Spectral mapping theorem for polynomials

Remarks

Proof idea: if $p$ is constant equal to $c$, both sides are $\set{c}$, using the nonemptiness of the spectrum. Otherwise fix $\mu \in \C$ and factor $p(z) - \mu = c (z - \lambda_1) \cdots (z - \lambda_k)$ with $c \ne 0$. Then $p(a) - \mu 1 = c (a - \lambda_1 1) \cdots (a - \lambda_k 1)$, a product of commuting factors. A product of commuting elements is invertible if and only if each factor is invertible. So $\mu \in \sigma(p(a))$ if and only if some root $\lambda_i$ of $p - \mu$ lies in $\sigma(a)$, that is, if and only if $\mu = p(\lambda)$ for some $\lambda \in \sigma(a)$.

Example: if $a$ is a $2 \times 2$ complex matrix with eigenvalues $1$ and $-1$, then $a^2$ has the single eigenvalue $1$, matching $p(\sigma(a)) = \set{1}$ for $p(z) = z^2$.

Going deeper

For a student

The theorem gives $\sigma(a^n) = \set{\lambda^n : \lambda \in \sigma(a)}$, which is the easy half of Gelfand's spectral radius formula. It extends to holomorphic functions in the holomorphic functional calculus, and to continuous functions of normal elements in the continuous functional calculus of C*-algebras.