let the matrix $$A=(a_{ij})_{n\times n}$$ where $$a_{ij}=\sqrt{i^2+j^2}$$


Find the difference $sign{(A)}$

can see this define:http://en.wikipedia.org/wiki/Sylvester's_law_of_inertia

My try: consider the $$|\lambda I-A|=\begin{vmatrix} \lambda-\sqrt{2}&-\sqrt{3}&\cdots&-\sqrt{1^2+n^2}\\ -\sqrt{3}&\lambda-\sqrt{8}&\cdots&-\sqrt{n^2+2^2}\\ \cdots&\cdots&\cdots\cdots\\ -\sqrt{n^2+1}&-\sqrt{n^2+2}&\cdots&\lambda-\sqrt{n^2+n^2} \end{vmatrix}$$ and I found this determinant is not easy,

maybe we can consider this characteristic polynomial.and this problem is from china hard linear algebra book problem Thank you

  • 5
    $\begingroup$ Is there any reason to expect there to be a nice answer? E.g., is this from a contest or homework? Even the eigenvalues for the $3\times 3$ case are horrible to write down. In any case, I think your $(2,1)$ and $(1,2)$ entries should be $-\sqrt{5}$, no? $\endgroup$
    – Casteels
    Dec 9, 2013 at 12:25
  • $\begingroup$ Hell0,I have edit this problem,sorry $\endgroup$
    – math110
    Dec 11, 2013 at 13:52
  • $\begingroup$ @Casteels Note that the question is only about the signature of the quadratic form, not the eigenvalues. There may be no nice closed form for the eigenvalues or the characteristic polynomial, but perhaps there is a clever argument that gives the signature. $\endgroup$ Dec 12, 2013 at 3:47
  • $\begingroup$ @EwanDelanoy,yes,I think we can consider the characteristic polynomial. $\endgroup$
    – math110
    Dec 12, 2013 at 6:11
  • $\begingroup$ @EwanDelanoy OP asked for the eigenvalues in the original question. $\endgroup$
    – Casteels
    Dec 12, 2013 at 8:37

2 Answers 2


As David Speyer's numerical experiments suggest, the quadratic form associated to this symmetric matrix is negative-definite on the hyperplane $c_1 + \cdots + c_n = 0$, and thus has signature $(1,n-1)$ because it is positive on the unit vectors. This is the special case $a_i = i^2$, $s = 1/2$ of the following result:

Proposition. Let $a_1,\ldots,a_n$ be distinct positive real numbers and $s \in (0,1)$. Then the quadratic form $$ Q(c_1,\ldots,c_n) = \sum_{i=1}^n\sum_{j=1}^n (a_i+a_j)^s c_i c_j $$ is negative-definite on the hyperplane $c_1 + \cdots + c_n = 0$.

Proof: We use the integral representation $$ a^s = \frac{s}{\Gamma(1-s)} \int_{x=0}^\infty (1-e^{-ax}) \, x^{-s} \frac{dx}{x}, $$ which holds for all $a>0$, and follows from the Gamma integral $\int_0^\infty e^{-ax} x^{-s} dx = \Gamma(1-s) \, a^{s-1}$ by integration by parts. It follows that $$ Q(c_1,\ldots,c_n) = \int_{x=0}^{\infty} \frac{s}{\Gamma(1-s)} (\,f(0)^2 - f(x)^2) \, x^{-s} \frac{dx}{x}$$ where $f(x) = \sum_{i=1}^n c_i e^{-a_i x}$. If $c_1+\cdots+c_n = 0$ then $f(0)=0$, and then the integrand is $-f(x)^2 \, x^{-s} dx/x$, which is everywhere $\leq 0$, and not identically zero unless $c_i=0$ for all $i$. Therefore $Q(c_1,\ldots,c_n) \leq 0$ with equality only at zero, QED.

  • $\begingroup$ That's already in the first paragraph: $Q$ is positive on any unit vector. So there are $1$-dimensional subspaces on which $Q$ is positive-definite, and once we've found an $n-1$ dimensional subspace on which it's negative-definite the signature must be $(1,n-1)$. $\endgroup$ Dec 15, 2013 at 6:11
  • $\begingroup$ Indeed, I missed that part, sorry for the stupid question. $\endgroup$ Dec 15, 2013 at 6:29
  • $\begingroup$ I think you dropped an integral sign from your last displayed equation; I edited it in for you. Some day, I have to learn the trick of seeing some function of $a$ and writing it as a useful integral of $e^{ax}$; it's the trick that broke open this question and mathoverflow.net/questions/84958 . Very slick! $\endgroup$ Dec 15, 2013 at 17:35
  • $\begingroup$ Thanks (though I'd have put the integral sign after the constant factor $s / \Gamma(1-s)$); I suppose in this forum your $\int_{x=0}^\infty$ is better than my $\int_0^\infty$. Yes, integral representations like this (also with elementary functions other than $e^{ax}$) are a useful technique; I've used this trick elsewhere here (#516263) and on Mathoverflow (e.g. #109185), and the auto-generated list of questions "Related" with this one includes the very similar #577197. $\endgroup$ Dec 15, 2013 at 23:27

Report on some basic experiments. For $n \leq 30$, there is one positive eigenvalue and all the others are negative.

I checked this with the following Mathematica command:

 mm[n_] := Table[Sqrt[i^2 + j^2], {i, 1, n}, {j, 1, n}]

 Table[Count[Sign[Eigenvalues[SetPrecision[mm[n], 50]]], 1], {n, 1, 30}]

SetPrecision tells Mathematica to treat the square roots as floating point numbers with $50$ decimal digits of accuracy. If you tell it to treat them as exact quantities, the computation times out; if you use the default accuracy it won't get the signs of the smallest eigenvalues right. The smallest eigenvalues here are around $10^{-30}$, so you need to be careful.

Probably the easiest way to prove this would be to exhibit an $n-1$ dimensional subspace on which this matrix is negative definite. I took my first guess, the span of the vectors $(1, -1,0,0,0,\ldots)$, $(0,1,-1,0,0,\ldots)$, $(0,0,1,-1,0,0,\ldots)$, ...

(* Change of basis matrix to the n-1 dimensional space. *)
ss[n_] := Table[If[i == j, 1, If[i == j + 1, -1, 0]], {i, 1, n}, {j, 1, n - 1}]

(* Quadratic form in the new basis. *)
qq[n_] := Transpose[ss[n]].mm[n].ss[n]

Table[PositiveDefiniteMatrixQ[SetPrecision[-qq[n], 50]], {n, 2, 30}]

For $n \leq 30$, the qudratic form is negative definite on this $n-1$ plane.

  • $\begingroup$ I think this can prove by hand.Thank you give this relsut $\endgroup$
    – math110
    Dec 12, 2013 at 6:10

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