TITLE: If $\lambda$ has algebraic multiplicity $m$ has $\lambda^k$ the same multiplicity?
QUESTION [6 upvotes]: Let $A\in M_n(\mathbb{C})$ and $\lambda$ a eigenvalue of $A$ with the algebraic multiplicity equal to $m$. 
I know that $\lambda^k$ is an eigenvalue of $A^k$. What can I say about its algebraic multiplicity?

REPLY [4 votes]: The algebraic multiplicity of $\lambda^k$ with respect to $A^k$ will be the sum of the algebraic mulitplicities of eigenvalues $\lambda_i$ of $A$ such that $\lambda_i^k=\lambda^k$.
The proof uses the Jordan canonical form.