When do the powers of a square complex matrix \(A \in M_n(\mathbb{C})\) tend to zero (\(\lim_{k \to \infty} A^k = 0\))?
Solution
One has \[ A^k\longrightarrow0 \quad\Longleftrightarrow\quad \rho(A)<1, \] where \(\rho(A)\) is the spectral radius.
Put \(A=PJP^{-1}\) in Jordan normal form. Then \[ A^k=PJ^kP^{-1}, \] so \(A^k\to0\) if and only if every Jordan block tends to zero.
For a Jordan block \[ J_\lambda=\lambda I+N, \qquad N^d=0, \] one has \[ J_\lambda^k = \sum_{j=0}^{d-1}\binom{k}{j}\lambda^{k-j}N^j. \] If \(|\lambda|<1\), then for each fixed \(j\), \[ \binom{k}{j}|\lambda|^{k-j}\to0, \] because exponential decay dominates polynomial growth. Hence \(J_\lambda^k\to0\).
If \(|\lambda|\ge1\), the diagonal entries of \(J_\lambda^k\) are \(\lambda^k\), which do not tend to zero. Thus a Jordan block tends to zero exactly when its eigenvalue has modulus \(<1\).
Therefore \(A^k\to0\) exactly when every eigenvalue of \(A\) lies in the open unit disk, equivalently when \(\rho(A)<1\).