Diagonalizable matrices over $\mathbb{C}$

Question

1. Consider $A=\begin{bmatrix}
   0 & -2 & 1 \\
   1 & 2 & -1 \\
   3 & -1 & -3
   \end{bmatrix}$
   . Is $A$ diagonalizable over $\mathbb{C}$?

2. Does there exist a $3\times 3$ matrix with rational coefficients with no eigenvectors over $\mathbb{Q}$ which is not diagonalizable over $\mathbb{C}$? Find an example of such a matrix, or prove non exists.

For 1), the characteristic polynomial is $\lambda^3+\lambda^2-8\lambda-7$. If I can factor this, and if the eigenvalues are distinct, then $A$ is diagonalizable. However, I don't think I can factor this on my own. I wonder if there is any other easier ways to determine whether this is diagonalizable.

For 2), I don't think such matrix exist, because if it's over $\mathbb{C}$, we can always make the characteristic polynomial split. Is this correct?

Thanks in advance for your help!

Answer 1

1. If $p=\chi_A$ is reducible over $\mathbb{Q}$, then it has a rational root (because of degree $3<2+2$). By Gauß lemma, it will then have an integral root. Since $p$ is monic, this integral root $\xi$ must devide the constant coefficient $7$, which leaves $\xi\in\{\pm 1, \pm 7\}$ as candidates. None works out. Thus $p$ is irreducible over $\mathbb{Q}$ and has $3$ distinct roots. Thus $A$ is diagonizable.

2. See the comments.