Define a series of matrices$$H_N= \begin{bmatrix} 1&1/N&1/N&\cdots&1/N\\ 1/N&2&1/N&\cdots&1/N\\ 1/N&1/N&3&\cdots&1/N\\ \vdots&\vdots&\vdots&&\vdots\\ 1/N&1/N&1/N&\cdots&N \end{bmatrix}$$ My question is, when $N\to+\infty$, would the eigenvalues of $H$ be different from $\{1,\ldots,N\}$ ? The answer is not obvious, as, for the matrix series $$G_N=\begin{bmatrix} 1&1/N&1/N&\cdots&1/N\\ 1/N&1&1/N&\cdots&1/N\\ 1/N&1/N&1&\cdots&1/N\\ \vdots&\vdots&\vdots&&\vdots\\ 1/N&1/N&1/N&\cdots&1 \end{bmatrix}$$ You can verify that $G_N$ has an eigenvalue of $2$.

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    $\begingroup$ 1) Your statement regarding $G_N$ having an eigenvalue of $2$ isn't quite right: Rather, it has an eigenvalue $2-1/N$ which converges to $2$. 2) Judging from numerical evidence in Mathematica, the answer looks to be "yes." I'll see if I can find a good argument as to why... $\endgroup$ Aug 17, 2016 at 3:50

1 Answer 1


Nearly a year passed and now I finally got the answer for my problem: In the limit $N\to\infty$, the eigenvalues of $H_N$ won't be different from $\{1,2,\ldots,N\}$.

Proof: The matrix element of $H_N$ is given by $H_{ij}=j\delta_{ij}+1/N$, where we have added an unimportant $1/N$ for the diagonal elements (and in the following, we won't write the subindex $N$ explicitly). We can directly solve the eigenvalue equation $H_{ij}\psi_j=\lambda\psi_i$ in this case, giving $$i \psi_i+\frac{1}{N}\sum^N_{j=1}\psi_j=\lambda\psi_i~~\Rightarrow~~ \psi_i=-\frac{1}{N(i-\lambda)}\sum^N_{j=1}\psi_j,$$ summing over $i$ on both sides, we get $$\sum^N_{i=1}\psi_i=-\sum^{N}_{i=1}\frac{1}{N(i-\lambda)}\sum^N_{j=1}\psi_j$$ If $\sum^N_{j=1}\psi_j=0$, then we have $(i-\lambda)\psi_i=0$ for all $i$, which is not possible for a nonzero eigenvector. It follows that $$\sum^{N}_{i=1}\frac{1}{(\lambda-i)}=N.$$ It is easy to prove that this equation has $N$ distinct roots $1<\lambda_1<2<\lambda_2<3<\ldots <N<\lambda_N<N+1$. We now prove that $\lambda_j\to j$ in the limit $N\to+\infty$. Let $\lambda_j=j+\Delta_j$, if $$\lim_{N\to\infty}\Delta_j=\Delta>0,$$ then there exist $N_0$ such that for all $N>N_0$, $\Delta/2<\Delta_j<1$, which lead to $$\sum^N_{i=1}\frac{1}{j+\Delta_j-i}<\sum^j_{i=1}\frac{1}{j+\Delta_j-i}=\sum^{j-1}_{i=0}\frac{1}{\Delta_j+i}<\frac{1}{\Delta/2}+\frac{1}{\Delta/2+1}+\ln N,$$ which is obviously much smaller than $N$ for sufficiently large $N$, a contradiction. Thus $$\lim_{N\to\infty}\Delta_j=0~~\Leftrightarrow \lim_{N\to\infty}\lambda_j=j$$ for all $j\in N_+$.


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