When can we bring a matrix to its Jordan form within a subfield of $\mathbb C$? 
Can the following matrix $A$ be brought into Jordan form over the field of rational numbers?
  $$ A=\begin{pmatrix}-3&-1&-1\\6&4&1\\6&5&0\end{pmatrix} $$


My solution:
By direct computation, we have the Jordan form of $A$ is $$ J(A)=\begin{pmatrix}-1&0&0\\1&-1&0\\0&0&3\end{pmatrix}. $$
Moreover, we get $v_3=(1,-3,-3)$ is an eigenvector of $3$ and $v_2=(1,-1,-1)$ is an eigenvalue of $1$. And we can find a vector $v_1=(0,0,-1)$ such that $v_1A=-v_1+v_2$. Hence, if we set $$P=\begin{pmatrix}0&1&1\\0&-1&-3\\-1&-1&-3\end{pmatrix}\in M_3(\mathbb Q)$$ then $P^{-1}AP=J(A)$.

However, in general, is it true that if all the entries and eigenvalues of a matrix $A$ are in a subfield, say $F$, of $\mathbb C$, then we can always bring $A$ into its Jordan form within the field $F$? 
 A: Yes that's true. In general, you can state the following:

Let $F$ be a field, and let $A \in M_{n \times n}(F)$ be a given matrix. $A$ has a Jordan canonical form over $F$ if and only if the characteristic polynomial of $A$ splits over $F$ (i.e all eigenvalues of $A$ belong to $F$).

By "$A$ has a JCF over $F$" I mean there exist matrices $P,J \in M_{n \times n}(F)$, with $J$ being a Jordan matrix, such that $A = PJP^{-1}$.
The "only if" part of the statement is trivial. The "if" part of the statement is non-trivial and the proof can be found in any decent algebra/linear algebra book. You may also find the Theorem stated in the language of linear transformations as follows:

Let $V$ be a vector space over a field $F$, and let $T:V \to V$ be a given linear transformation. $T$ has a Jordan canonical form if and only if the characteristic polynomial of $T$ splits over $F$ (i.e all the eigenvalues of $T$ belong to $F$).

By "$T$ has a Jordan canonical form", I mean there exists a basis $\beta$ of $V$ such that the matrix of $T$ relative to $\beta$, $[T]_{\beta}$, is a Jordan matrix.
