Properties of a matrix with the minimal polynomial $m_A(t) = t^3+2t^2+t+1$? Note: A is a 9 by 9 matrix over the rationals. 
The question asks if $A$ is diagonizable over the rationals, which it isn't because it's easy to check that $A$ has no rational roots and therefore rational eigenvalues.
$A$ is diagonizable over the complex numbers because in that field there exist 3 distinct eigenvalues for $A$.
Next the quesiton asks me to prove that $A^3 - 5A^2 +4A +3I$ is an invertible matrix. I tried abusing the fact that $A^3 + 2A^2 +A +I = 0$, but didn't really get anywhere. Is solving this as simple as showing that none of the eigenvalues are $0$?
Lastly, the question asks me for the trace and determinant of $A$. I know how to get to them from the characteristic polynomial of $A$, but I have no idea how to get there from the minimal polynomial.
Any help would be appreciated. 
 A: 
Is solving this as simple as showing that none of the eigenvalues are 0?

Yes it is. The fact that a matrix is invertible iff $0$ is not an eigenvalue is a fact which is independent of the field. All invertibility requires is that the determinant is a unit. Since we are in a field, that's equivalent to requiring non-zero determinant, which is in turn equivalent to requiring that $0$ is not an eigenvalue.
Let $B = A^3 - 5A^2 + 4A + 3I = -7A^2 + 4A + 2I$. Let $f(x) = -7x^2 + 4x + 2$. Note that the eigenvalues of $B$ are precisely $f(\lambda)$ where $\lambda$ is an eigenvalue of $A$. Therefore $B$ will be singular if and only if $f$ and $m_A$ share a root, which we can easily check that they do not. Therefore $B$ is invertible.

Lastly, the question asks me for the trace and determinant of A. I know how to get to them from the characteristic polynomial of A, but I have no idea how to get there from the minimal polynomial.

This is a good point. Indeed, the minimal polynomial does not typically determine the trace or determinant of a matrix. In this case however, we can actually recover the characteristic polynomial from the minimal polynomial.
Note that your minimal polynomial splits over $\mathbb{C}$ as
$$m_A(x) = (x-\lambda)(x-\mu)(x-\overline{\mu})$$
where $\lambda$ is a real irrational number and where $\mu$ and $\overline{\mu}$ are complex conjugate root pairs. Your characteristic polynomial must be composed of these irreducible terms as well.
$$\chi_A(x) = (x-\lambda)^a[(x-\mu)(x-\overline{\mu})]^b$$
where $a,b\ge 1$ and the degree of the polynomial is $9$. Note that the complex conjugate terms must appear together due to the conjugate root theorem.
But importantly, your characteristic polynomial is a rational polynomial. The only possibility for these terms to combine into a polynomial with rational terms is for the characteristic polynomial to comprise of $3$ copies of the minimal polynomial (you can check case by case, but there is also a more rigorous argument involving a bit of field theory if you're interested):
$$\chi_A(x) = [m_A(x)]^3$$
All of this is just a consequence that your minimal polynomial is irreducible as a rational polynomial. 
For a perhaps more familiar example, if your matrix were complex and your minimal polynomial was $(x-\lambda)$, then you automatically know that your characteristic polynomial is $(x-\lambda)^n$. This is the same sort of idea.
Knowing your characteristic polynomial, you can easily recover your determinant and your trace.
