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Let $A$ be a Hilbert matrix,

$$a_{ij}=\frac{1}{1+i+j}$$

We have the result $\| A \| \leq \pi$. I am using the subordinate norm of the Euclidean norm, i.e.,

$$\| A \| = \sup\{\langle Ax,y\rangle:\quad x,y\in\mathbb{R}^n,\quad\Vert x\Vert_2\leq 1,\quad\Vert y\Vert_2\leq 1\}$$

This inequality can be proved using Hilbert's Inequality. Look here.

Question: Do we have an equality? I found nothing on the Internet about such equality.

Rodrigo de Azevedo
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  • You can take a look at this paper : Trick and Treats with the Hilbert matrix http://www.cecm.sfu.ca/~jborwein/Expbook/Updates/Already%20Included/choi.pdf

    An another way to prove the result is to use Fubini's theorem.

     – Krokop Jun 17 '14 at 22:51

2 Answers2

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Certainly not (if you're talking about a finite matrix, rather than an operator on $\ell^2$). $\pi$ is transcendental, and the norm of a finite matrix with rational entries is an algebraic number.

Robert Israel
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  • I think the OP means either the norm of the infinite-dimensional Hilbert matrix or supremum of all the finite-dimensional norms. – anomaly Jun 17 '14 at 22:24
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    Note that the OP mentioned ${\mathbb R}^n$, but said nothing about $n \to \infty$. Still, you may be right. – Robert Israel Jun 17 '14 at 22:26
  • Do you have any reference about your result please? – Krokop Jun 17 '14 at 22:37
  • The one about the norm being an algebraic number? The norm is obtained by solving an optimization problem. Just verify that the Karush-Kuhn-Tucker conditions for this problem are equations and inequalities involving algebraic functions, so the solution comes from solving a system of algebraic equations. – Robert Israel Jun 18 '14 at 06:33
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    Alternatively, $|A|$ is the largest singular value of $A$, which is equal to the square root of the largest eigenvalue of $A^\ast A$. Since the characteristic polynomial of $A^\ast A$ has rational coefficients, it follows that $|A|$ is an algebraic number. – user1551 Jun 18 '14 at 11:46
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According to this paper, the answer is yes. I'm not familiar with the theorem of Nehari mentioned there; if you assume that, though, you can check that the value of $g(n)$ does produce the Hilbert matrix as indicated and has $L^\infty$ norm $\pi$. (Here $\hat g$ is the Fourier transform of $g$). (The paper refers to the infinite-dimensional Hilbert matrix as an operator $A : \ell^2 \to \ell^2$, but it's not hard to show that a positive result in that case shows that $\pi$ is also the supremum of the corresponding Hilbert matrices $A_n : {\mathbb{R}}^n \to {\mathbb{R}}^n$, if you're dealing with the finite-dimensional case.)

Of course, if you're asking whether any particular $A: {\mathbb{R}}^n \to {\mathbb{R}}^n$ has norm exactly $\pi$, the answer is no; see the comment above.

anomaly
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