Add a comment. Venkataramana Venkataramana I'd just start with "In case your semigroup is made of invertible matrices, Your current statement seems to hold in general, in terms of the invertible matrices in the semigroup which can be far from enough, e. Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password. Post as a guest Name. Email Required, but never shown. Featured on Meta. Now live: A fully responsive profile.
Related 1. Question feed. MathOverflow works best with JavaScript enabled. Accept all cookies Customize settings. Walk through homework problems step-by-step from beginning to end. Hints help you try the next step on your own. Unlimited random practice problems and answers with built-in Step-by-step solutions. Practice online or make a printable study sheet. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more.
MathWorld Book. According to 3. A square matrix is reducible iff the associated directed graph has smaller strongly connected components. So you may use a strong component algorithm to solve your problem. The following are equivalent: a A is irreducible. It is Theorem 6. Go check it if you need a complete proof. The best place to look is this wiki link. If it is non-negative, then one needs to check other things.
We can use the test cf. The best is to use the "strong component algorithm" cf user Sign up to join this community. The best answers are voted up and rise to the top.
Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. How does one show a matrix is irreducible and reducible?
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