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!

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

$\newcommand{\nrm}[1]{\left\lVert{#1}\right\rVert}\newcommand{\norm}{\nrm{\bullet}}$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 $\sqrt[n]{\nrm{A^n}_2}$


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