Proving a fact about non-nilpotent matrices

Question

Let $A$ be a square matrix such that all its eigenvalues are less than or equals 1 in absolute value. If A is not nilpotent, then prove that $$ \text{There exists an } i_0 \text{ such that rank }(A^{i_0-1})=\text{ rank }.(A^{i_0}) $$

I am relatively new to the theory of nilpotent matrices. Can anyone guide me in the proof?

Answer 1

Hint:

$Ker(A^n)\subset Ker(A^{n+1})$ this implies $dim(ker(A^n)\leq dim(ker A^{n+1})$. The sequence $(dimKer A^n)$ is an increasing sequence, which takes it value in $\{0,1,...,m\}$, this implies there exists $N$ such that $n>N$ implies that $dimker A^n=dim ker A^N$ since we have the equality $dim ker B+rank B=m$, we deduce that for $n>N, rank A^n=m-dim ker A^n=m-dim kerA^N=rank A^N$.