Certain matrices of interesting determinant











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Let $M_n$ be the $ntimes n$ matrix with entries
$$binom{i}{2j}+binom{j}{2i}, qquad text{for $1leq i,jleq n$}.$$




QUESTION. Is this true? There is some evidence. The determinant $det(M_{2n+1})=0$ and
$$det(M_{2n})=(-1)^nbinom{2n}n^22^{n(n-3)}.$$







linear-algebra determinants elementary-proofs







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    The odd case is easy because $M_{2n-1}$ has an $n times n$ block of zeros at bottom right. In the even case, $M_{2n}$ still has that $n times n$ block of zeros, so its determinant is $(-1)^n$ times the product of the determinants of the $n times n$ blocks at top right and bottom left, which are equal because they are each other's transpose. So it comes down to a determinant of $i choose 2j$ for $i in [n-1,2n]$ and $j in [1,n]$, which must be known.
    – Noam D. Elkies
    4 hours ago






  • 1




    @NoamD.Elkies typo: $iin [n+1,2n]$
    – Fedor Petrov
    3 hours ago















up vote
5
down vote

favorite












Let $M_n$ be the $ntimes n$ matrix with entries
$$binom{i}{2j}+binom{j}{2i}, qquad text{for $1leq i,jleq n$}.$$




QUESTION. Is this true? There is some evidence. The determinant $det(M_{2n+1})=0$ and
$$det(M_{2n})=(-1)^nbinom{2n}n^22^{n(n-3)}.$$







linear-algebra determinants elementary-proofs







share|cite|improve this question


















  • 5




    The odd case is easy because $M_{2n-1}$ has an $n times n$ block of zeros at bottom right. In the even case, $M_{2n}$ still has that $n times n$ block of zeros, so its determinant is $(-1)^n$ times the product of the determinants of the $n times n$ blocks at top right and bottom left, which are equal because they are each other's transpose. So it comes down to a determinant of $i choose 2j$ for $i in [n-1,2n]$ and $j in [1,n]$, which must be known.
    – Noam D. Elkies
    4 hours ago






  • 1




    @NoamD.Elkies typo: $iin [n+1,2n]$
    – Fedor Petrov
    3 hours ago













up vote
5
down vote

favorite









up vote
5
down vote

favorite











Let $M_n$ be the $ntimes n$ matrix with entries
$$binom{i}{2j}+binom{j}{2i}, qquad text{for $1leq i,jleq n$}.$$




QUESTION. Is this true? There is some evidence. The determinant $det(M_{2n+1})=0$ and
$$det(M_{2n})=(-1)^nbinom{2n}n^22^{n(n-3)}.$$







linear-algebra determinants elementary-proofs







share|cite|improve this question













Let $M_n$ be the $ntimes n$ matrix with entries
$$binom{i}{2j}+binom{j}{2i}, qquad text{for $1leq i,jleq n$}.$$




QUESTION. Is this true? There is some evidence. The determinant $det(M_{2n+1})=0$ and
$$det(M_{2n})=(-1)^nbinom{2n}n^22^{n(n-3)}.$$







linear-algebra determinants elementary-proofs




linear-algebra determinants elementary-proofs






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asked 4 hours ago









T. Amdeberhan

16.9k228124




16.9k228124








  • 5




    The odd case is easy because $M_{2n-1}$ has an $n times n$ block of zeros at bottom right. In the even case, $M_{2n}$ still has that $n times n$ block of zeros, so its determinant is $(-1)^n$ times the product of the determinants of the $n times n$ blocks at top right and bottom left, which are equal because they are each other's transpose. So it comes down to a determinant of $i choose 2j$ for $i in [n-1,2n]$ and $j in [1,n]$, which must be known.
    – Noam D. Elkies
    4 hours ago






  • 1




    @NoamD.Elkies typo: $iin [n+1,2n]$
    – Fedor Petrov
    3 hours ago














  • 5




    The odd case is easy because $M_{2n-1}$ has an $n times n$ block of zeros at bottom right. In the even case, $M_{2n}$ still has that $n times n$ block of zeros, so its determinant is $(-1)^n$ times the product of the determinants of the $n times n$ blocks at top right and bottom left, which are equal because they are each other's transpose. So it comes down to a determinant of $i choose 2j$ for $i in [n-1,2n]$ and $j in [1,n]$, which must be known.
    – Noam D. Elkies
    4 hours ago






