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. 


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.
