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).

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# Eigenvalues <1 imply 0 as a limit

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