  • 1




    @NoamD.Elkies typo: $iin [n+1,2n]$
    – Fedor Petrov
    3 hours ago








5




5




The odd case is easy because $M_{2n-1}$ has an $n times n$ block of zeros at bottom right. In the even case, $M_{2n}$ still has that $n times n$ block of zeros, so its determinant is $(-1)^n$ times the product of the determinants of the $n times n$ blocks at top right and bottom left, which are equal because they are each other's transpose. So it comes down to a determinant of $i choose 2j$ for $i in [n-1,2n]$ and $j in [1,n]$, which must be known.
– Noam D. Elkies
4 hours ago




The odd case is easy because $M_{2n-1}$ has an $n times n$ block of zeros at bottom right. In the even case, $M_{2n}$ still has that $n times n$ block of zeros, so its determinant is $(-1)^n$ times the product of the determinants of the $n times n$ blocks at top right and bottom left, which are equal because they are each other's transpose. So it comes down to a determinant of $i choose 2j$ for $i in [n-1,2n]$ and $j in [1,n]$, which must be known.
– Noam D. Elkies
4 hours ago




1




1




@NoamD.Elkies typo: $iin [n+1,2n]$
– Fedor Petrov
3 hours ago




@NoamD.Elkies typo: $iin [n+1,2n]$
– Fedor Petrov
3 hours ago










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Noam Elkies in the comments reduces the problem to evaluating the determinant of $binom{(n+1)+i}{2j+2}$, $i,j=0,1,dots,n-1$. It looks that in general the determinant $binom{N+i}{c_j+j}$, $i,j=0,dots,n-1$ for integers $0leqslant c_0leqslant c_1leqslant dots leqslant c_{n-1}$ may be calculated by the following trick.



Consider the Young diagrams $lambda,mu$ with rows length $c_0leqslant c_1leqslant dots leqslant c_{n-1}$ and $N,N,dots,N$ ($n$ rows in total)respectively. Count the skew Young tableaux of the shape $musetminus lambda$. On one hand, the number of such tableaux is known (see Okounkov, A., Olshanski, G.: Shifted Schur functions. St. Petersburg Math. J. 9(2), 73–146 (1997), Theorem 8.1) to be equal to $$frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N+i)!}cdot prod_{i=0}^{n-1} (c_i+i)!cdot detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}.$$
On the other hand, it is the same as the number of usual Young tableaux of the shape $(N-c_0,dots,N-c_{n-1})$, which may be evaluated by Hook Length Formula as
$$
frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
$$
Therefore
$$
detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}=\
prod_{i=0}^{n-1}frac{(N+i)!}{(c_i+i)!(N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
$$
It remains to substitute $N=n+1$ and $c_i=i+2$.
We get $$prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i<j}(j-i)=prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i=0}^{n-1} i!$$
and the answer rewrites as
$$
frac{1!2!dots (n-1)! (n+1)!dots (2n)!}
{2!4!dots (2n)! 2! 4!dots (2n-2)!}2^{n(n-1)/2}.
$$
Replace $2!4!dots (2n)!$ to $$2cdot 1!cdot 4cdot 3!cdot 6cdot 5!cdot ldotscdot (2n)cdot(2n-1)!=\=2^nn! 1!3!dots (2n-1)!.$$ We get $frac{(2n)!}{n!n!} 2^{n(n-3)/2}$
as supposed.






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    Noam Elkies in the comments reduces the problem to evaluating the determinant of $binom{(n+1)+i}{2j+2}$, $i,j=0,1,dots,n-1$. It looks that in general the determinant $binom{N+i}{c_j+j}$, $i,j=0,dots,n-1$ for integers $0leqslant c_0leqslant c_1leqslant dots leqslant c_{n-1}$ may be calculated by the following trick.



