Denote $ \sigma(P)$ to be the spectrum of a matrix $P.$ Let $ \omega \in \sigma(P^k).$ If $\omega = \lambda^ k$ then show that $ \lambda \in \sigma(P).$
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It sounds like you're trying to get at the Spectral Mapping Theorem, which says that $\sigma(P^k) = \sigma(P)^k$; equivalently, if $\omega \in \sigma(P^k)$, then there exists $\lambda \in \sigma(P)$ with $\lambda^k = \omega$.
To prove this, let $\lambda_1, \dots, \lambda_k$ be the roots (not necessarily distinct) of the polynomial $X^k - \omega$, so that $$ X^k - \omega= (X - \lambda_1) \cdots (X - \lambda_k). $$ Substituting the matrix $P$ into this equation yields $$ P^k - \omega I = (P - \lambda_1 I) \cdots (P - \lambda_k I). $$ Since $\omega \in \sigma(P^k)$, the left-hand side is singular, so at least one of the factors on the right must be singular. Hence at least one of $\lambda_1, \dots, \lambda_k$ must be in $\sigma(P)$.