Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $n\ge 3$ be a natural number, and for $1\leq k\leq n-2$ consider the $n\times n$ matrix

$$A_{n,k}=\begin{pmatrix} 1^{k}& 2^{k}& \cdots& n^{k}\\ (n+1)^k& (n+2)^k& \cdots& (2n)^k\\ \vdots& \vdots& & \vdots\\ (n^2-n+1)^k& (n^2-n+2)^k& \cdots& n^{2k} \end{pmatrix}_{\large{.}}$$

It seems that $\det (A_{n,k})=0$. I can prove this for $n=3$ and $k=1$, and for all $n\ge 4$ and $k=1,2.$ But my ways do not generalize to higher values of $k$.

Is there a simple proof that $\textbf{$\det(A_{n,k})=0$?}$

share|cite|improve this question
What is the exact Formula for $(A_{n,k})_{ij}$ ? – AlexR Aug 22 '13 at 19:57

Notice that every row of the matrix is a linear combination of rows $(1^s,2^s,\dots,n^s)$, $0\leq s\leq k$ (just expand using binomial formula). So they must be linearly dependent.

share|cite|improve this answer

Here's a different way to look at it and generalize it. Let's start with an easier problem: If instead we take a matrix of the first $n^2$ Fibonacci numbers (n>2) we also get a matrix with determinant 0. Why is this true? The third column will just be the sum of the first two.

Well if we replace the Fibonacci numbers by any sequence $A_i$ defined recursively where each term is a fixed linear combination of the previous m terms, then the same argument shows that a matrix filled with the first $n^2$ (with $n>m$) elements of this sequence will have determinant 0.

Now back to your problem, it would be enough to see that the sequence $1^k, 2^k, 3^k,...$ satisfies a fixed linear recurrence relation based on the previous k+1 terms. Well it just so happens that $a^k = {{k+1}\choose{1}}(a-1)^k - {{k+1}\choose{2}}(a-2)^k + {k+1\choose 3}(a-3)^k - ... \pm (a - k-1)^k $ for all $a$. (I didn't just pull that recurrence out of nowhere, there is a lot of great structure in the set of sequences satisfying some linear recurrence that one can exploit to find such relations)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.