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 $n\to \infty$.

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.
