# Let $\omega \in \sigma(P^k).$ If $\omega = \lambda^ k$ then $\lambda \in \sigma(P).$

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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Are you sure about the wording? Say I take the identity matrix. The spectrum of its square is still $\{1\}$, and yet $-1$ is not an eigenvalue of the original matrix. – Dylan Moreland Feb 18 '12 at 18:44
Dylan's right. Since $(P^k-\lambda^kI)=(A-\lambda I)(A^{k-1}+\lambda A^{k-2} + \ldots + \lambda^{k-1}I)$, you need to know that the determinant of the second piece does not vanish. In his example, $I^2 - (-1)^2 I = (I+I)(I-I)$. – Parsa Feb 18 '12 at 20:42
This 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. – M Turgeon Feb 18 '12 at 21:52

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 $\sigma(P)$.