# Eigenvalues <1 imply 0 as a limit

1. Sep 4, 2011

### slamminsammya

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).

2. Sep 5, 2011

### HallsofIvy

If A does not have a "full set" of eigenvectors, it can still be put into "Jordan Normal Form", with the eigenvalues along the diagonal and "1"s in some of the positions just above the normal form. It can be shown that the nth power of such a matrix have the n powers of the eigenvalues on the diagonal and constants, times lower powers of the eigenvalues above the diagonal.

For example, the nth power of
$$\begin{bmatrix}a & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & a\end{bmatrix}$$
is
$$\begin{bmatrix}a^n & na^{n-1} & n(n-1)a^{n-2}\\ 0 & a^n & na^{n-1} \\ 0 & 0 & a^n\end{bmatrix}$$.