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).$
Let $ \omega \in \sigma(P^k).$ If $\omega = \lambda^ k$ then $ \lambda \in \sigma(P).$
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linear-algebra
matrices
functional-analysis
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0This seems highly improbable, since if you're over the complex numbers (say), *every* $\omega\in\sigma(P^k)$ is of the form $\omega=\lambda^k$. And so, I don't think Dylan's example is merely an exception, i.e. I don't think that adding hypotheses on $P$ will yield an interesting result. – 2012-02-18
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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 $\sigma(P)$.
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0@ Dave, thanks, its exactly what I wanted. I don't know. Should I edit the question (thus the comments will be not valid) or just leave it like this. I think I should just leave it as its! – 2012-02-22