2

I have the following matrix

$$H=\begin{bmatrix} 1 & \frac{1}{2} & \cdots & \mbox{ad}\ +\infty\\ \frac{1}{2} & \frac{1}{3} & \cdots & \mbox{ad}\ +\infty\\ \vdots & \vdots & \ddots\\ \ & \ & \mbox{ad}\ +\infty \end{bmatrix}$$

which is a Hilbert matrix of order $\infty$. My problem is to find the largest eigenvalue $\lambda_{max}$ of $H$ and find an eigenvector corresponding to $\lambda_{max}$. I do not think the conventional way of finding eigenvalues and eigenvectors is going to help me here. Otherwise, I am not sure how to proceed to solve this question. Please help.

Rodrigo de Azevedo
  • 23,223
Samrat Mukhopadhyay
  • 16,568
  • Although this question is over 3 years old and I can't fully answer it, I want to at least point out that the operator norm of $A$ (interpreting $A$ as an operator on $\ell_2$) is $\frac{\pi}{\sqrt6}$ as is readily verified so since $A$ additionally is hermitian, all the eigenvalues of $A$ have to be contained in the real interval $[-\frac{\pi}{\sqrt6},\frac{\pi}{\sqrt6}]$. – Frederik vom Ende Nov 08 '17 at 12:57
  • The operator norm is going to be $\max_{x: |x|_2 = 1} x^T H x$, right? Can you kindly elaborate how you evaluate the operator norm to be $\pi/\sqrt{6}$? – Samrat Mukhopadhyay Nov 09 '17 at 05:45
  • Sorry, I misremembered. The operator norm of $H$ is in fact $\pi$ (and not $\pi/\sqrt6$ as I first stated), see here. Still, that means that possible eigenvalues of $H$ have to be contained in the real interval $[-\pi,\pi]$.

    Sadly $H$ is not compact, see this paper, Example (8) - because then we'd know that either $\pi$ or $-\pi$ has to be an eigenvalue of $H$ (c.f. Theorem VIII.§3.3 in the book "Introduction to Hilbert Space" by Berberian).

     – Frederik vom Ende Nov 09 '17 at 09:40
  • 1
    Thanks for the reference. – Samrat Mukhopadhyay Nov 09 '17 at 15:43

1 Answers1

1

$H$ does not have $\pi$ as an eigenvalue (interpreting $H$ as operating on $\ell^2$). See for instance Theorem 323 of Hardy, Littlewood and Polya, "Inequalities" where it is shown that for a non-zero $x$ in $\ell^2$, $x^THx<\pi\|x\|^2$ strictly. But there are eigenvalues arbitrarily close to $\pi$ since if $x_r=1/\sqrt r$ for $r=1,\ldots,N$ and $x_r=0$ for $r>N$ then $x^THx/\|x\|^2$ tends to $\pi$ as $N\to\infty$.

David
  • 11
  • What I get is that the maximal eigenvalue must be larger than or equal to $\pi$. Is your argument actually showing that there are eigenvalues arbitrarily close to $\pi$? – Bananach Dec 31 '22 at 15:12