Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to show that: \begin{equation} \det\left[ \binom{2n}{n+i-j} \right]_{i,j=0}^{n-1}=\prod_{i=0}^{n-1} \frac{\binom{2n+i}{n}}{\binom{n+i}{n}} \end{equation}

I have tried playing with the algebra for some time. For example, if we fix $i$ and consider a particular row vector, we have: \begin{equation} \left[ \begin{array}{c} \binom{2n}{n+i} & \binom{2n}{n+i-1} & \binom{2n}{n+i-2} & \dots & \binom{2n}{i+1} \end{array} \right] \end{equation} Which equals \begin{equation} \left[ \begin{array}{c} \frac{(2n)!}{(n+i)!(n-i)!} & \frac{(2n)!}{(n+i-1)!(n-i+1)!} & \frac{(2n)!}{(n+i-2)!(n-i+2)!} & \dots & \frac{(2n)!}{(i+1)!(2n-i-1)!} \end{array} \right] \end{equation} It seems that our goal should be to factor out $\binom{2n+i}{n} / \binom{n+i}{n}$ and leave a matrix whose determinant evaluates to 1. Clearly: \begin{equation} \binom{2n+i}{n} / \binom{n+i}{n} = \frac{(2n+i)!}{n!(n+i)!} \cdot \frac{n!(i!)}{(n+i)!} = \frac{(2n+i)!}{(n+i)!} \cdot \frac{i!}{(n+i)!} \end{equation} However, I am unsure of how to proceed.

For those interested, the determinant given enumerates plane partitions contained within an $n\times n \times n$ cube, or equivalently, rhombic tilings of a regular hexagon with side length $n$.

share|improve this question
    
I don't have a complete answer, but I can point out that your matrix is of a special type (symmetric Toeplitz). Maybe that will help find some general algorithm which is applicable. –  Sidious Lord Aug 19 '12 at 10:21
add comment

2 Answers 2

up vote 2 down vote accepted

It seems that our goal should be to factor out ... and leave a matrix whose determinant evaluates to 1.

Actually, one can factor out something to leave a (generalized) Vandermonde matrix.

Using column operations one can see that $$ \det\binom{2n}{n+i-j}=\det\binom{2n+j}{n+i}. $$ Now $\binom{n}{k}=\frac{n^{\downarrow k}}{k!}$, where $n^{\downarrow k}:=n(n-1)\ldots(n-k+1)$, so the determinant we want to compute is just $$ \det\frac{(2n+j)^{\downarrow n+i}}{(n+i)!}= \prod_i\frac1{(n+i)!}\cdot\prod_j(2n+j)^{\downarrow n}\cdot \det(n+j)^{\downarrow i} $$ (here we used that $a^{\downarrow k+l}=a^{\downarrow k}\cdot(a-k)^{\downarrow l}$).

Observe that $$ \det(n+j)^{\downarrow i}=\det(n+j)^i=\prod j!, $$ so the answer is $$ \prod_j\frac{(2n+j)!j!}{((n+j)!)^2}= \prod_j\frac{\binom{2n+j}n}{\binom{n+j}n}. $$

(Perhaps, all this is also contained in Krattenthaler's paper quoted above. But anyway.)

share|improve this answer
add comment

Krattenthaler, in this article, proves a more general formula, of which the OP's determinant is a special case. Given the $n\times n$ matrix $\mathbf A$ with elements

$$\mathbf a_{j,k}=\binom{p+q}{p+j-k}, \quad 1\leq j,k\leq n$$

then

$$\begin{align*} \det\mathbf A&=\prod_{j=1}^n \prod_{k=1}^p \prod_{\ell=1}^q \frac{j+k+\ell-1}{j+k+\ell-2}\\ &=\prod_{j=1}^n \frac{(p+q+j-1)!(j-1)!}{(p+j-1)!(q+j-1)!} \end{align*}$$

where the triple product formula is attributed to MacMahon. A number of proofs for this determinantal identity are given in the linked article. For your particular special case,

$$\prod_{j=0}^{n-1} \frac{(2n+j)!j!}{((n+j)!)^2}=\prod_{j=0}^{n-1} \frac{\frac{(2n+j)!}{(n+j)!n!}}{\frac{(n+j)!}{n!j!}}=\prod_{j=0}^{n-1} \frac{\binom{2n+j}{n}}{\binom{n+j}{n}}$$

See this article as well.

share|improve this answer
3  
Another source is Section 3.3 in Bressoud's book Proofs and Confirmations. –  Hans Lundmark Aug 19 '12 at 15:22
    
Thanks J.M. and Hans, I will look into these today. –  01000100 Aug 19 '12 at 22:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.