# Spectrum of the product of two bounded operators

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 $$n\to \infty$$.

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

EDIT

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.

• Thank you for your answer, I know the jacobson Lemma. However I want to prove that if $T$ is normal then $\sigma(ST)=\sigma(TS)$ – Schüler Dec 7 '18 at 15:30
• If $T$ is not invertible, then $0 \in \sigma(T)$. But why $0$ is in the approximate point spectrum of $T$? Thanks a lot. – Schüler Dec 7 '18 at 15:35
• Please see my edit. – Schüler Dec 7 '18 at 15:47
• 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. – Aweygan Dec 7 '18 at 15:54
• Yes, but I haven't had time to reply. I'll take a look at it later – Aweygan Feb 6 '19 at 20:50