Determine all real numbers that can be eigenvalues of operator A.

Question

Problem:

Let $A$ be some linear operator such that: $$ (A^{2006}-I)^{2006}-I=0. $$ Determine all real numbers that can be eigenvalues of operator $A$.

Question: How to solve this problem?

My attempt:

$(A^{2006}-I)^{2006}-I=0 \implies (A^{2006}-I)^{2006}=I \implies det((A^{2006}-I)^{2006})=1 \implies [det(A^{2006}-I)]^{2006} = 1$

$A^{2006} - I = (A-x_1I)...(A-x_{2006}I) \implies [det(A-x_1I)...det(A-x_{2006}I)]^{2006}=1$

Let's say that $A$ can have $n$ real eigenvalues and let's call them $\lambda_j$ ($j=1,...n$). That means $det(A-\lambda_jI)=0$. From this follow that no eigenvalue can be $x_j$ ($j=1,...2006$). That means that real numbers which satisfy equation $x^{2006}=1$ can not be eigenvalues. But it would seem to me that my "solution" is incomplete, for how do I know that there are no more restrictions on eigenvalues? And in my attempt I prejudiced that there is finate number $n$ of eigenvalues, how do I know there is not infinitley many eigenvalues?

Thank you for any help.

Answer 1

$\textbf{Hint:}$ Suppose $\lambda$ is an eigenvalue of $A$. Is $p(\lambda)$ an eigenvalue of $p(A)$? ($p(\cdot)$ denotes a polynomial.)

Answer 2

If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $\pm 1$. We can have $x=0$ or $x^{2006}=2, x=\pm2^{1/2006}$