# Help with this determinant

I need help to calculate this determinant: $$\begin{vmatrix} 1^k & 2^k & 3^k & \cdots & n^k \\ 2^k & 3^k & 4^k & \cdots & (n+1)^k\\ 3^k & 4^k & 5^k & \cdots & (n+2)^k\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ n^k & (n+1)^k & (n+2)^k & \cdots & (2n-1)^k \end{vmatrix}$$ Where $2\leq n$ and $0\leq k \leq n-2.$ I did the $2\times 2$, $3\times 3$ and $4\times 4$ cases but I couldn't find a pattern to follow.
I did those cases by making zeros in the first column (except the $1^k$ in the first row) and then using the Laplace expansion.
The $2\times 2$ case equals: $3^k -4^k$, the $3\times 3$ case equals: $15^k -16^k -20^k -27^k +24^k +24^k$

• Welcome to MathSE. Its always advisable to show what you have done so far or any thoughts on the problem. – Naive Dec 10 '17 at 17:45
• en.wikipedia.org/wiki/Circulant_matrix – openspace Dec 10 '17 at 17:54
• You can't find a path probably because you use numbers. Go directly to the symbolic calculation, i.e. start with $$\begin{pmatrix} n^k & (n+1)^k \\ (n+1)^k & (n+2)^k \end{pmatrix}$$ Then the $3x3$ case and so on. Maybe you'll find a path. Openspace's link does not work because this is not a circular matrix. – Von Neumann Dec 10 '17 at 18:27
• This is somewhat similar question in the case $k=n-1$: Can we determine the determinant?. – Martin Sleziak Dec 10 '17 at 18:51
• If $0 \le k \le n-2$, then the determinant is $0$. – achille hui Dec 10 '17 at 19:13

For any $n \ge 1$, let $u_1, \ldots, u_n$ and $v_1,\ldots ,v_n$ be any elements in any commutative ring.

For any $0 \le k \le n-2$, consider the $n \times n$ matrix $A = (a_{ij})$ whose entry at row $i$ and column $j$ has the form: $$a_{ij} = (u_i + v_j)^k$$

By binomial theorem, it can be decomposed as

$$a_{ij} = \sum_{\ell=0}^k \binom{k}{\ell} u_i^\ell v_j^{k-\ell}$$

This means $A$ can be rewritten as a sum of $k+1$ matrix products of column vectors

$$A = \sum_{\ell=0}^k \binom{k}{\ell} U_\ell V_{k-\ell}^T \quad\text{ where }\quad U_\ell = \begin{bmatrix}u_1^\ell \\ u_2^\ell \\ \vdots \\ u_n^\ell\end{bmatrix}\text{ and } V_\ell = \begin{bmatrix}v_1^\ell \\ v_2^\ell \\ \vdots \\ v_n^\ell\end{bmatrix}. \tag{*1}$$

Since matrix product of a pair of column vectors has rank at most $1$ and matrix rank are sub-additive, i.e. ${\rm rank}(P + Q) \le {\rm rank}(P) + {\rm rank}(Q)$, we get

$${\rm rank}(A) \le \sum_{\ell=0}^k {\rm rank}(U_\ell V_\ell^T) \le \sum_{\ell=0}^k 1 = k+1 < n$$

As a result, matrix $A$ doesn't have full rank and hence $\det A = 0$.

For the problem at hand, take $u_i = i$ and $v_j = j-1$. When $0 \le k \le n - 2$, the determinant at hand vanishes.

Update

If one doesn't want to use the concept of rank, there is another approach.

Construct three auxiliary $n \times n$ matrices by

• $U$, whose $\ell^{th}$ column is the column vector $U_{\ell-1}$ for $1 \le \ell \le k+1$ and zero otherwise.
• $V$, whose $\ell^{th}$ column is the column vector $V_{k-\ell+1}$ for $1 \le \ell \le k+1$ and zero otherwise.
• $D$, a diagonal matrix whose $\ell^{th}$ diagonal entry is $\binom{k}{\ell-1}$ for $1 \le \ell \le k+1$ and zero otherwise.

