# Some determinants which are closely related to recurrences

Let the sequence $$(a(n,k))_{ n \in \mathbb{Z}}$$ satisfy $$\sum_{j=0}^k c(k,j)a(n-j,k)=0$$ with $$c(k,j)=c(k,k-j)$$ and $$c(k,0)=1$$ and with initial values $$a(-n,k)=0$$ for $$1\leq n\leq{k-1}$$ and $$a(0,k)=1.$$

For example for the binomial coefficients $$c(k,j)=\binom{k}{j},$$ we get $$a(n,k)=\binom{-k}{n}.$$

Let $$A_k(n)$$ be the matrix whose entries are $$a(n+i+j-k+2,k)$$ for $$0 \leq i,j \leq {k-2}.$$

Computer experiments suggest that $$\det{A_k(n)}=(-1)^{kn+1+\binom{k+1}{2}}a(n,k).$$

A similar result seems to hold if $$c(k,j)=-c(k,k-j).$$

Any idea how to prove this? Is this a known result or a special case of a more general result?

The question concerns the determinant of a Hankel matrix, or a fixed element of a Hankel matrix transform of a shifted sequence $$a(n,k)$$ for a fixed $$k$$, although I do not see how this fact alone can be useful. I give a standalone proof below.

Let $$k$$ be fixed. For the sake of simplicity, let's denote $$c_j := c(k,j)$$ and $$a_n := a(n,k)$$.

First, notice that the sequence $$(a_n)$$ has the characteristic polynomial $$C(x) := \sum_{j=0}^k c_j x^{k-j},$$ while $$C(-x)$$ gives the characteristic polynomial for the sequence $$((-1)^n a_n)$$.

Now, let $$b_t(n)$$ for $$t\in\{0,\dots,k-1\}$$ be the determinant of the matrix obtained from the $$(k-1)\times k$$ matrix: $$\big(a_{n+i+j-k}\big)_{1\leq i\leq k-1\atop 0\leq j\leq k-1}$$ by removing the column with $$j=t$$. Then $$\det A_k(n) = b_0(n) = b_{k-1}(n+1)$$.

From the recurrence for $$a_n$$, it follows that $$\begin{bmatrix} b_0(n) \\ b_1(n)\\ \vdots\\ b_{k-1}(n) \end{bmatrix} = M\cdot \begin{bmatrix} b_0(n-1) \\ b_1(n-1)\\ \vdots\\ b_{k-1}(n-1) \end{bmatrix},$$ where $$M := \begin{bmatrix} (-1)^{k-1} c_{k-1} & (-1)^{k-1} c_k & 0 & \dots & 0\\ (-1)^{k-2} c_{k-2} & 0 & (-1)^{k-1} c_k & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -c_1 & 0 & 0 & \dots & (-1)^{k-1} c_k\\ 1 & 0 & 0 & \dots & 0 \end{bmatrix}.$$ This is almost a companion matrix and its characteristic polynomial equals the reciprocal of $$C((-1)^k x)$$, which is same as $$C((-1)^k x)$$ thanks to the symmetry. In particular, $$b_0(n)$$ has a characteristic polynomial $$C(x)$$ when $$k$$ is even, $$C(-x)$$ when $$k$$ is odd.

It remains to notice that $$b_0(n)=(-1)^{(k-1)(k-2)/2} a_n$$ for $$n=-(k-1),\dots, 0$$, implying that $$\det A_k(n) = b_0(n) = (-1)^{kn + (k-1)(k-2)/2} a_n$$. QED

UPDATE. More generally, we can drop the symmetricity requirement $$c_j = c_{k-j}$$ and keep only $$c_0 = c_k$$. Then $$\det A_k(n) = (-1)^{kn + (k-1)(k-2)/2} a_{-n-k},$$ where $$a_n$$ is extended to large negative indices by the same recurrence (see also this answer). This formula implies the original result, since for symmetric $$c_j$$, we have $$a_{-n-k}=a_n$$.

Similarly, if we have $$c_0 = -c_k$$, then $$\det A_k(n) = (-1)^{(k+1)n + (k+1)k/2} a_{-n-k}.$$

• @ Max Alekseyev: Thank you for this very nice proof! – Johann Cigler Jan 8 at 7:40