I trying to prove a statement, which boils down to showing that the determinant of a specific matrix is nonzero. I use the convention that $\binom{n}{k} = 0$
if $k > n$ or $k < 0$. Let $k,l$ be natural numbers such that $k \le l$. Then the $n\times n$-Matrix $A$ is defined to have the entries
$a_{ij} = \binom{l}{k +i – j}$. So it looks like
$A = \left( \begin{array}{cccccc} \binom{l}{k} & \cdots & \binom{l}{0} & 0 & \cdots & 0 \\
\vdots & \ddots & & \ddots & \ddots & \vdots \\
\binom{l}{l} & & \ddots & & \ddots & 0
\\
0& \ddots && \ddots & & \binom{l}{0}\\
\vdots&\ddots&\ddots&&\ddots&\vdots\\
0&\cdots&0& \binom{l}{l} & \cdots & \binom{l}{k}
\end{array}\right)$.
Clearly the cases $k = l$ and $k = 0$ are trivial, since $A$ is then triangular. My first idea was to use the formula
$\binom{r}{s} = \binom{r-1}{s-1} + \binom{r}{s}$ and add columns/rows to each other. But that does not work out that well…
So if anyone has any ideas, or this matrix is known to be invertible, I would be very thankful.
Best Answer
Answer: The determinant of $A$ is $\dfrac{H\left(k\right) H\left(l-k\right) H\left(n\right) H\left(l+n\right)}{H\left(l-k+n\right) H\left(n+k\right) H\left(l\right)}$, where we are using the notation $H\left(m\right)$ for the hyperfactorial $\left(m-1\right)! \left(m-2\right)! \cdots 1! 0!$ of a nonnegative integer $m$.
(There are various other ways to express the answer, but the above is the most compact.)
1st proof. Theorem 8 in my note A hyperfactorial divisibility says that for every nonnegative integers $a$, $b$ and $c$, we have \begin{align*} & \dfrac{H\left( a\right) H\left( b\right) H\left( c\right) H\left( a+b+c\right) }{H\left( b+c\right) H\left( c+a\right) H\left( a+b\right) }\\ & =\det\left( \left( \dbinom{a+b+i-1}{a+i-j}\right) _{1\leq i\leq c,\ 1\leq j\leq c}\right) =\det\left( \left( \dbinom{a+b}{a+i-j}\right) _{1\leq i\leq c,\ 1\leq j\leq c}\right) . \end{align*} Applying this to $a = k$, $b = l-k$ and $c = n$ (noticing that $b$ is nonnegative because $k \leq l$), we obtain \begin{align*} &\dfrac{H\left(k\right) H\left(l-k\right) H\left(n\right) H\left(l+n\right)}{H\left(l-k+n\right) H\left(n+k\right) H\left(l\right)} \\ & =\det\left( \left( \dbinom{l+i-1}{k+i-j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) =\det\left( \left( \dbinom{l}{k+i-j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) . \end{align*} But the matrix on the right(most) side of this equality is exactly your $A$; so the Answer is proven.
2nd proof (sketched). The question has a combinatorial interpretation in terms of nonintersecting lattice paths on a rectangular grid, or semistandard Young tableaux, or Schur functions. These three objects are essentially different languages for the same argument; I will use the third since it is the easiest to write about.
I'll assume that you are familiar with the concept of symmetric functions (see, e.g., Chapter 2 of Darij Grinberg and Victor Reiner, Hopf algebras in combinatorics, or Mark Wildon, An involutive introduction to symmetric functions, or Chapter 7 of Richard Stanley, Enumerative Combinatorics, volume 2). In particular, I'll use the notations common in this subject, such as $s_\lambda$ for a Schur function and $e_i$ for an elementary symmetric function.
Set $b = l-k$; this is a nonnegative integer since $k \leq l$.
Let $\lambda$ be the partition $\left(n^b\right)$, which is to be understood as a shorthand for $\left(\underbrace{n,n,\ldots,n}_{b\text{ terms}}\right)$. The transpose (aka conjugate) $\lambda^t$ of this partition $\lambda$ is then $\left(b^n\right)$ (understood similarly). The Jacobi-Trudi formula (the one that uses elementary symmetric functions) thus says that \begin{equation} s_\lambda = \det \left( \left( e_{b-i+j} \right) _{1\leq i\leq n,\ 1\leq j\leq n} \right) . \end{equation} Now, substitute $\underbrace{1,1,\ldots,1}_{l \text{ entries}},0,0,0,\ldots$ for the countably many indeterminates into this equality. This substitution transforms each elementary symmetric function $e_p$ into $e_p\left(\underbrace{1,1,\ldots,1}_{l \text{ entries}},0,0,0,\ldots\right) = \dbinom{l}{p}$. Thus, the equation transforms into \begin{equation} s_\lambda\left(\underbrace{1,1,\ldots,1}_{l \text{ entries}},0,0,0,\ldots\right) = \det \left( \left( \dbinom{l}{b-i+j} \right) _{1\leq i\leq n,\ 1\leq j\leq n} \right) . \end{equation} Since each $i$ and $j$ satisfy $\dbinom{l}{b-i+j} = \dbinom{l}{k+i-j}$ (indeed, this follows from the symmetry of Pascal's triangle, because $l - \left(b-i+j\right) = k+i-j$), this rewrites as \begin{equation} s_\lambda\left(\underbrace{1,1,\ldots,1}_{l \text{ entries}},0,0,0,\ldots\right) = \det \left( \left( \dbinom{l}{k+i-j} \right) _{1\leq i\leq n,\ 1\leq j\leq n} \right) = \det A . \end{equation} It remains to compute the left hand side.
But the combinatorial definition of $s_\lambda$ shows that $s_\lambda\left(\underbrace{1,1,\ldots,1}_{l \text{ entries}},0,0,0,\ldots\right)$ is just the number of semistandard Young tableaux of shape $\lambda$ with entries in $\left\{1,2,\ldots,l\right\}$. For this number, there is a formula (known as Weyl's character formula in type A; see, e.g., https://mathoverflow.net/questions/106606/new-formula-for-counting-ssyts ), which can be written as \begin{equation} s_\lambda\left(\underbrace{1,1,\ldots,1}_{l \text{ entries}},0,0,0,\ldots\right) = \dfrac{1}{H\left(l\right)} \prod_{1 \leq i < j \leq l} \left(\left(\lambda_i - i\right) - \left(\lambda_j - j\right)\right) , \end{equation} where $\lambda_p$ is the $p$-th entry of $\lambda$ (so, in our case, $\lambda_p = b$ if $p \leq n$, and otherwise $\lambda_p = 0$). It takes a while to massage the right hand side so as to look like the Answer given above (the first step is to observe that the only factors in the product that are distinct from $1$ are those with $1 \leq i \leq n$ and $n < j \leq l$, so that all other factors can be discarded), but it isn't difficult (you just need to check that two large products have the same factors).