What is the term for a matrix with spectral radius less than 1, with all eigenvalues of modulus = 1 associated with a 1x1 Jordan block? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T06:07:13Z http://mathoverflow.net/feeds/question/29816 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29816/what-is-the-term-for-a-matrix-with-spectral-radius-less-than-1-with-all-eigenval What is the term for a matrix with spectral radius less than 1, with all eigenvalues of modulus = 1 associated with a 1x1 Jordan block? Jeremy Levick 2010-06-28T18:46:07Z 2010-06-28T19:04:32Z <p>I am looking at the matrices described in the title: matrices where all eigenvalues lie in the unit disc, and with the eigenvalues of modulus 1 having 1x1 Jordan blocks. My question is, is there a term for such matrices? </p> <p>A related question is, since we lose the Jordan normal form in infinite dimensions, what is the appropriate infinite-dimensional analog of this condition (if there is one)?</p> http://mathoverflow.net/questions/29816/what-is-the-term-for-a-matrix-with-spectral-radius-less-than-1-with-all-eigenval/29818#29818 Answer by Helge for What is the term for a matrix with spectral radius less than 1, with all eigenvalues of modulus = 1 associated with a 1x1 Jordan block? Helge 2010-06-28T19:04:32Z 2010-06-28T19:04:32Z <p>To the second question: You can check that your condition is equivalent to $$\limsup_{n \to \infty} \|A^n \| \leq 1.$$ If there is an eigenvalue > 1, it is clear that the above fails. If an eigenvalue = 1 has non-trivial Jordan block, you have $\|A^n\| \gtrsim n$, so the condition is violated.</p> <p>As for a name, in my opinion "subunitary" would seem appropriate. But that is just a guess and not based on knowledge.</p>