Determinant and trace of a matrix 
Find $\det A$ and $\text{Tr} A$ for the matrix $A\in M_n(\mathbb{Q})$ such that $\sqrt[n]{p}$ is an eigenvalue of $A$, where $p$ is a prime number or a  positive integer such that the square root is irrational number.

My attempt is described below. From hypothesis we know that
$$ \det(A-\sqrt[n]{p}I)=0,
$$ hence $ \det(A+\sqrt[n]{p}I)=0$ because the characteristic polynomial of matrix $A$ has rational coefficients. Multiplying these two relation we get $ \det(A^2-\sqrt[n]{p^2}I)=0$ and so on. 
From here, I don't find something. How to proceed? Thanks.
 A: It seems that your question misses details, otherwise there are many possible solutions.
Indeed, suppose that $p$ is an integer at the power $n$, let's say $p=a^n$, with $a \in \mathbb{N}$. Then any diagonal matrix, with rational coefficients and $a$ on the diagonal, has $\sqrt[n]{p}$ as an eigenvalue. Then you can have any trace and determinant you want...
A: I see that $n$ appears in $\sqrt[n]{p}$ and in $M_n(\mathbb{Q})$, where in the last one I take it as the dimension (= number of columns) of the square matrix $A$. So the characteristic polynomial is a polynomial of degree $n$, with rational coefficients and with $\sqrt[n]{p}$ as a root: 
\begin{equation}
p(\lambda)= \lambda^n + c_{1} \lambda^{n-1} +\dots + c_{n}
\end{equation}
Now each coefficient $c_{i}$ will be a function involving sums of monomials of degree $i$ constructed with $\sqrt[n]{p}$ and the other (possibly complex) roots. For example one can show that $c_{n}=\det(A)$ and $c_{1} = \text{Tr}(A)$. Now the key argument is that to get a rational number involving products of  $\sqrt[n]{p}$, you must at least raise it to the power $n$! This shuold imply that every coefficient is $0$ except from $c_{n}=p$. So you have $c_{n}=\det(A)=p$ and $c_{1} = \text{Tr}(A)=0$. 
Now one should make this kind of argument sound by filling the gaps and proving every step. I hope that I've not made any mistakes and that this is helpful!
EDIT: Note that this argument depends on the fact that $\sqrt[n]{p}^k$ is irrational for all $k<n$
