Interesting Determinant

Let $x_1,x_2,\ldots,x_n$ be $n$ real numbers that satisfy $x_1<x_2<\cdots<x_n$. 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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Your guess would not account for the case $n=2$... –  Arturo Magidin May 14 '12 at 1:56
For $n=2$ is trivial. Maybe for $n>2$ –  tes May 14 '12 at 1:59
I try math induction, but don't know how to connect n by n matrix with (n+1) by (n+1) matrix –  tes May 14 '12 at 2:00
I'll just note that the matrices being considered by the OP, 1. are very similar to Cauchy matrices; 2. can be expressed as the difference of two type D matrices, and 3. have inverses that are cyclic tridiagonal (tridiagonal with extra elements in the upper right and lower left corners). –  Guess who it is. May 14 '12 at 2:45

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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Note that $x_i<x_{i+1}$ –  tes May 14 '12 at 2:03
@tes, you didn't use that hypothesis anywhere in your calculations, did you? So the formulas are correct no matter how the variables are related, aren't they? In particular, you're allowed to look at what happens if two are equal, as Robert has done, right? –  Gerry Myerson May 14 '12 at 2:24
I just want to prove that A is invertible –  tes May 14 '12 at 3:12
@robert: How you can conclude that the determinant must be a polynomial divisible by (x1−x2)(x2−x3)…(xn−1−xn)(xn−x1) –  tes May 14 '12 at 3:16
Somewhat apropos: Yueh, in these papers 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). –  Guess who it is. May 14 '12 at 18:04

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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Thank you very much –  tes May 14 '12 at 3:38
Q(x)=(x1−x2)(x2−x3)…(xn−xn−1), where is the x? –  tes May 14 '12 at 3:43
@tes Messed it up. Let me clear it. –  Dilawar May 14 '12 at 3:45
It is not so obvious that the constant is not $0$. –  Robert Israel May 14 '12 at 16:45