Eigenvalues of operator given polynomial I have a homework question that is really bothering me because I cannot manage to use any of the theorems we have studied so far, it seems. Could someone please give me a pointer please (no, Cayley-Hamilton is not the answer) ? I can't quite tie it to the characteristic polynomial, unless I am being stupid; this is over a generic field, not necessarily reals or complex.
Let $p(t) = t^2 + t + 1$, and let it be given that $p(T) = 0$ for an operator $T$. List all the eigenvalues of $T$.
 A: The relevant facts are as follows: if $p(t)$ is a polynomial and $T$ an operator such that $p(T) = 0$, then the minimal polynomial of $T$ divides $p$.  Moreover, every root of the minimal polynomial of $T$ is necessarily an eigenvalue of $T$.
Note, however, that the minimal polynomial of $T$ is not generally equal to its characteristic polynomial.
To answer this question: the roots of the polynomial $t^2 + t + 1$ are $\omega = -1/2 + i\sqrt{3}/2$ and $\overline{\omega}$.  If $T$ is a real-linear operator, then both $\omega$ and $\overline{\omega}$ must be eigenvalues of $T$, and $T$ can have no additional eigenvalues.  If, however, $T$ is an arbitrary complex-linear operator, then the eigenvalues of $T$ can be either $\omega, \overline{\omega},$ or both. 
A: As far as p(T)=0, this means that characteristic f(T) pol. divides it: f|p (I mean that p=f*g, g:pol.)  Thus your eigenvalue is one of the roots of p or both of them, as far as g=id.
A: Maybe the following works:
Suppose $Tx=\lambda x$
\begin{align*}
P(T)(x)&=T^2x+Tx+x\\
&=\lambda^2 x+ \lambda x + x\\
&=(\lambda^2+\lambda+1)x=0.
\end{align*}
So the eigenvalues are the solutions to $\lambda^2+\lambda+1$.
