The following question was posed on an old qualifying exam for linear algebra:(adsbygoogle = window.adsbygoogle || []).push({});

Suppose A is an n by n complex matrix, and that A has spectral radius <1 (the eigenvalue with largest norm has norm <1). Show that A^n approaches 0 as n goes to infinity.

The solution is easy when the eigenspace of A is equal to n, for one can show that A is similar to a diagonal matrix whose entries are the eigenvalues of A. I am having trouble showing that A must go to 0 in the case when the geometric multiplicity of the eigenvalues is less than n (when the eigenspace of A fails to span the space).

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Eigenvalues <1 imply 0 as a limit

**Physics Forums | Science Articles, Homework Help, Discussion**