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Let $x_1,x_2,\ldots,x_n$ be $n$ real numbers that satisfy $x_1. Define \begin{equation*} A=% \begin{bmatrix} 0 & x_{2}-x_{1} & \cdots & x_{n-1}-x_{1} & x_{n}-x_{1} \\ x_{2}-x_{1} & 0 & \cdots & x_{n-1}-x_{2} & x_{n}-x_{2} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ x_{n-1}-x_{1} & x_{n-1}-x_{2} & \cdots & 0 & x_{n}-x_{n-1} \\ x_{n}-x_{1} & x_{n}-x_{2} & \cdots & x_{n}-x_{n-1} & 0% \end{bmatrix}% \end{equation*}

Could you determine the determinant of $A$ in term of $x_1,x_2,\ldots,x_n$?

I make a several Calculation: For $n=2$, we get

\begin{equation*} A=% \begin{bmatrix} 0 & x_{2}-x_{1} \\ x_{2}-x_{1} & 0% \end{bmatrix}% \text{ and}\det (A)=-\left( x_{2}-x_{1}\right) ^{2} \end{equation*}

For $n=3$, we get

\begin{equation*} A=% \begin{bmatrix} 0 & x_{2}-x_{1} & x_{3}-x_{1} \\ x_{2}-x_{1} & 0 & x_{3}-x_{2} \\ x_{3}-x_{1} & x_{3}-x_{2} & 0% \end{bmatrix}% \text{ and}\det (A)=2\left( x_{2}-x_{1}\right) \left( x_{3}-x_{2}\right) \left( x_{3}-x_{1}\right) \end{equation*}

For $n=4,$ we get

\begin{equation*} A=% \begin{bmatrix} 0 & x_{2}-x_{1} & x_{3}-x_{1} & x_{4}-x_{1} \\ x_{2}-x_{1} & 0 & x_{3}-x_{2} & x_{4}-x_{2} \\ x_{3}-x_{1} & x_{3}-x_{2} & 0 & x_{4}-x_{3} \\ x_{4}-x_{1} & x_{4}-x_{2} & x_{4}-x_{3} & 0% \end{bmatrix} \\% \text{ and} \\ \det (A)=-4\left( x_{4}-x_{1}\right) \left( x_{2}-x_{1}\right) \left( x_{3}-x_{2}\right) \left( x_{4}-x_{3}\right) \end{equation*} Finally, I guess that the answer is $\det(A)=2^{n-2}\cdot (x_n-x_1)\cdot (x_2-x_1)\cdots (x_n-x_{n-1})$. But I don't know how to prove it.

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    Thanks J.M. I will learn it2012-05-14

2 Answers 2

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Clearly the determinant is $0$ if $x_i = x_{i+1}$ (because two adjacent rows are identical) or $x_1 = x_n$ (last row is $-$ first row). So the determinant must be a polynomial divisible by $(x_1 - x_2)(x_2 - x_3) \ldots (x_{n-1} - x_n)(x_n - x_1)$. But the determinant has degree $n$, so it is a constant times this product. To determine what the constant is, you might try a special case: $x_i = i$.

EDIT: Thanks to J.M.'s remark, you can show that in that special case the inverse of your matrix $A_n$ looks like this:

$ \pmatrix{ -\frac{1}{2}+\frac{1}{2n-2} & \frac{1}{2} & 0 & 0 & \ldots & 0 & \frac{1}{2n-2}\cr \frac{1}{2} & -1 & \frac{1}{2} & 0 & \ldots & 0 & 0\cr 0 & \frac{1}{2} & -1 & \frac{1}{2} & \ldots & 0 & 0\cr \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \cr 0 & 0 & 0 & 0 & \ldots & -1 & \frac{1}{2}\cr \frac{1}{2n-2} & 0 & 0 & 0 & \ldots & \frac{1}{2} & -\frac{1}{2} + \frac{1}{2n-2}\cr}$ where the elements on the main diagonal are all $-1$ except for the first and last, those just above and below the diagonal are all $1/2$, the top right and bottom left are $1/(2n-2)$, and everything else is $0$.

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    Somewhat apropos: Yueh, in [these](http://www.math.nthu.edu.tw/~amen/2006/05-03-03-6.pdf) [papers](http://www.math.nthu.edu.tw/~amen/2005/040903-7.pdf) discusses tridiagonal matrices similar to the inverse Robert obtained in this answer. I suspect another proof of the determinant's evaluation can be done based on these (e.g. by expressing the determinant as a product of trigonometric functions of angles in progression).2012-05-14
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Expanding Robert solution.

Let $det(A) = P(x)$. Let the polynomial on the right is a multi-variable polynomial $P(x)$.

If $x_1 = x_2$ then $det(A) = 0$ i.e $P(x) = 0$ i.e. $(x_1 - x_2)$ is a factor of $P(x)$.

If $x_2 = x_3$ then $det(A) = 0$ i.e $P(x) = 0$ i.e. $(x_2 - x_3)$ is a factor of $P(x)$.

etc. We calculate possible factors of $P(x)$. Have we calculated all possible factors of $P(x)$?

Let $Q(x) = (x_1 - x_2) (x_2 - x_3) \ldots (x_{n} - x_{1}) $

What we know about the degree of $P(x)$? It is $n$, equal to that of $Q(x)$. Thus $Q(x)$ multiplied by some constant factor should give us $P(x)$ i.e. we already have all possible factors of $P(x)$.

A Robert has already mentioned, we should calculate this constant factor.

It follows that if for any $i$, $x_i = x_{i+1}$, then $P(x) =0$ i.e. $det(A) = 0$. Since you alretady have constraints such as $x_1 >x_2 \ldots x_n$, $det(A) \ne 0$.

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    It is not so obvious that the constant is not $0$.2012-05-14