Conditions for the allowance of eigenvalue exponentiation

Suppose that there is a matrix $A$, and the corresponding eigenvalues $\lambda_i$ and the eigenvector $v_i$.

1) What is the condition for all eigenvalues to have distinct eigenvectors, and what is the condition for all eigenvalues to share same eigenvectors?

2) Suppose that we do the operation like $A^kv$ where $v$ is some eigenvector. Then what is the condition for $A^kv = \lambda^k v$ when $\lambda$ is eigenvalue for all natural number $k$?

3) If we restrict $k$ in 2) to be 2 and 3, how would conditions differ?

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1. Different eigenvalues always have different eigenvectors. If $Av = \lambda_1 v$, there can't be some other eigenvalue $\lambda_2$ such that $Av = \lambda_2 v$ (for the same $v$) as well!
2. If $v$ is an eigenvector of $A$, then $Av = \lambda v$, so then by repeatedly applying $A$, it is always true that $A^k v = \lambda^k v$, for every positive integer $k$.
Assume:$$A\alpha=\lambda_1\alpha$$$$A\beta=\lambda_2\beta$$.
if $$k_1\alpha+k_2\beta=0$$,then$$A(k_1\alpha+k_2\beta)=\lambda_1k_1\alpha+\lambda_2k_2\beta=0$$. multiply $k_1\alpha+k_2\beta=0$ with $\lambda_1$,and minus $\lambda_1k_1\alpha+\lambda_2k_2\beta=0$,we get $k_2(\lambda_1-\lambda_2)\beta=0$. so $k_2=0$ then $k_1=0$. then $\alpha$ and $\beta$ are linear indepandent