0
$\begingroup$

Show that spectral radius does not depend on equivalent norm using Gelfand's formula, i.e $$||.||_1 \sim ||.||_2 \Rightarrow \lim\limits_{n\to \infty}\sqrt[n]{||A^n||_1} = \lim\limits_{n\to \infty}\sqrt[n]{||A^n||_2}$$

I don't have any idea to solve this problem. Can anyone give me some hints? Thank you in advance!

$\endgroup$
  • $\begingroup$ Hint: For any $c>0, c^{1/n} \to 1$ as $n\to \infty$ $\endgroup$ – Prahlad Vaidyanathan Aug 26 '16 at 10:08
1
$\begingroup$

$\newcommand{\nrm}[1]{\left\lVert{#1}\right\rVert}\newcommand{\norm}{\nrm{\bullet}}$The definition of being equivalent norms is $$\norm_1\sim\norm_2\iff \exists c,b>0,\ c\norm_1\le \norm_2\le b\norm_1$$

Now, use $h>0\implies\lim_{n\to\infty}\sqrt[n]{h}=1$ to evaluate $\limsup\limits_{n\to\infty}$ and $\liminf\limits_{n\to\infty}$ of $\sqrt[n]{\nrm{A^n}_2}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.