How is the spectrum of a product of operators related to the spectrum of each term in the product? I will use the usual notation of $\sigma(A)$ to denote the spectrum of an operator $A$.
Is there a relationship between the spectrum of bounded operators (on complex Hilbert space) $A$ and $B$ and the spectrum of the product $AB$? A seemingly too good to be true hope would be that the spectrum of $AB$ is $\sigma(A)\cdot \sigma(B)$, but I highly doubt that is true. On this note, is there a relationship between the spectrum of $A$ and $B$ and the spectrum of the sum $A+B$? Is is somehow related to $\sigma(A) +\sigma(B)$.
It is easy to show the that $\sigma(AB) = \sigma(A)\sigma(B)$ and $\sigma(A+B) \subset \sigma(A)+\sigma(B)$ in the setting of commutative $C^*$ algebras, but the general case does not seem straightforward to me.
 A: No, these statements are already false for $2 \times 2$ matrices. The correct statements for commutative C*-algebras are also subtle: the problem is that the spectrum does not come in any prescribed order, so how do you know which elements of the spectrum to add / multiply with which? It's certainly not true that you get all sums / products (which is the standard interpretation of the notation you're using, and which makes the claim $\sigma(AB) = \sigma(A) \sigma(B)$ false). 
You might at least hope that the spectrum of $A + B$ or $AB$ is bounded by the spectrum of $A$ and the spectrum of $B$ in some way, but this isn't true either, and again there's already a counterexample for $2 \times 2$ matrices (take $A$ and $B$ to be nilpotent). In the $2 \times 2$ case I think in general the only constraints are that $\text{tr}(A + B) = \text{tr}(A) + \text{tr}(B)$ and $\det(AB) = \det(A) \det(B)$. 
If $A$ and $B$ are self-adjoint you at least have bounds: in this case the C*-norm coincides with the spectral radius, so $\sigma(A + B)$ lies in a ball of radius 
$$\| A + B \| \le \| A \| + \| B \| = \nu(A) + \nu(B)$$ 
around the origin. $AB$ is no longer self-adjoint, but $\sigma(AB)$ still lies in a ball of radius 
$$\sqrt{\| AB (AB)^{\ast}\| } \le \| A \| \| B \| = \nu(A) \nu(B)$$
around the origin. 
For the special case of sums of self-adjoint matrices or products of unitary matrices much more can be said; see the Knutson-Tao theorem and its "quantum" analogue. 