The decomposition of $(*1)$ can be rewritten as $A = UDV^T$.

When $k < n - 2$, $U$ contains a zero column and hence

$$\det U = 0\quad\implies\quad\det A = \det U \det D \det V = 0$$

As a side benefit, this approach allow us to compute the determinant at hand when $k = n - 1$.

When $u_i = i$ and $v_j = j - 1$, $U$ is a Vandermonde matrix with determinant $$\det U = \prod_{1\le i < j \le n}(u_i - u_j) = \prod_{1\le i < j \le n}(i-j) = (-1)^{n(n-1)/2} \prod_{\ell=1}^{n-1} \ell!$$ $V$ can be obtained from a Vandermonde matrix but reversing the order of its columns. It is easy to see its determinant equals to $$\det V = \prod_{1 \le i < j \le n }(v_j - v_i) = \prod_{1 \le i < j \le n}(j-i) = \prod_{\ell=1}^{n-1} \ell!$$ The determinant of $D$ is easy, it is just the product of a bunch of binomial coefficients. Combine these, when $k = n - 1$, the determinant at hand becomes

$$\left. \det A \right|_{k=n-1} = (-1)^{\frac{n(n-1)}{2}} \prod_{\ell=0}^{n-1}\binom{n-1}{\ell} \times \left(\prod_{\ell=1}^{n-1}\ell!\right)^2 = (-1)^{\frac{n(n-1)}{2}} ((n-1)!)^n$$

Another proof:

Consider the polynomials $P_1(X)=(X+1)^k,\; P_2(X)=(X+2)^k,\dots,\;P_n(X)=(X+n)^k$, which belong to the vector space $\mathbf R_k[X]$ of polynomials of degree at most $k$. This space has dimension $k+1$, hence, if $n>k+1$ (i.e. $k\le n-2$), these polynomials are linearly dependent, so there's a linear relation with coefficients $\lambda_i\enspace (1\le\lambda\le n)$, not all $0$, such that the polynomial $$Q(X)=\sum_{i=1}^{n}\lambda_iP_i(X)=0.$$ Now observe row n° $i$ in the determinant is just $\;\bigl(P_1(i-1), P_2(i-1), \dots P_n(i-1)\bigr)$, so $Q(i-1)=0$ for all $i=1,\dots,n.\,$ If $C_i$ denotes the $i$-th column of the determinant, this shows we have the following linear relation between columns: $$\sum_{i=1}^{n}\lambda_i\,C_i=0,$$ hence the determinant is $0$, under the assumption $k\le n-2$.

Another proof.

This involves some tricks on manipulating polynomials.

$\mathit{Proof}.\blacktriangleleft$ Consider $$f(z) = \begin{vmatrix} (z+1)^k & (z+2)^k & (z+3)^k & \cdots & (z+n)^k\\ 2^k & 3^k & 4^k & \cdots & (n+1)^k\\ 3^k & 4^k & 5^k & \cdots & (n+2)^k \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ n^k & (n+1)^k & (n+2)^k & \cdots & (2n-1)^k \end{vmatrix}.$$ Then the computational definition determinant, i.e. $$\det(\boldsymbol A) = \sum_{\sigma \in \mathfrak S_n} \mathrm{sgn}(\sigma) \prod_1^n a_{\sigma(j), j}$$ implies that $f(z)$ is a polynomial of degree $k$. The goal is to compute $f(0)$. Note that $f(1) = f(2) = \cdots = f(n-1) = 0$, since the $1^{\mathrm {st}}$ row coincides with the $(j+1 )^{\mathrm {st}}$ row, and a determinant with duplicated rows is $0$. Therefore $f(z)$ is a real polynomial with $n-1$ zeros while its degree is $k \leqslant n-2$. Thus the polynomial is just the zero polynomial, and the original determinant is $f(0) = 0$. $\blacktriangleright$