- #1
slamminsammya
- 14
- 0
The following question was posed on an old qualifying exam for linear algebra:
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).
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).