spectral radius and numerical radius of matrix

Question

Let $\mathcal{H}$ be a complex finite dimensional Hilbert space ($\dim\mathcal{H}=d$). Let $A\in M_d(\mathbb{C})$.

How can I show without using spectral considerations that $$\lim_{n\to\infty}\|A^n\|^{\frac{1}{n}} \le\displaystyle\sup_{\|x\|=1}|\langle Ax,x\rangle|?? $$

Thank you for your help!!

This question is a motivation to the following one:

It is well known that the spectral radius of a bounded linear operator $T$ acting on a complex infinite dimensional Hilbert space is given by $$r(T)=\lim_{n\to\infty}\|T^n\|^{\frac{1}{n}}.$$

I want to prove using the above formula (rather than the definition of the spectral radius) that $$r(T)=\lim_{n\to\infty}\|T^n\|^{\frac{1}{n}}\le \sup_{\|x\|=1}|\langle Tx.x\rangle|.$$

On $\mathbb{R}^2$, with the usual Hilbert structure, let $A$ such that for $u=(x_1,x_2)$, $Au=(-x_2,x_1)$. What is $<Au,u>$ ? — Kelenner, Oct 30 '17 at 14:54
I have assumed that $\mathcal{H}$ is a complex Hilbert space. — Student, Nov 04 '17 at 16:50

score 3 · Answer 1 · edited Oct 08 '19 at 15:30

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Let me denote $\rho:=\lim \|A^n\|^{1/n}$. Then $|\lambda|\le\rho$ for all eigenvalues. Moreover, there is an eigenvalue $\lambda$ of $A$ such that $|\lambda|=\rho$. Using the corresponding unit eigenvector, $$ \rho \le \sup_{\|x\|=1}| \langle x,Ax\rangle| $$ follows.

I am not sure what the question is about. For instance, the inequality $$ \|A^n\|^{1/n} \le \sup_{\|x\|=1}| \langle x,Ax\rangle| $$ is not true for all $n$ in general. To see this, take the nilpotent matrix $$ A =\pmatrix{ 0 & 1 & 0 & 0 \\ \vdots & \ddots & \ddots &0\\ \vdots&\ddots&\ddots& 1 \\ 0 & \dots &\dots &0}. $$ Then $\|A^n\|=1$ for all $n<d$ and $\sup_{\|x\|=1} |\langle x,Ax\rangle|<1$.

edited Oct 08 '19 at 15:30

ViktorStein

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answered Nov 13 '17 at 15:57

daw

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I think the asker does not want to use spectral considerations. I think they are interested in a proof that uses only elementary properties of a Hilbert space and not of its operator algebra. I think this because they have written: \ "...I want to prove using the above formula (rather than the definition of the spectral radius) that ....". – s.harp Nov 13 '17 at 16:38
Thank you for your answer but I doesn't want to use spectral considerations. – Student Nov 13 '17 at 17:32
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then my answer shows that the inequality only holds in the limit, which excludes the first take on an elementary proof. – daw Nov 13 '17 at 18:56

score 2 · Accepted Answer · edited May 10 '18 at 10:24

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In Michel Crouzeix - Numerical range and functional calculus in Hilbert space, we see that

Using the inequality $2$ with $P(z)=z^n$ we get the desired result.

edited May 10 '18 at 10:24

Pang

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answered Nov 16 '17 at 05:58

Schüler

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This is the link: https://perso.univ-rennes1.fr/michel.crouzeix/publis/Final4.pdf – Schüler Nov 16 '17 at 12:43
@daw both links in the comments are broken, too. – ViktorStein Oct 08 '19 at 15:18

spectral radius and numerical radius of matrix

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