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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?

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    $\begingroup$ The vector $v_0=(1,\ldots,1)^T$ is an eigenvector with eigenvalue $0$ since the rows sum to zero. For $N=4$ the other eigenvectors are $(-1,1,0,0)^T$, $(-\frac12,-\frac12,1,0)^T$ and $(-\frac13,-\frac13,-\frac13,1)^T$. So maybe we should consider the vectors $v_k=(-\frac1k,\ldots,-\frac1k,1,0,\ldots,0)^T$ where the entry $1$ is at position $k+1$ for $k\in [1,N-1]$ for general $N$. $\endgroup$ – Philipp Lampe Feb 26 at 11:24
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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 $j=1,\dots,N-1$.

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