Prove that if $A \in M_{n \times n}(C)$ is diagonalizable and $L = \lim_{m \to \infty} A^m$ exists, then either $L = I_n$ or $\operatorname{rank}(L) < n$.
Since $A$ is diagonalizable, there exists an invertible matrix $Q$ such that $Q^{-1} A Q = D$ is a diagonal matrix. This implies that $L = Q \lim_{m \to \infty} D^m Q^{-1}$. Note that the diagonal entries of $D$ consist of eigenvalues of $A$ contained in $\{\lambda \in C: \lambda =1 , |\lambda|<1\}$ since the limit of $A^m$ exists. This implies that when the multiplicity of $1$ as an eigenvalue of $A$ is $n$, $D = I_n$, so $L = I_n$. On the other hand, when the eigenvalues of $A$ include $|\lambda| < 1$, some diagonal entries of $D^m$ converges to $0$, so $\lim_{m \to \infty} D^m = D'$ has rank less than $n$. I know that $\operatorname{rank}(QD'Q^{-1}) \le \operatorname{rank}(Q) =n$, but I am struggling to prove that $\operatorname{rank}(QD'Q^{-1}) < \operatorname{rank}(Q) =n$ when $Q$ is full rank, but $D$ is non full rank. I would appreciate if you give some help.
For only proving the strict inequality case, Take $Rank(D') < n $ , this possible iff there must atleast one $|\lambda| \lt 1 $
Now, since, $Rank(Q) + Rank(D')+Rank(Q^{-1}) - n -n \le Rank(QD'Q^{-1}) \le min \{Rank(Q),Rank(D'),Rank(Q^{-1})\} $
$\implies Rank(D') \le Rank(QD'Q^{-1}) \le Rank(D') (\text{ as ,$ Rank(Q) = n = Rank(Q^{-1})$})$
$\implies Rank(QD'Q^{-1}) = Rank(D') \lt n $