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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?

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    I'm curious what you mean by 'fancier'...? – 2017-01-09
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    Are you sure it only has one real root? – 2017-01-09
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    Hint: $A$ can't have eigenvalue $0$ because $(3I+A-A^2)A = 2I$, and hence $A$ is invertible. – 2017-01-09
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    Oh yeah, good point. – 2017-01-09
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    Apparently I made a couple of mistakes. – 2017-01-09
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    So taking everything into account the answer is probably $\binom{n+2}{2}$ right? – 2017-01-09
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    Yes, it is the number of non-negative integer solutions to $a+b+c=n$. Then the possible determinants would be $2^a\phi^b\left(\frac{-1}{\phi}\right)^c = 2^a(-1)^{c}\phi^{b-c}$, where $\phi=\frac{1+\sqrt{5}}{2}$. If you knew $A$ had only rational coefficients, then you'd have $b=c$. – 2017-01-09
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    @ThomasAndrews thanks.${}{}{}{}{}{}$ – 2017-01-09

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