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From my linear algebra class assignments:

Evaluate the determinant of the matrix shown below $$\begin{pmatrix} 1^{2023} & 2^{2023} & \cdots & 2025^{2023} \\ 2^{2023} & 3^{2023} & \cdots & 2026^{2023} \\ \vdots & \vdots & \ddots & \vdots \\ 2025^{2023} & 2026^{2023} & \cdots & 4049^{2023}\end{pmatrix}$$ where the bases of entries in each row and column form an arithmetic progression with $d=1$.

My initial attempt was to do row reduction, but soon realized it's not feasible because the entries have rather weird exponent $2023$. So I turn to other constructive methods because this matrix looks carefully crafted with the progression pattern hidden in it. But so far I still could not crack it, and from the beauty test of hardy I conceive the answer would be very simple like $1$ or $0$ cuz the entries only involve integers. So could anyone provide some illuminating ideas?

Rodrigo de Azevedo
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user3835
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    As a general tip with problems like this, the numbers $2023$ and $2025$ can safely be assumed to be irrelevant and can be replaced with two variables. Then you can set those variables to smaller numbers, like $2$ or $3$, where you can figure the answer out by hand and guess the pattern. Then you can try to prove the pattern. – Qiaochu Yuan Apr 16 '25 at 05:53
  • "... the rows and columns form arithmetic progression with $d=1$" No, they don't. The entires are of the form $a_i^{2023}$, and the $a_i$ in each row and column form arithmetic progressions with $d=1$, but the $a_i^{2023}$ most definitely do not form an arithmetic progression. – Gerry Myerson Apr 16 '25 at 06:15
  • Sorry for the misleading expression, I have edited it to the base of entries. – user3835 Apr 16 '25 at 06:30
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    Another one of the same kind: https://math.stackexchange.com/q/2417289/42969 – Martin R Apr 16 '25 at 08:47
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    I may add that the first answer in math.stackexchange.com/q/2417289 settled this kind of problem completely because the exponent $2023$ and the size $2025$ in the matrix are not essential, as long as the size is at least bigger than the exponent by $2$. You sort of just replace the first row by polynomials $$x^{2023},(x+1)^{2023},\cdots,(x+2024)^{2023}$$, expand the determinant and notice that the result is a polynomial of degree $2023$ vanishing on $2024$ numbers $2,3,\cdots,2025$.Therefore, it's a zero polynomial by polynomial method. – user3835 Apr 16 '25 at 09:28

2 Answers2

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If $k\ge0$ and $n\ge k+2$, then the $n\times n$ determinant $$\left|\matrix{1^k&2^k&3^k&\cdots&j^k&\cdots&n^k\cr 2^k&3^k&4^k&\cdots&(j+1)^k&\cdots&(n+1)^k\cr 3^k&4^k&5^k&\cdots&(j+2)^k&\cdots&(n+2)^k\cr \vdots&&&&\vdots&&\vdots\cr (k+2)^k&(k+3)^k&(k+4)^k&\cdots&(j+k+1)^k&\cdots&(n+k+1)^k\cr \vdots&&&&\vdots&&\vdots\cr}\right|$$ is zero. Proof. The polynomials $$x^k\,,\ (x+1)^k\,,\ (x+2)^k\,,\ldots,\ (x+k)^k$$ are linearly independent. These polynomials therefore form a basis for the $(k+1)$-dimensional vector space $P_k$ (polynomials of degree up to $k$), and hence $$(x+k+1)^k=\alpha_0x^k+\alpha_1(x+1)^k+\alpha_2(x+2)^k+\cdots+\alpha_k(x+k)^k$$ for some scalars $\alpha_j$. So row $k+2$ is a linear combination of the first $k+1$ rows, and the determinant is zero.

David
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  • Thanks for your response, I am still grappling on your statement that the list of polynomials $$x^{k},(x+1)^{k},(x+2)^{k},\cdots,(x+k)^{k}$$ form a basis of the vector space $P_{k}$ and trying to formulate a rigorous proof. – user3835 Apr 16 '25 at 07:31
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    Let $\sum_{j=0}^k \alpha_j(x+j)^k=\bf0$. Differentiating $k$ times (and cancelling a common constant each time), then substituting $x=1$ (or in fact any constant value) gives a $k+1$ by $k+1$ system of equations for the $\alpha_j$. The coefficients form a Vandermonde matrix with non-zero determinant. See if you can fill in the details. – David Apr 16 '25 at 23:34
  • Yeah, I get your point, thanks again! – user3835 Apr 17 '25 at 02:27
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For $(n+1)$ order matrix $\left.A=\left( \begin{array} {cccc}(a_0+b_0)^m & (a_0+b_1)^m & \cdots & (a_0+b_n)^m \\ (a_1+b_0)^m & (a_1+b_1)^m & \cdots & (a_1+b_n)^m \\ \vdots & \vdots & & \vdots \\ (a_n+b_0)^m & (a_n+b_1)^m & \cdots & (a_n+b_n)^m \end{array}\right.\right);$

According to the Binomial Theorem, we can got

$\left.A=\left( \begin{array} {ccccc}a_0^m & C_m^1a_0^{m-1} & \cdots & C_m^m \\ a_1^m & C_m^1a_1^{m-1} & \cdots & C_m^m \\ \vdots & \vdots & & \vdots \\ a_n^m & C_m^1a_n^{m-1} & \cdots & C_m^m \end{array}\right.\right) \begin{pmatrix} 1 & 1 & \cdots & 1 \\ b_0 & b_1 & \cdots & b_n \\ \vdots & \vdots & & \vdots \\ b_0^m & b_1^m & \cdots & b_n^m \end{pmatrix}=BC.$

$B$ is $(n+1)\times(m+1)$ matrix, $C$ is $(m+1)\times(n+1)$ matrix.

If $n>m$,then

$\left.A=\left( \begin{array} {ccccc}a_0^m & C_m^1a_0^{m-1} & \cdots & C_m^m & 0 & \cdots & 0 \\ a_1^m & C_m^1a_1^{m-1} & \cdots & C_m^m & 0 & \cdots & 0 \\ \vdots & \vdots & & \vdots & \vdots & & \vdots \\ a_n^m & C_m^1a_n^{m-1} & \cdots & C_m^m & 0 & \cdots & 0 \end{array}\right.\right) \begin{pmatrix} 1 & 1 & \cdots & 1 \\ b_0 & b_1 & \cdots & b_n \\ \vdots & \vdots & & \vdots \\ b_0^m & b_1^m & \cdots & b_n^m \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & 0 \end{pmatrix}=\widetilde{B}\widetilde{C}.$

$|A|=|\widetilde{B}\widetilde{C}|=|\widetilde{B}| \cdot |\widetilde{C}|=0\cdot 0=0.$

If $n=m$,then by using of Vandermonde matrix, we got

$|A|=|B| \cdot |C|=(-1)^{\frac{n(n+1)}{2}}C_{n}^{1}C_{n}^{2}\cdots C_{n}^{n}\prod\limits_{0\leq j<i\leq n}(a_{i}-a_{j})(b_{i}-b_{j}).$

In your question,$n+1=2025,n=2024>2023=m$, So $|A|=0.$

anyon
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