eigenvalues of a symmetric matrix I have a special $N\times N$ matrix with the following form. It is symmetric and zero row (and column) sums.
$$K=\begin{bmatrix}
k_{11} & -1 & \frac{-1}{2} & \frac{-1}{3} & \frac{-1}{4} & \ldots &
\frac{-1}{N-2} & \frac{-1}{N-1} & \\
-1 & k_{22} & \frac{-1}{2} & \frac{-1}{3} & \frac{-1}{4} & \ldots &
\frac{-1}{N-2} & \frac{-1}{N-1} & \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots &
\\
\frac{-1}{N-2} & \frac{-1}{N-2} & \frac{-1}{N-2} & \frac{-1}{N-2} & \frac{-1}{N-2} & \ldots & k_{N-1,N-1} & \frac{-1}{N-1} & \\
\frac{-1}{N-1} & \frac{-1}{N-1} & \frac{-1}{N-1} & \frac{-1}{N-1} & \frac{-1}{N-1} & \ldots & \frac{-1}{N-1} & 1 & \\
\end{bmatrix}
$$
where
$K_{ii}=\sum_{j=1, j\ne i}^{N}{(-k_{ij})}$  for $i=1, 2,3,\ldots , N
$
For example if N=4, we have:
$$K = \begin{bmatrix}
11/6 & -1 & -1/2 & -1/3 & \\
-1 & 11/6 & -1/2 & -1/3 & \\
-1/2 & -1/2 & 4/3 & -1/3 & \\
-1/3 & -1/3 & -1/3 & 1 & \\
\end{bmatrix}
$$
How can I find an explicit equation for its eigenvalues?  

Phillip Lampe seems to be correct. Here are the eigenvalues and eigenvectors computed by hand:
Let $k_1 = 2 + \tfrac12 + \cdots + \tfrac{1}{N-1}$, then:
$\lambda_0 = 0$ with eigenvector all ones (by construction).
$\lambda_1 = k_{1}$ with eigenvector $\begin{bmatrix}-1& 1&  0&\cdots& 0\end{bmatrix}^T$
$\lambda_2 = k_1-1$ with eigenvector $\begin{bmatrix}-\tfrac12& -\tfrac12& 1& 0 &\cdots& 0\end{bmatrix}^T$
$\lambda_3 = k_1 -1- \tfrac12$ with eigenvector $\begin{bmatrix}-\tfrac13& -\tfrac13& -\tfrac13& 1& 0&\cdots& 0\end{bmatrix}^T$
$\lambda_4 = k_1 - 1-\tfrac12 - \tfrac13$ with eigenvector $\begin{bmatrix}-\tfrac14& \cdots& -\tfrac14& 1& 0&\cdots &0\end{bmatrix}^T$
and so on until
$\lambda_{N-1} =  k_1 -1-\tfrac12-\cdots-\tfrac{1}{N-2} = 1 + \tfrac{1}{N-1} = \tfrac{N}{N-1}$ with eigenvector $\begin{bmatrix}-\tfrac1{N-1}& \cdots& -\tfrac{1}{N-1}& 1\end{bmatrix}^T$.
So in short: The eigenvalues are $0$ and the values
$\lambda_j = 1+\sum_{i=j}^{N-1}\tfrac1i$ for $j=1,\dots,N-1$.
