The spectrum of a bounded linear operator Suppose $X$ is a Banach space. For $T\in L(X,X)$, let its spectrum
be $\sigma(T)$.
Show that $\lambda\in\sigma(T)\Rightarrow\lambda^{n}\in\sigma(T^{n}),\ \forall n\in\mathbb{N}$.
Show that the converse is true for $\mathbb{C}$ but not for $\mathbb{R}$.
Thank you. 
 A: The converse is not "for all $n$, if $\lambda^n \in \sigma(T^n)$ then $\lambda \in \sigma(T)$" (that would be false), it is "if $\lambda^n \in \sigma(T^n)$ for all $n$, then $\lambda \in \sigma(T)$".   Well, that's trivially true (take $n=1$).  Somewhat less trivial (for either $\mathbb C$ or $\mathbb R$) is that you can have $\lambda^n \in \sigma(T^n)$ for all $n > 1$ and still
$\lambda \notin \sigma(T)$.  For example, this will be the case if $\lambda = -1$ and
$\sigma(T) =\{e^{i\theta}: -\pi/2 \le \theta \le \pi/2\}$.
EDIT: What is true over $\mathbb C$ but not $\mathbb R$ is that for every $\mu \in \sigma(T^n)$ there is $\lambda \in \sigma(T)$ such that $\lambda^n = \mu$. 
A: Consider $\mathbb{C}$-case. We will need two following simple observations. 
1) Let $a=a_1\cdot\ldots\cdot a_n$, for some elements $a$,$a_1,\ldots,a_n$ of algebra $A$ and assume that $a_1,\ldots,a_n$ commute. Then 
$$
a\text{ is invertible }\Longleftrightarrow a_1,\ldots, a_n\text{ are invertible.}
$$
2) One can easily check that 
$$
T^n-\lambda^nI=(T-\lambda I)\left(\sum\limits_{k=0}^{n-1}\lambda^{n-1-k}T^k\right)=\left(\sum\limits_{k=0}^{n-1}\lambda^{n-1-k}T^k\right)(T-\lambda I)
$$
Now we have implication
$$
\lambda^n\notin\sigma_\mathbb{C}(T^n)\Longleftrightarrow
T^n-\lambda^nI\text{ is invertible }\Longrightarrow
$$
$$
T-\lambda I\text{ is invertible }\Longleftrightarrow
\lambda\notin\sigma_\mathbb{C}(T)
$$
Thus, $\lambda\in\sigma_\mathbb{C}(T)\Longrightarrow\lambda^n\in\sigma_\mathbb{C}(T^n)$. As Robert Israel pointed out the reverse implication holds only if we are given $\lambda^n\in\sigma_\mathbb{C}(T^n)$ for all $n\in\mathbb{N}$.
Consider $\mathbb{R}$-case. Define bounded linear operators
$$
T:\mathbb{R}^2\to\mathbb{R}^2:(x_1,x_2)\mapsto(-x_2,x_1)
$$
$$
T^2:\mathbb{R}^2\to\mathbb{R}^2:(x_1,x_2)\mapsto(-x_1,-x_2)
$$
This are rotations on the angles $90^\circ$ and $180^\circ$ respectively. Matrices of this operators in standard basis are
$$
[T]=\begin{pmatrix}0&&-1\\1&&0\end{pmatrix}\qquad
[T^2]=\begin{pmatrix}-1&&0\\0&&-1\end{pmatrix}
$$
You can easily check that $\sigma_\mathbb{R}(T)=\varnothing$ whereas $\sigma_\mathbb{R}(T^2)=\{-1\}$.
