Let $X$ be a Banach space over the field $\mathbb C$ of complex numbers, suppose that $A:X\to X$ is a bounded linear operator. Does there hold that $\sigma(A^3)=\{\lambda^3;\lambda\in\sigma(A)\}?$ Here $I:X\to X$ is the identity from $X$ to itself and we use the notation that $$\rho(A)=\{\lambda\in\mathbb C;\lambda I-A\ \hbox{has bounded inverse}\},\sigma(A)=\mathbb C\setminus\rho(A).$$
For the question above, I have some slight trials at this time.
For any $\lambda\in\sigma(A),$ let us show that $\lambda^3\in\sigma(A^3).$ Arguing via contradiction and assuming that $\lambda^3\in\rho(A^3),$ that is, $\lambda^3I-A^3:X\to X$ is injective and also surjective. We note that $$\lambda^3 I-A^3=(\lambda I-A)(\lambda^2I+\lambda A+A^2)=(\lambda^2I+\lambda A+A^2)(\lambda I-A),$$ then $\lambda^2I+\lambda A+A^2:X\to X$ is injective and also surjective, and $\lambda I-A$ is injective, but it must not be surjective since $\lambda\in\sigma(A).$ For any operator $T:X\to X,$ we denote by $R(T)$ the range of $T.$ Since $R(\lambda I-A)\subsetneqq X$ and $(\lambda^2 I+\lambda A+A^2)|_{R(\lambda I-A)}:R(\lambda I-A)\to X$ is surjective, we conclude that $\lambda^2 I+\lambda A+A^2:X\to X$ should not be injective, a contradiction.
On the other hand, if $\mu\in\sigma(A^3),$ let $\omega_1,\omega_2,\omega_3\in\mathbb C$ be the complex solutions of the equation $z^3=\mu.$ Then we obtain $$\mu I-A^3=\prod_{i=1}^3(\omega_iI-A),$$ and there must be some $i\in\{1,2,3\}$ such that $\omega_i I-A:X\to X$ does not possess bounded inverse, so $\omega_i\in\sigma(A).$
For above proof of my question, is it true? Any comments are welcome and thanks in advance!
Let $\lambda \in \mathbb{C}$ be given with $\lambda\ne 0$. Let $\lambda_1,\lambda_2,\lambda_3$ be the unique roots of $p(z)=z^3-\lambda$. Then $$ A^3-\lambda I = (A-\lambda_1 I)(A-\lambda_2 I)(A-\lambda_3 I). $$ $A^{3}-\lambda I$ is non-invertible iff one of the factors on the right is non-invertible. That is, $\lambda\in\sigma(A^3)$ iff $\{ \lambda_1, \lambda_2,\lambda_3 \} \cap\sigma(A) \ne \emptyset$. Equivalently, $\lambda\in\sigma(A^3)$ iff $\lambda \in \sigma(A)^3$. The case of $\lambda=0$ is easily handled separately.