    Consider the Young diagrams $lambda,mu$ with rows length $c_0leqslant c_1leqslant dots leqslant c_{n-1}$ and $N,N,dots,N$ ($n$ rows in total)respectively. Count the skew Young tableaux of the shape $musetminus lambda$. On one hand, the number of such tableaux is known (see Okounkov, A., Olshanski, G.: Shifted Schur functions. St. Petersburg Math. J. 9(2), 73–146 (1997), Theorem 8.1) to be equal to $$frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N+i)!}cdot prod_{i=0}^{n-1} (c_i+i)!cdot detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}.$$
    On the other hand, it is the same as the number of usual Young tableaux of the shape $(N-c_0,dots,N-c_{n-1})$, which may be evaluated by Hook Length Formula as
    $$
    frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
    $$
    Therefore
    $$
    detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}=\
    prod_{i=0}^{n-1}frac{(N+i)!}{(c_i+i)!(N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
    $$
    It remains to substitute $N=n+1$ and $c_i=i+2$.
    We get $$prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i<j}(j-i)=prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i=0}^{n-1} i!$$
    and the answer rewrites as
    $$
    frac{1!2!dots (n-1)! (n+1)!dots (2n)!}
    {2!4!dots (2n)! 2! 4!dots (2n-2)!}2^{n(n-1)/2}.
    $$
    Replace $2!4!dots (2n)!$ to $$2cdot 1!cdot 4cdot 3!cdot 6cdot 5!cdot ldotscdot (2n)cdot(2n-1)!=\=2^nn! 1!3!dots (2n-1)!.$$ We get $frac{(2n)!}{n!n!} 2^{n(n-3)/2}$
    as supposed.






    share|cite|improve this answer



























      up vote
      4
      down vote













      Noam Elkies in the comments reduces the problem to evaluating the determinant of $binom{(n+1)+i}{2j+2}$, $i,j=0,1,dots,n-1$. It looks that in general the determinant $binom{N+i}{c_j+j}$, $i,j=0,dots,n-1$ for integers $0leqslant c_0leqslant c_1leqslant dots leqslant c_{n-1}$ may be calculated by the following trick.



      Consider the Young diagrams $lambda,mu$ with rows length $c_0leqslant c_1leqslant dots leqslant c_{n-1}$ and $N,N,dots,N$ ($n$ rows in total)respectively. Count the skew Young tableaux of the shape $musetminus lambda$. On one hand, the number of such tableaux is known (see Okounkov, A., Olshanski, G.: Shifted Schur functions. St. Petersburg Math. J. 9(2), 73–146 (1997), Theorem 8.1) to be equal to $$frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N+i)!}cdot prod_{i=0}^{n-1} (c_i+i)!cdot detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}.$$
      On the other hand, it is the same as the number of usual Young tableaux of the shape $(N-c_0,dots,N-c_{n-1})$, which may be evaluated by Hook Length Formula as
      $$
      frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
      $$
      Therefore
      $$
      detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}=\
      prod_{i=0}^{n-1}frac{(N+i)!}{(c_i+i)!(N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
      $$
      It remains to substitute $N=n+1$ and $c_i=i+2$.
      We get $$prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i<j}(j-i)=prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i=0}^{n-1} i!$$
      and the answer rewrites as
      $$
      frac{1!2!dots (n-1)! (n+1)!dots (2n)!}
      {2!4!dots (2n)! 2! 4!dots (2n-2)!}2^{n(n-1)/2}.
      $$
      Replace $2!4!dots (2n)!$ to $$2cdot 1!cdot 4cdot 3!cdot 6cdot 5!cdot ldotscdot (2n)cdot(2n-1)!=\=2^nn! 1!3!dots (2n-1)!.$$ We get $frac{(2n)!}{n!n!} 2^{n(n-3)/2}$
      as supposed.






      share|cite|improve this answer

























        up vote
        4
        down vote










        up vote
        4
        down vote









        Noam Elkies in the comments reduces the problem to evaluating the determinant of $binom{(n+1)+i}{2j+2}$, $i,j=0,1,dots,n-1$. It looks that in general the determinant $binom{N+i}{c_j+j}$, $i,j=0,dots,n-1$ for integers $0leqslant c_0leqslant c_1leqslant dots leqslant c_{n-1}$ may be calculated by the following trick.



