If $T$ is not invertible normal operator and $S$ is a bounded operator. Why $TS$ and $ST$ have the same spectrum?

Proof: Assume that $T$ is not invertible normal operator, then $0 \in \sigma(T)$. Since $0$ is in the approximate point spectrum of $T$, it is clear that $0 \in \sigma(ST).$ Since $T$ is normal, it holds $\|Tx \| = \|T^*x \|$ for any vector $x$. Hence $0$ is in the approximate point spectrum of $T^*$ and hence $0 \in \sigma(S^*T^*) = \sigma((TS)^* = \overline{\sigma(TS)}$. Hence $0 \in \sigma(TS)$.

Recall the following definition:

Definition: Let $T$ be a bounded linear operator of a complex Hilbert space $\mathcal{H}$. The approximate point spectrum of $T$ is the set of all values $\lambda \in \mathbb{C}$ such that there exists a sequence of unit vectors $(x_n)_n\subset \mathcal{H}$ so that $\|(T-\lambda)x_n\|\to 0$ as $n\to \infty$.

  • $\begingroup$ I think you should give more detail what exactly you dont understand, because there are multiple steps in the second case. $\endgroup$ – supinf Dec 7 '18 at 15:36
  • $\begingroup$ @supinf I don't understand why If $T$ is not invertible then $0$ is in the approximate point spectrum of $T$? because $0 \in \sigma(T)$. $\endgroup$ – Schüler Dec 7 '18 at 15:40
  • $\begingroup$ Your definition of approximate point spectrum includes the point spectrum. $\endgroup$ – DisintegratingByParts Dec 8 '18 at 7:25

Without the assumption of normality, we can at least prove the following:

If $T,S\in \mathcal{B}(\mathcal{H})$, then $\{0\}\cup\sigma(ST)=\{0\}\cup\sigma(TS)$.

(This holds more generally in unital Banach algebras.) The second case is saying that $0$ is already in $\sigma(ST)$ and $\sigma(TS)$.


This follows from the general statement:

If $T\in \mathcal{B}(\mathcal{H})$ is normal, then $\sigma(T)=\sigma_{ap}(T)$.

Indeed, fix $\lambda\in\sigma(T)$. If $\lambda$ is an eigenvalue, we are done, so assume it is not, i.e., assume $\ker(T-\lambda)=\{0\}$. Since $T$ is normal, we have $\ker(T^*-\overline\lambda)=\{0\}$, and thus $$\overline{\operatorname{Range}(T-\lambda)}=\ker(T^*-\overline\lambda)^\perp=\mathcal{H}.$$ Hence $\operatorname{Range}(T-\lambda)$ is a proper dense subspace of $\mathcal{H}$. Thus $T-\lambda$ is not bounded below, and the result follows.

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  • $\begingroup$ Thank you for your answer, I know the jacobson Lemma. However I want to prove that if $T$ is normal then $\sigma(ST)=\sigma(TS)$ $\endgroup$ – Schüler Dec 7 '18 at 15:30
  • $\begingroup$ If $T$ is not invertible, then $0 \in \sigma(T)$. But why $0$ is in the approximate point spectrum of $T$? Thanks a lot. $\endgroup$ – Schüler Dec 7 '18 at 15:35
  • $\begingroup$ Please see my edit. $\endgroup$ – Schüler Dec 7 '18 at 15:47
  • $\begingroup$ I see. I'm about to go to class, so I can't edit my answer just yet. If nobody answers by the time I'm out I'll edit to answer your question. $\endgroup$ – Aweygan Dec 7 '18 at 15:54
  • $\begingroup$ Yes, but I haven't had time to reply. I'll take a look at it later $\endgroup$ – Aweygan Feb 6 '19 at 20:50

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