Example of a linear operator T with only trivial T-invariant subspaces I am trying to construct an example of a linear operator $T : \mathbb{Q}^3 \rightarrow \mathbb{Q}^3$ for which the only $T$-invariant subspaces are the whole space and the zero subspace.  
If we first look at an example from the 2x2 case let $T$ be the linear operator on $\mathbb{R}^2$ represented in the standard ordered basis by
$$
A = \left( \begin{array}{ccc}
0 & -1  \\
1 & 0  \end{array} \right)  
$$
Then if $W$ is any other invariant subspace not equal to $\{0\}$ or the whole space then $W$ must have dimension $1$ and so is spanned by some nonzero vector $\alpha$.  But $W$ invariant under $T$ implies that $\alpha$ is a eigenvector, but $A$ has no real real eigenvalues.
If we try to apply the above logic to a 3x3 matrix then I am stuck on what to do if I assume the dimension of the invariant subspace is 2.  
Question: In any case is it still clear that if $A$ represents some linear operator $T : \mathbb{Q}^3 \rightarrow \mathbb{Q}^3$ then for $T$ to have no nontrivial invariant subspaces should A not have any real eigenvalues?

Over the reals, you won't find any examples in dimension 3 or any odd dimension because every operator in such a space has an eigenvector (since every real polynomial of odd degree has a real root).
Over the rationals, you only need to find a polynomial of degree 3 with rational coefficients having no rational root and take its companion matrix. The simplest one I can think of is $x^3-x-1$. 
