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A friend of mine who does not use a computer raised the following question, in connection with a problem he has been working on for a long time. He would like to know that the following $n\times n$ matrix has exactly one eigenvalue larger than 1 and all the rest less than 1. It is symmetric (so all eigenvalues are real) and all diagonal entries are 1, so I will describe only the entries below the diagonal. The $k$th row's entries are $1/k,2/k,\ldots,(k-1)/k$.

In low dimensions, I have used an online calculator to do this. In $2\times 2$ they are obviously $1/2$ and $3/2$. I did the $3\times3$ case, but didn't write down the answer. In $4\times 4$, they are approximately $0.21,0.41,0.85,2.54$.

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    $\begingroup$ Downvoted for clear lack of effort. You didn't typeset the matrix in Latex, and you don't want to recompute the result for a 3x3 matrix? $\endgroup$
    – Federico Poloni
    Nov 10, 2016 at 17:51
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    $\begingroup$ @FedericoPoloni I think this is needlessly harsh. Michael Barr is not some apathetic undergraduate $\endgroup$
    – Yemon Choi
    Nov 10, 2016 at 18:39
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    $\begingroup$ the conjecture seems false: for $n=5$ I find the eigenvalues, 3.06664, 1.00351, 0.501031, 0.272797, 0.15602, so there are two eigenvalues greater than 1. Just to be sure we are talking about the same matrix: $$\left( \begin{array}{ccccc} 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \frac{1}{2} & 1 & \frac{2}{3} & \frac{2}{4} & \frac{2}{5} \\ \frac{1}{3} & \frac{2}{3} & 1 & \frac{3}{4} & \frac{3}{5} \\ \frac{1}{4} & \frac{2}{4} & \frac{3}{4} & 1 & \frac{4}{5} \\ \frac{1}{5} & \frac{2}{5} & \frac{3}{5} & \frac{4}{5} & 1 \\ \end{array} \right)$$ $\endgroup$
    – Carlo Beenakker
    Nov 10, 2016 at 19:39
  • $\begingroup$ For $n=20$, there are three eigenvalues outside the unit disk; for $n=50$, there are six and for $n=100$, there are eight. $\endgroup$
    – Anthony Quas
    Nov 11, 2016 at 7:48
  • $\begingroup$ Yes that's the right matrix. Thanks. BTW, they add up, as the must to 5, which is a good check. $\endgroup$
    – Michael Barr
    Nov 11, 2016 at 14:33

1 Answer 1

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These are some hints in case you still need informations on the eigenvalues of your matrices. It is not difficult to check that your matrices are inverses of tri-diagonal matrices $M_n$, coming from a Jacobi operator. Precisely, the non-zero entries in $M_n$ are $$ m^{(n)}_{k,k+1}=-{k(k+1)\over 2k+1},\qquad m^{(n)}_{k,k}:={4k^3\over 4k^2-1}\qquad \text{ for } 1\le k <n$$ while $$m^{(n)}_{n,n}={n^2\over 2n-1}.$$ Hence, their characteristic polynomials turn out to be
$$\operatorname{det}(x-M_n)=P_n+{n^2\over 2n+1}\,P_{n-1}$$ where $P_n$ are orthogonal polynomials satisfying the three-term linear recurrence: $$ \begin{cases} P_0=1\\ P_1=x-{4\over 3} \\ P_{n}=\Big(x-{4n^3\over4n^2-1}\Big)\,P_{n-1}-{n^2(n-1)^2\over(2n-1)^2}\,P_{n-2}\,. \end{cases} $$ I believe with a little more work they should be identified as a known sequence (e.g. Wilson polynomials for some choice of parameters).

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    $\begingroup$ Your matrices $A_n$ may also be written in the form $D_nB_nD_n$, where $D_n:=\mathrm{diag}(1,1/2,1/3,1/4,..,1/n)$ and $B_n$ with entries $(\min\{i,j\})^2$. This $B_n^{-1}$ has a tridiagonal form like $M_n:=A_n^{-1}$, actually quite simpler. $\endgroup$
    – Pietro Majer
    Nov 11, 2016 at 15:37

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