How find this matrix $A=(\sqrt{i^2+j^2})$ eigenvalue let the matrix 
$$A=(a_{ij})_{n\times n}$$
where
$$a_{ij}=\sqrt{i^2+j^2}$$
Question:
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
 A: 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.
A: 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.
