show that the characteristic polynomial of this matrix has negative coefficients Let $n\geq 2$,  $A$ be the $n\times n$ matrix $A=(a_{ij})$ where $a_{ij}=\max(i,j)$.
Can anybody show that the characteristic polynomial
$P(x)=\det(xI-A)$ has all its coefficients negative except the leading one?
I have checked this for $2 \leq n \leq 20$.
 A: Here is a sketch. The calculations that need to be filled in are easy to do, but please let me know if you'd like me to edit it for more detail. 
We have:

$$A^{-1}=\begin{bmatrix} -1 & 1 & 0 & 0 & 0 & \cdots & 0& 0 & 0\\1 & -2 & 1 & 0&0&\cdots & 0& 0 & 0\\ 0 & 1 & -2 & 1 & 0 & \cdots & 0& 0 & 0\\\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\0 & 0 & 0 &0  & 0 & \cdots & 1& -2 & 1\\0 & 0 & 0 &0  & 0 & \cdots & 0& 1 & \frac{1-n}{n}
\end{bmatrix}.$$

The important point here is that $A^{-1}$ is a tridiagonal matrix and so the recursion to find the determinant is nice. In particular, we have $$\det(xI-A^{-1}) = \left(x+\frac{n-1}{n}\right)f_{n-1}(x)-f_{n-2}(x),$$ where $f_m(x)$ is the characteristic polynomial of the $m\times m$ leading principle submatrix. Note that all of these have the same form and so the recursive formula for the determinant yields the following, which is also straightforward to check by induction.

For $1\leq m\leq n-1$,  we have $$f_m(x) = \sum_{k=0}^m{m+k\choose m-k}x^k.$$

A little work now shows:

$$\det(xI-A^{-1}) = x^n + \sum_{k=1}^{n-1}\left({n-1+k\choose n-k}-\frac{1}{n}{n-1+k\choose n-1-k}\right)x^k - \frac{1}{n}.$$ 

All coefficients besides the constant one are positive. This implies your desired fact using the well-known relation between the characteristic polynomial of $A$ and $A^{-1}$.
A: Maybe something can be done with this.  It looks to me like 
$$P(t) = \dfrac{\sqrt{4t+1}-1-2n}{2\sqrt{4t+1}} \left(t+\frac{1}{2} + \frac{\sqrt{4t+1}}{2}\right)^n + \dfrac{\sqrt{4t+1}+1+2n}{2\sqrt{4t+1}} \left(t+\frac{1}{2} - \frac{\sqrt{4t+1}}{2}\right)^n$$
