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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.

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up vote 3 down vote accepted

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 $\sigma_\mathbb{R}(T^2)=\{-1\}$.

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Thank you, but I cannot see how the rotation of matrix is related to this question. Can you clarify a little? – user16859 Apr 19 '12 at 22:47
@user16859: $\mathbb{R}^2$ is a real Banach space, and a rotation matrix is an operator on it... – Martin Wanvik Apr 19 '12 at 22:50
To elaborate, the equivalence should be $T^n-\lambda^nI\text{ is invertible }\Longleftrightarrow T-\lambda I\text{ is invertible and } \sum\limits_{k=0}^{n-1}\lambda^{n-1-k}T^{k} \text{ is invertible}.$ – copper.hat Apr 20 '12 at 0:18
No, $T^2: (x_1, x_2) \mapsto (-x_1, -x_2)$, $[T^2] = \pmatrix{-1 & 0\cr 0 & -1\cr}$ and $\sigma(T^2) = \{-1\}$. – Robert Israel Apr 20 '12 at 0:21
@user16859 Just apply observation 1) where $a=T^n-\lambda^n I$, $a_1=\sum\limits_{k=0}^{n-1}\lambda^{n-1-k}I$ and $a_2=T-\lambda I$ – Norbert Apr 22 '12 at 15:11

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 $\lambda^n = \mu$.

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I agree with your ``EDIT''. In fact, $\sigma(T^{n})=[\sigma(T)]^{n}$ for the complex case. That's why the converse is true for the complex. The proof is not trivial. Do you agree? Also @Norbert – user16859 Apr 22 '12 at 15:49
Which converse? The proof that $\sigma(T^n) = (\sigma(T))^n$ for the complex case is very simple: $T^n - \mu I = \prod_{\lambda:\ \lambda^n = \mu} (T - \lambda I)$, and in any ring with identity a product of commuting factors is invertible iff each factor is invertible. – Robert Israel Apr 22 '12 at 18:29
Thank you. I basically proved that. – user16859 Apr 22 '12 at 19:36

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