} prod _{j=1}^{n-1}dfrac{n}{omega_j-1}
    =dfrac{n^n(n+1)}{2} dfrac{1}{prod _{j=1}^{n-1}(omega_j-1)}
    $$
    Now $prod _{j=1}^{n-1}(omega_j-1)=h(1)$ for
    $$
    h(z)=prod _{j=1}^{n-1}(omega_j-z)
    =(-1)^{n-1}prod _{j=1}^{n-1}(z-omega_j)
    =(-1)^{n-1}dfrac{z^{n}-1}{z-1}
    =(-1)^{n-1}(1+z+z^2+z^3+cdots+z^{n-1})
    $$
    and so $h(1)=(-1)^{n-1}n$. Thus
    $$
    Delta_n
    =dfrac{n^n(n+1)}{2} dfrac{1}{(-1)^{n-1}n}
    =(-1)^{n-1}dfrac{n^{n-1}(n+1)}{2}
    $$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      These matrices are actually circulant matrices, whose determinant is well known.



      For the matrices at hand, the determinant is
      $
      Delta_n=prod _{j=0}^{n-1}f(omega _{j})
      $      ,
      where $f(z)=1+2z+3z^{2}+dots +nz^{n-1}$ and $w_j$ is the $j$-th root of unity.



      We have $f(omega_0)=f(1)=1+2+3+cdots +n = dfrac{n(n+1)}{2}$.



      We also have $f(z)=g'(z)$, where $g(z)=1+z+z^2+z^3+cdots+z^n=dfrac{z^{n+1}-1}{z-1}$. Therefore,
      $$
      g'(z)=dfrac{n z^{n+1} - (n+1) z^n + 1}{(z-1)^2}
      $$
      and so, for $jne0$,
      $$
      f(omega_j)=g'(omega_j)=dfrac{n omega_j - n}{(omega_j-1)^2}=dfrac{n}{omega_j-1}
      $$
      Thus,
      $$
      Delta_n=prod _{j=0}^{n-1}f(omega _{j})
      =dfrac{n(n+1)}{2} prod _{j=1}^{n-1}dfrac{n}{omega_j-1}
      =dfrac{n^n(n+1)}{2} dfrac{1}{prod _{j=1}^{n-1}(omega_j-1)}
      $$
      Now $prod _{j=1}^{n-1}(omega_j-1)=h(1)$ for
      $$
      h(z)=prod _{j=1}^{n-1}(omega_j-z)
      =(-1)^{n-1}prod _{j=1}^{n-1}(z-omega_j)
      =(-1)^{n-1}dfrac{z^{n}-1}{z-1}
      =(-1)^{n-1}(1+z+z^2+z^3+cdots+z^{n-1})
      $$
      and so $h(1)=(-1)^{n-1}n$. Thus
      $$
      Delta_n
      =dfrac{n^n(n+1)}{2} dfrac{1}{(-1)^{n-1}n}
      =(-1)^{n-1}dfrac{n^{n-1}(n+1)}{2}
      $$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        These matrices are actually circulant matrices, whose determinant is well known.



        For the matrices at hand, the determinant is
        $
        Delta_n=prod _{j=0}^{n-1}f(omega _{j})
        $        ,
        where $f(z)=1+2z+3z^{2}+dots +nz^{n-1}$ and $w_j$ is the $j$-th root of unity.



        We have $f(omega_0)=f(1)=1+2+3+cdots +n = dfrac{n(n+1)}{2}$.



        We also have $f(z)=g'(z)$, where $g(z)=1+z+z^2+z^3+cdots+z^n=dfrac{z^{n+1}-1}{z-1}$. Therefore,
        $$
        g'(z)=dfrac{n z^{n+1} - (n+1) z^n + 1}{(z-1)^2}
        $$
        and so, for $jne0$,
        $$
        f(omega_j)=g'(omega_j)=dfrac{n omega_j - n}{(omega_j-1)^2}=dfrac{n}{omega_j-1}
        $$
        Thus,
        $$
        Delta_n=prod _{j=0}^{n-1}f(omega _{j})
        =dfrac{n(n+1)}{2} prod _{j=1}^{n-1}dfrac{n}{omega_j-1}
        =dfrac{n^n(n+1)}{2} dfrac{1}{prod _{j=1}^{n-1}(omega_j-1)}
        $$
        Now $prod _{j=1}^{n-1}(omega_j-1)=h(1)$ for
        $$
        h(z)=prod _{j=1}^{n-1}(omega_j-z)
        =(-1)^{n-1}prod _{j=1}^{n-1}(z-omega_j)
        =(-1)^{n-1}dfrac{z^{n}-1}{z-1}
        =(-1)^{n-1}(1+z+z^2+z^3+cdots+z^{n-1})
        $$
        and so $h(1)=(-1)^{n-1}n$. Thus
        $$
        Delta_n
        =dfrac{n^n(n+1)}{2} dfrac{1}{(-1)^{n-1}n}
        =(-1)^{n-1}dfrac{n^{n-1}(n+1)}{2}
        $$






        share|cite|improve this answer











        $endgroup$



        These matrices are actually circulant matrices, whose determinant is well known.



        For the matrices at hand, the determinant is
        $
        Delta_n=prod _{j=0}^{n-1}f(omega _{j})
        $        ,
        where $f(z)=1+2z+3z^{2}+dots +nz^{n-1}$ and $w_j$ is the $j$-th root of unity.



        We have $f(omega_0)=f(1)=1+2+3+cdots +n = dfrac{n(n+1)}{2}$.



        We also have $f(z)=g'(z)$, where $g(z)=1+z+z^2+z^3+cdots+z^n=dfrac{z^{n+1}-1}{z-1}$. Therefore,
        $$
        g'(z)=dfrac{n z^{n+1} - (n+1) z^n + 1}{(z-1)^2}
        $$
        and so, for $jne0$,
        $$
        f(omega_j)=g'(omega_j)=dfrac{n omega_j - n}{(omega_j-1)^2}=dfrac{n}{omega_j-1}
        $$
        Thus,
        $$
        Delta_n=prod _{j=0}^{n-1}f(omega _{j})
        =dfrac{n(n+1)}{2} prod _{j=1}^{n-1}dfrac{n}{omega_j-1}
        =dfrac{n^n(n+1)}{2} dfrac{1}{prod _{j=1}^{n-1}(omega_j-1)}
        $$
        Now $prod _{j=1}^{n-1}(omega_j-1)=h(1)$ for
        $$
        h(z)=prod _{j=1}^{n-1}(omega_j-z)
        =(-1)^{n-1}prod _{j=1}^{n-1}(z-omega_j)
        =(-1)^{n-1}dfrac{z^{n}-1}{z-1}
        =(-1)^{n-1}(1+z+z^2+z^3+cdots+z^{n-1})
        $$
        and so $h(1)=(-1)^{n-1}n$. Thus
        $$
        Delta_n
        =dfrac{n^n(n+1)}{2} dfrac{1}{(-1)^{n-1}n}
        =(-1)^{n-1}dfrac{n^{n-1}(n+1)}{2}
        $$







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        edited Jan 8 at 13:00

























        answered Jan 8 at 11:27









        lhflhf

        163k10169393




        163k10169393






























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