Are the only eigenvalues of an operator $T^k$ the $\lambda_n^k$?

Question

If I have an operator T which has eigenvalues $(\lambda_n)_{n \in \mathbb{N}}$, I understand that $\lambda_n^k$ are eigenvalues for $T^k$. But how to show that these are the only eigenvalues? Is this even true?

Answer 1

Over a field which is not algebraically closed, this is false. Consider $$A=\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}\in M_2(\mathbb R).$$ Then $A^2=-I$, which has only the eigenvalues $-1$, but $A$ has no eigenvalues.

But over an algebraically closed field, this is true for operators on finite-dimensional spaces by the Jordan decomposition: squaring the matrix applies to each Jordan block separately.