Relation between eigenvalues of $A^k$ and eigenvalues of $A$ I need to prove that if $A$ is an $n \times n$ matrix, then $\lambda $ is an eigenvalue of $A$ if and only if $\lambda^k $ is an eigenvalue of $A^k$ for any positive integer $k \geq 1$. I am assuming $\lambda \in \mathbb R$. Otherwise I think the propositions is false.
The first part ($\implies$) is very easy by induction; but I am having difficulties showing it for the second part ($\Leftarrow$). 
Any suggestions?
 A: First, I think this result is true for $\lambda \in \mathbb{C}$ also.This state is also true  for any $p(A)$.
So, here I am showing how to prove the $(\Leftarrow)$ part.
What I have is- $p(\lambda_j)$ is a eigenvalue of $p(A)$ and assume that $p$ is a $k$ degree polynomial.
$(p(A)-p(\lambda_j)I_n) = (A-\alpha_1)^{a_1} \dots (A-\alpha_i)^{a_i}$ such that $\sum_{i=1}^n a_i =k$
$(p(A)-p(\lambda_j)I_n)x = (A-\alpha_1)^{a_1} \dots (A-\alpha_i)^{a_i}x$ assuming $x$ is the eigenvector corresponding to the eigenvalue $\lambda_j$.
So,
$(p(A)-p(\lambda_j)I_n)x = (A-\alpha_1I_n)^{a_1} \dots (A-\alpha_iI_n)^{a_i}x = \theta_n$
$\implies \det(p(A)-p(\lambda_j)I_n)=0$ which implies $\det((A-\alpha_l I_n)^{a_l}) = 0$ (where $l$ is an integer between $1$ to $i$).
$\det((A-\alpha_l I_n)^{a_l}) = 0 \implies \det(A-\alpha_l I_n) = 0.$ From this we can conclude that $\alpha_l$ is an eigenvalue of $A$ and $x$ is the corresponding eigenvector. So, $p(A)x = p(\alpha_l)x=p(\lambda_j)x$. So, $\lambda_j = \alpha_l$.
Hence, it is proved.
A: The eigenvalues and vectors of $A$ are related by :
$$Au_k=\lambda_k u_k$$
For any $n$ a positive integer greater than one, we can say :
$$A^nu_k=A^{n-1}Au_k=A^{n-1}(\lambda_k u_k)=\lambda_k A^{n-1}u_k$$
From this we easily see that :
$$A^nu_k=\lambda_k^n u_k$$
So the eigenvalues of $A^n$ are $\lambda_k^n$ and the eigenvectors are the same as those of $A$.
Second part.
Given only $A^nu_k=\lambda_k^n u_k$ try and prove that $\lambda_k$ is an eigenvalue of $A$
Let $Au_k=\mu_k u_k$ for some scalar $\mu_k$.
As we already know that $A^nu_k=\mu_k^nu_k$ we need only consider if $\mu_k^n = \lambda_k^n$
Strictly speaking this can only tell us that the eigenvalues of $A$ are any valid value of $(\lambda_k^n)^\frac{1}{n}$, which is not necessarily $\lambda_k$.
For example if $n$ was even, then we could have $\pm\lambda_k$ as eigen values of $A$.
So going from the power to the root is not always possible and the second part fails.