        Consider the Young diagrams $lambda,mu$ with rows length $c_0leqslant c_1leqslant dots leqslant c_{n-1}$ and $N,N,dots,N$ ($n$ rows in total)respectively. Count the skew Young tableaux of the shape $musetminus lambda$. On one hand, the number of such tableaux is known (see Okounkov, A., Olshanski, G.: Shifted Schur functions. St. Petersburg Math. J. 9(2), 73–146 (1997), Theorem 8.1) to be equal to $$frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N+i)!}cdot prod_{i=0}^{n-1} (c_i+i)!cdot detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}.$$
        On the other hand, it is the same as the number of usual Young tableaux of the shape $(N-c_0,dots,N-c_{n-1})$, which may be evaluated by Hook Length Formula as
        $$
        frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
        $$
        Therefore
        $$
        detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}=\
        prod_{i=0}^{n-1}frac{(N+i)!}{(c_i+i)!(N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
        $$
        It remains to substitute $N=n+1$ and $c_i=i+2$.
        We get $$prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i<j}(j-i)=prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i=0}^{n-1} i!$$
        and the answer rewrites as
        $$
        frac{1!2!dots (n-1)! (n+1)!dots (2n)!}
        {2!4!dots (2n)! 2! 4!dots (2n-2)!}2^{n(n-1)/2}.
        $$
        Replace $2!4!dots (2n)!$ to $$2cdot 1!cdot 4cdot 3!cdot 6cdot 5!cdot ldotscdot (2n)cdot(2n-1)!=\=2^nn! 1!3!dots (2n-1)!.$$ We get $frac{(2n)!}{n!n!} 2^{n(n-3)/2}$
        as supposed.






        share|cite|improve this answer














        Noam Elkies in the comments reduces the problem to evaluating the determinant of $binom{(n+1)+i}{2j+2}$, $i,j=0,1,dots,n-1$. It looks that in general the determinant $binom{N+i}{c_j+j}$, $i,j=0,dots,n-1$ for integers $0leqslant c_0leqslant c_1leqslant dots leqslant c_{n-1}$ may be calculated by the following trick.



        Consider the Young diagrams $lambda,mu$ with rows length $c_0leqslant c_1leqslant dots leqslant c_{n-1}$ and $N,N,dots,N$ ($n$ rows in total)respectively. Count the skew Young tableaux of the shape $musetminus lambda$. On one hand, the number of such tableaux is known (see Okounkov, A., Olshanski, G.: Shifted Schur functions. St. Petersburg Math. J. 9(2), 73–146 (1997), Theorem 8.1) to be equal to $$frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N+i)!}cdot prod_{i=0}^{n-1} (c_i+i)!cdot detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}.$$
        On the other hand, it is the same as the number of usual Young tableaux of the shape $(N-c_0,dots,N-c_{n-1})$, which may be evaluated by Hook Length Formula as
        $$
        frac{(Ncdot n-sum c_i)!}{prod_{i=0}^{n-1} (N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
        $$
        Therefore
        $$
        detleft(binom{N+i}{c_j+j}right)_{i,j=0}^{n-1}=\
        prod_{i=0}^{n-1}frac{(N+i)!}{(c_i+i)!(N-c_{n-i-1}+i)!}cdot prod_{i<j} (c_j+j-c_i-i).
        $$
        It remains to substitute $N=n+1$ and $c_i=i+2$.
        We get $$prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i<j}(j-i)=prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}prod_{i=0}^{n-1} i!$$
        and the answer rewrites as
        $$
        frac{1!2!dots (n-1)! (n+1)!dots (2n)!}
        {2!4!dots (2n)! 2! 4!dots (2n-2)!}2^{n(n-1)/2}.
        $$
        Replace $2!4!dots (2n)!$ to $$2cdot 1!cdot 4cdot 3!cdot 6cdot 5!cdot ldotscdot (2n)cdot(2n-1)!=\=2^nn! 1!3!dots (2n-1)!.$$ We get $frac{(2n)!}{n!n!} 2^{n(n-3)/2}$
        as supposed.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 38 mins ago

























        answered 2 hours ago









        Fedor Petrov

        46.9k5110217




        46.9k5110217






























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