I am trying to proof the following, of which I am sure it is true. I would be pleased if you could give me hints, how to approach such a problem.

Given a finite set of matrices $\mathcal{A}=\{A_j\in\mathbb{R}^{s\times s},j=1,\ldots,J\}$. Define the set $$ \mathcal{M}=\{ (A_{i_n}\cdots A_{i_1})_n\in\mathcal{A}^\mathbb{N} \ : \ %i_j\in\{1,\ldots,J\},\ \rho(A_{i_n}\cdots A_{i_1})<1 \text{ and }\ A_{i_{n-1}}\cdots A_{i_1}\in\mathcal{M} \}. $$

Thus, $\mathcal{M}$ consists of sequences of products, all of whose entries have spectral radius then less then one.

Show that there exists $C>0$ such that for all $x\in\mathbb{R}^s$ and all sequences $(A_{i_n}\cdots A_{i_1})_{n\in\mathbb{N}}\in\mathcal{M}$

$$ \sup_{n\in\mathbb{N}} \|A_{i_n}\cdots A_{i_1}x\|<C. $$

EDIT: After more thoughts about the problem, and making some numerical experiments, the above would follow from the following generalization of Gelfands formula, which I suppose is true.

Let $(i_n)_n\in\{1,\ldots,J\}$. Then for $(B_n)_n=(A_{i_n}\cdots A_{i_1})_n$, $$ \lim_{n\rightarrow\infty} \left(\frac{\rho(B_n)}{\|B_n\|}\right)^{1/n}=1 $$

Clearly $\rho(M_n)\leq\|M_n\|$, but the other direction I could not proof yet.


Your Answer

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

Browse other questions tagged or ask your own question.