So, the problem goes:
If A diagonalizable matriz with eigenpairs ($\lambda_i$, $v_i$) (i=1,...,n) such that |$\lambda_i$|$\geq$|$\lambda_{i+1}$|.
Knowing that $\exists$ k such that |$\lambda_k$|$\gt$|$\lambda_{k+1}$|. Defining $U_k$=span{$v_{k+1}$, ... ,$v_n$}
a) Show that Null{$A^m$}$\subseteq$$U_k$ $\forall$ m$\gt$0
b) If T=span{$v_1$,...,$v_k$} prove that dim($A^m$T)=k $\forall$ m$\gt$0
What i have accomplished so far is the following:
a)
z$\in$Nul{$A^m$} $\Rightarrow$ $A^m$z=0 $\Rightarrow$ $A^m$$\sum_{i=1}^n \alpha_iv_i$=0 $ $$\Rightarrow$ $\sum_{i=1}^n \alpha_i\lambda^mv_i$=0
$\Rightarrow$ $\sum_{i=1}^k \alpha_i\lambda^mv_i$ +$\sum_{i=(k+1)}^n \alpha_i\lambda^mv_i$=0
I know that since A diagonalizable it has a basis of eigenvectors that is obviously linearly independent but I can´t seem to think about something to proceed from here. As for b) i have little clue on what to do. All answers are welcome. Thanks guys!
The eigenvalues of $A^m$ are $\lambda_1^m, \ldots, \lambda_n^m$. The nullspace is $\text{Null}(A^m) = \text{span}\{v_i : \lambda_i^m = 0\} = \text{span}\{v_i : \lambda_i = 0\}$. Since $|\lambda_1| \ge \cdots \ge |\lambda_n|$, this nullspace must either be $\{0\}$ or of the form $\text{span}\{v_j, v_{j+1}, \ldots, v_n\}$. If it is $\{0\}$, then (a) holds automatically. Otherwise we must have $j \ge k+1$ since $|\lambda_{k+1}| \ge 0$, so (a) holds.
$A^m(c_1 v_1 + \cdots + c_k v_k) = c_1 \lambda_1^m v_1 + \cdots + c_k \lambda_k^m v_k$. Since $|\lambda_1^m| \ge \cdots \ge |\lambda_k^m| > 0$, we actually have $A^m T = T$.