Rank of a matrix of binomial coefficients

Question

This question arose as a side computation on error correcting codes. Let $k$, $r$ be positive integers such that $2k-1 \leqslant r$ and let $p$ a prime number such that $r < p$. I would like to find the rank of the following $k \times k$ matrix with coefficients in the field $F_p$ $$ \begin{pmatrix} \binom {r}{k} & \binom {r}{k+1} & \dotsm & \binom {r}{2k-1}\\ \vdots & \\ \binom {r}{2} & \binom {r}{3} & \dotsm & \binom {r}{k+1} \\ \binom {r}{1} & \binom {r}{2} & \dotsm & \binom {r}{k} \end{pmatrix} $$ where all binomial coefficients are taken modulo $p$. I conjecture the rank should be $k$ but I have no formal proof. I am aware of Lucas's theorem, but it didn't help so far.

Answer 1

Not true, unfortunately.

Indeed, by adding the $2$nd row to the $1$st, $3$rd to $2$nd, ... , $k$th to $(k-1)$th etc, you end up with the matrix $$ \begin{pmatrix} \binom {r+1}{k} & \binom {r+1}{k+1} & \dotsm & \binom {r+1}{2k-1}\\ \vdots & \vdots & \ddots & \vdots \\ \binom {r+1}{2} & \binom {r+1}{3} & \dotsm & \binom {r+1}{k+1} \\ \binom {r}{1} & \binom {r}{2} & \dotsm & \binom {r}{k} \end{pmatrix} $$ which has the same rank.

In particular, for $r=p-1$ we get $\pmod p$ $$ \begin{pmatrix} 0 & 0 & \dotsm & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dotsm & 0 \\ * & * & \dotsm & * \end{pmatrix}. $$ I doubt this has rank $k$.

Moreover, it is easy to see with this argument that the rank is at most $p-r$.

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Update: For $k\ge p-r$, the rank is exactly $p-r$. Sketch: make the above transformations with first $k-1$ rows, then with first $k-2$ rows, etc. Then do the same to columns. You will end up with $$ \begin{pmatrix} \binom {r+k-1}{k} & \binom {r+k}{k+1} & \dotsm & \binom {r+2k-2}{2k-1}\\ \vdots & \vdots & \ddots & \vdots \\ \binom {r+2}{3} & \binom {r+3}{4} & \dotsm & \binom {r+k+1}{k+2} \\ \binom {r+1}{2} & \binom {r+2}{3} & \dotsm & \binom {r+k}{k+1} \\ \binom {r}{1} & \binom {r+1}{2} & \dotsm & \binom {r+k-1}{k} \end{pmatrix} $$ Now $\pmod p$ the non-zero elements of this matrix form a lower-left triangle of size $p-r$ (I hope this is clear, as I don't know how LaTeX such thing). Hence, the rank is $p-r$.

It is also possible to do this for $k>p-r$, which gives only lower bound $2p-2r-k$ for the rank; not a lot of help, probably.

Answer 2

Write $A(r,k)$ for the matrix you describe. Then $$ \det A(r,k) =\prod_{i,j=1}^k \frac{r-k-1+i+j}{i+j-1}. $$ This follows from equation (2.17) in Krattenthaler's Advanced determinant calculus (with $a=n=k$, $b=r-k$). For $p\geq r+k$ every factor in the product has numerator coprime to $p$, so the reduction of $A(r,k)$ has full rank.