Let $A$ be a bounded operator on complex Hilbert space $H$ such that $$(1+A^6)(1+A^2+A^4)=0.$$ Let $k\in\mathbb{C}$ be an element of the spectrum $\sigma(A)$.
How do I show that $k^{12}=1$?
What I know:
The spectrum of $A$ is $\sigma(A)=\{\lambda\in\mathbb{C}:A-\lambda I\text{ is not invertible }\}$. So there exists a non-zero vector $x\in H$ in the kernel of $A-kI$, otherwise said $Ax=kx$.
Then we compute $(1+A^6)(1+A^2+A^4)=1+A^2+A^4+A^6+A^{8}+A^{10}=0$. This gives $(1+k^2+k^4+k^6+k^{8}+k^{10})x=0$. How do I get the statement from this?
Edit: Following @DanielFischer's hint (thank you), we multiply by $(1-A^2)$ to get $$(1-A^2)(1+A^2+A^4+A^6+A^{8}+A^{10})=1-A^{12}=0.$$ This gives $k^{12}=1$.
Note that a priori you do not know that the spectrum of $A$ consists of real eigenvalues. That is, if $A - \lambda I$ is not invertible, in the infinite dimensional setting it doesn't always mean you can find $0 \neq x \in H$ such that $Ax = \lambda x$. However, you are given that $A$ satisfies some polynomial equation $p(A) = 0$ with
$$ p(x) = (1 + x^6)(1 + x^2 + x^4). $$
Since
$$ (1 - x^2)(1 + x^6)(1 + x^2 + x^4) = 1 - x^{12}$$
we also know that $g(A) = 0$ with $g(x) := 1 - x^{12}$. By the spectral mapping theorem, we must have $p(\sigma(A)) = \sigma(p(A)) = \{ 0 \}$ which implies that
$$\sigma(A) \subseteq \{ z \in \mathbb{C} \, | \, p(z) = 0 \} = \{ z \in \mathbb{C} \, | \, z^{12} = 1 \}. $$