4
$\begingroup$

Let $x_1,x_2,\ldots,x_n$ be $n$ real numbers that satisfy $x_1

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.

  • 5
    Please remember that in order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many users find the use of the imperative ("Prove", "Show", etc) to be rude when asking for help. Please consider rewriting your post.2012-05-14
  • 0
    Your guess would not account for the case $n=2$...2012-05-14
  • 0
    For $n=2$ is trivial. Maybe for $n>2$2012-05-14
  • 0
    I try math induction, but don't know how to connect n by n matrix with (n+1) by (n+1) matrix2012-05-14
  • 1
    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](http://math.stackexchange.com/questions/44511), and 3. have inverses that are cyclic tridiagonal (tridiagonal with extra elements in the upper right and lower left corners).2012-05-14
  • 0
    Thanks J.M. I will learn it2012-05-14

2 Answers 2

7

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 $0$.

  • 0
    Note that $x_i2012-05-14
  • 1
    @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?2012-05-14
  • 0
    I just want to prove that A is invertible2012-05-14
  • 0
    @robert: How you can conclude that the determinant must be a polynomial divisible by (x1−x2)(x2−x3)…(xn−1−xn)(xn−x1)2012-05-14
  • 0
    @Robert How did you come up with this solution? It is a standard practice in linear algebra? I must say it is truly ingenious if one is not aware of it. Although once told, it looks pretty obvious.2012-05-14
  • 0
    @Dilawar : Please explain to me I'm still don't understand Robert solution2012-05-14
  • 0
    @tes If $x_1 = x_2$ then determinant of your matrix goes to 0. Imagine a polynomial of determinant. Surely this polynomial should also go to zero if $x_1 = x_2$ i.e. you have a factor $(x_1 - x_2)$. Following the same logic, you also conclude that $(x_2 - x_2)$ ... are also factors. Since degree of polynomial is $n$, you conclude that there can not be any more factors contributing to the power of x. What is left is a constant factor.2012-05-14
  • 0
    @Dilawar: But in what variable?2012-05-14
  • 0
    @tes: The determinant is a polynomial in the matrix elements, and thus in $x_1, \ldots, x_n$. If it is zero whenever $x_1 = x_2$, it must be divisible (as a polynomial in $x_1$ over the field of rational functions in $x_2,\ldots,x_n$) by $x_1 - x_2$, and it's not hard to see that the quotient must be a polynomial in $x_1, \ldots, x_n$. Repeat for for each of the other factors.2012-05-14
  • 0
    @RobertIsrael: Thanks, i will try to work on it2012-05-14
  • 1
    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
0

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 $x_1 >x_2 \ldots x_n$, $det(A) \ne 0$.

  • 0
    Q(x)=(x1−x2)(x2−x3)…(xn−xn−1), where is the x?2012-05-14
  • 0
    @tes Messed it up. Let me clear it.2012-05-14
  • 1
    It is not so obvious that the constant is not $0$.2012-05-14