We have to determine as a function of $n$, the number of possible values for $\det(A)$ given that $A$ is an $n\times n$ real matrix with $A^3-A^2-3A+2I=0$.
I think the solution is just to notice that this polynomial has one real root $a$ and two complex roots $b,\overline b$. Clearly all of the complex roots of $A$ must be $0,a,b$ or $\overline b$. Since the determinant is the product of the complex roots with multiplicity we just have to find all the suitable combinations for the eigenvalue's multiplicities.
The first option is when $0$ is an eigenvalue.
Otherwise there are $\lfloor\frac{n}{2}\rfloor$ options for the multiplicity of $a$, and this determines the multiplicities of $b$ and $\overline b$ which must be equal.
Hence the answer is $\lfloor \frac{n}{2} \rfloor+1$. Is this correct? How would you make it fancier?
