Determinant of a symmetric Hankel matrix of order $2018$

Question

Find out the determinant of the following matrix $$\begin{bmatrix}1^{2016} & 2^{2016} & \dots & 2018^{2016}\\2^{2016} & 3^{2016} & \dots & 2019^{2016} \\ \vdots & \vdots & \ddots & \vdots \\ 2018^{2016} & 2019^{2016} & \dots & 4035^{2016}\end{bmatrix}$$

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Through examples of order $2\times 2$ as $\displaystyle \begin{bmatrix}1^0 & 2^0\\2^0 & 3^0\end{bmatrix}$ and $4\times 4$ as $$\displaystyle \begin{bmatrix}1^2 & 2^2 & 3^2 & 4^2\\2^2 & 3^2 & 4^2 & 5^2\\3^2 & 4^2 & 5^2 & 6^2\\4^2 & 5^2 & 6^2 & 7^2\end{bmatrix}$$ I found that in both case the answer is $0$. But I want to know the procedure to find such determinant.

Answer 1

So, what we do is the following : The above matrix $M$, seems to be given by $M_{ij} = (i+j-1)^{2016}$.

Expanding this binomially, we get $$M_{ij} = \sum_{x = 0}^{2016} \binom {2016}x (i-1)^{2016-x} j^{x}$$

Now, define the $2017$ column vectors $C_a$, $0 \leq a \leq 2016$, by $C_a = [1^a,2^a,...,2018^a]^T$. I claim that every column of $M_{ij}$ is spanned by these column vectors.

This happens, because given a column $M_{i}$, we can see from the above formula that: $$ M_i = \sum_{x=0}^{2016}\left(\binom{2016}{x}(i-1)^{2016-x}\right)C_x $$

For every $i$. I urge you to check this yourself.

This shows that $M$ has rank $2017$, since $0 \leq a \leq 2016$. However, the matrix you have given is square of dimension $2018$. It follows that it has non-trivial kernel, hence is not injective, hence has determinant zero.

Answer 2

Suppose $p_1, p_2...,p_n $ are polynomials of degree $m-1 $ with atleast one has degree exactly $m-1$ . Let us consider a matrix $A$ whose $(i,j)-$th matrix has entry $p_i(u_j)$ , where $u_1,u_2,\dots,u_n$ are $ n$ distinct number

let $p_i(x) = a_{i,1} + a_{i,2} x + \dots +a_{i,m}x^{m-1} $

Now we can easily show that $A=\displaystyle \begin{bmatrix}a_{1,1} & a_{1,2} \dots &a_{1,m}\\ . & \dots & .\\. & \dots & .\\a_{n,1} & a_{1,2} \dots &a_{n,m}\end{bmatrix}\displaystyle \begin{bmatrix}1 & 1 & ..... & 1\\u_1 & u_2 & \dots & u_n\\. & .& \dots & .\\. & .& ..... & .\\u_1^{m-1} & u_2^{m-1} & \dots & u_n^{m-1}\end{bmatrix}$,

This show that in particular if max $\deg(p_i) = m_i < n$ , (this mean , the first matrix is of size $n \times m$ and$ m < n$ ) , then $\operatorname{rank} (A) < n$ and so determinant is $0$

For the given matrix $p_i(x) = (i-1 +x)^{2016}$ and points are $ 1,\dots, 2018$. Since the matrix is of size $2018 \times 2018$, the determinant is $0$

Answer 3

For the Hankel matrix $H$, the $(i,j)$-th element can be expressed (using the Binomial theorem) as

$H(i,j)=f(i+j-1)=(i+j-1)^{2016}$ $=i^{2016} + {2016 \choose 1}i^{2015}(j-1)^1+{2016 \choose 2}i^{2014}(j-1)^2+\ldots+{2016 \choose 2016}(j-1)^{2016}$

$=\underbrace{\begin{bmatrix}i^{2016} & {2016 \choose 1}i^{2015} & {2016 \choose 2} i^{2014}& \ldots & {2016 \choose 2016}i^{0}\end{bmatrix}}_{u_i^T}\underbrace{\begin{bmatrix}1 \\ (j-1) \\ (j-1)^2 \\ \ldots \\ (j-1)^{2016}\end{bmatrix}}_{v_j}$

$=u_i^Tv_j$ (where $u_i$ and $v_j$ are $2017$-dimensional vectors)

Define the matrices $U\in \mathbb R^{2017\times 2}$ and $V\in \mathbb R^{2\times 2017}$ as:

$U=\begin{bmatrix} u_1^ T\\ u_2^ T\\ \vdots \\ u_{2018}^ T \end{bmatrix}$ and $V=\begin{bmatrix} v_1 & v_2 & \dots & v_{2018} \end{bmatrix}$

so that

$UV$

$=\begin{bmatrix} 1^{2016} & {2016 \choose 1}1^{2015} & {2016 \choose 2} 1^{2014}& \ldots & {2016 \choose 2016}1^{0} \\ 2^{2016} & {2016 \choose 1}2^{2015} & {2016 \choose 2} 2^{2014}& \ldots & {2016 \choose 2016}2^{0} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 2018^{2016} & {2016 \choose 1}2018^{2015} & {2016 \choose 2} 2018^{2014}& \ldots & {2016 \choose 2016}2018^{0} \end{bmatrix}.\begin{bmatrix}1 & 1 & 1 & \ldots & 1 \\ 0 & 1 & 2 & \ldots & 2018 \\ 0^2 & 1^2 & 2^2 & \ldots & 2018^2 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0^{2016} & 1^{2016} & 2^{2016} & \ldots & 2018^{2016} \end{bmatrix}$

$=H$

Now, $\mathrm{rank}(H)= \mathrm{rank}(UV) \leq \min(\mathrm{rank}(U), \mathrm{rank}(V)) = 2017$ (from here) and $H \in \mathbb R^{2018\times 2018} \implies H$ is not a full-rank matrix (with non-trivial nullspace) $\implies \det(H)=0$.