The $n$-order tridiagonal matrix is defined by $A(i,i)=0, A(i,i+1)=n-i, A(i+1,i)=i$, i.e.,
$$A=\begin{pmatrix}0&n-1\\1&0&n-2\\&2&0&n-3\\&&\ddots&\ddots&\ddots\\&&&n-2&0&1\\&&&&n-1&0\end{pmatrix}$$
All eigenvalues of $A$ are $-(n-1),-(n-3),\ldots,(n-3),(n-1)$, however, I could not figure out a simple way to show that.