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T. Amdeberhan
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Usually, I like working with determinants related to the Vandermonde matrix, i.e. $$\det(x_j^{i-1})=\prod_{i<j}(x_j-x_i).$$ However, I run into some unusual matrix and its determinant. Define the $(2n)\times (2n)$ matrix $\mathbb{M}_{2n}$ with entries $$\mathbb{M}_{2n}(i,j)= \begin{cases} (x_j-x_i)(x_i-x_n) \qquad \text{if $i<j$}, \\ (x_j-x_i)(x_j-x_n) \qquad \text{if $i\geq j$}. \end{cases}$$$$\mathbb{M}_{2n}(i,j)= \begin{cases} (x_j-x_i)(x_i-x_{2n}) \qquad \text{if $i<j$}, \\ (x_j-x_i)(x_j-x_{2n}) \qquad \text{if $i\geq j$}. \end{cases}$$

QUESTION. Is this true? $$\det\mathbb{M}_{2n}=(x_1-x_2)^2(x_2-x_3)^2\cdots(x_{2n-1}-x_{2n})^2(x_{2n}-x_1)^2.$$

Usually, I like working with determinants related to the Vandermonde matrix, i.e. $$\det(x_j^{i-1})=\prod_{i<j}(x_j-x_i).$$ However, I run into some unusual matrix and its determinant. Define the $(2n)\times (2n)$ matrix $\mathbb{M}_{2n}$ with entries $$\mathbb{M}_{2n}(i,j)= \begin{cases} (x_j-x_i)(x_i-x_n) \qquad \text{if $i<j$}, \\ (x_j-x_i)(x_j-x_n) \qquad \text{if $i\geq j$}. \end{cases}$$

QUESTION. Is this true? $$\det\mathbb{M}_{2n}=(x_1-x_2)^2(x_2-x_3)^2\cdots(x_{2n-1}-x_{2n})^2(x_{2n}-x_1)^2.$$

Usually, I like working with determinants related to the Vandermonde matrix, i.e. $$\det(x_j^{i-1})=\prod_{i<j}(x_j-x_i).$$ However, I run into some unusual matrix and its determinant. Define the $(2n)\times (2n)$ matrix $\mathbb{M}_{2n}$ with entries $$\mathbb{M}_{2n}(i,j)= \begin{cases} (x_j-x_i)(x_i-x_{2n}) \qquad \text{if $i<j$}, \\ (x_j-x_i)(x_j-x_{2n}) \qquad \text{if $i\geq j$}. \end{cases}$$

QUESTION. Is this true? $$\det\mathbb{M}_{2n}=(x_1-x_2)^2(x_2-x_3)^2\cdots(x_{2n-1}-x_{2n})^2(x_{2n}-x_1)^2.$$

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T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

A determinant of perfect square polynomials

Usually, I like working with determinants related to the Vandermonde matrix, i.e. $$\det(x_j^{i-1})=\prod_{i<j}(x_j-x_i).$$ However, I run into some unusual matrix and its determinant. Define the $(2n)\times (2n)$ matrix $\mathbb{M}_{2n}$ with entries $$\mathbb{M}_{2n}(i,j)= \begin{cases} (x_j-x_i)(x_i-x_n) \qquad \text{if $i<j$}, \\ (x_j-x_i)(x_j-x_n) \qquad \text{if $i\geq j$}. \end{cases}$$

QUESTION. Is this true? $$\det\mathbb{M}_{2n}=(x_1-x_2)^2(x_2-x_3)^2\cdots(x_{2n-1}-x_{2n})^2(x_{2n}-x_1)^2.$$