Eigenvalues of bounded linear operator

Question

Let $X$ be a Banach space, and $T$ be a bounded linear map from $X$ to $X$. Suppose that $a$ is an eigenvalue of $T^n$ for some postive integer $n$, Show that some $n$-th root of $a$ is an eigenvalue of $T$.

Answer 1

The key fact here is the Spectral Mapping Theorem, which says that $$ \sigma(T^n)=\sigma(T)^n. $$

Note that the theorem above is way more general: it shows that $\sigma(f(T))=f(\sigma(T))$ for any holomorphic (on $\sigma(T)$) function $f$.

As mentioned above by Ben, there is a straightforward way in the concrete case where $f(z)=z^n$.

If you let $\omega_1,\ldots,\omega_n$ be the $n^{\rm th}$ complex roots of $a$, then $(x-\omega_1)\cdots(x-\omega_n)=x^n-a$ (by the Fundamental Theorem of Algebra). Then this equality can be also written for $T$: $$\tag{*} T^n-aI=(T-\omega_1I)\cdots(T-\omega_nI). $$ Since $a$ is in the spectrum of $T^n$, we have that $T^n-aI$ is not invertible. Then at least one factor from the right hand side of $(*)$ is not invertible: this means that there exists at least one $k$ with $T-\omega_kI$ not invertible, and so $\omega_k$ is in the spectrum of $T$.

Note that the argument above does not require "eigenvalue", and it works for all elements of the spectrum.