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

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.
