No proper nontrivial T invariant subspaces Let Q be the field of rational numbers. Give an example of a linear operator T: $Q^3 \rightarrow Q^3$ having the property that the only T-invariant subspaces are the whole space and the zero subspace. 
My initial instinct is to go for rotations in three dimensions, but I can't seem to prove that that has no nontrivial T-invariant subspaces. Since the characteristic polynomial has degree three, and eigenvalues come in complex pairs, wouldn't that imply that there is a least one real eigenvalue, so that there must be at least one nontrivial T-invariant subspace, namely the eigenspace?
edit: ok, based on John B's comment, $ \begin{bmatrix}0&-2&0\\1&0&-2\\-1&-1&0\end{bmatrix}$ 
has characteristic polynomial $x^3 -4$, which has no rational roots. So this operator has no one dimensional T-invariant subspaces. How about the case for two dimensional subspaces? 
 A: You can construct an example using some abstract algebra.  The polynomial $f(X) = X^3 - 2$ is irreducible in $\mathbb{Q}[X]$ by the Eisenstein criterion, and so the quotient ring
$$R = \mathbb{Q}[X]/(f(X))$$
is a field as well as a  three dimensional vector space over $\mathbb{Q}$.  If $h(X) \in \mathbb{Q}[X]$, let $\overline{h(X)}$ denote its image in $R$.  The map $T:R \rightarrow R$ given by $\overline{h(X)} \mapsto  \overline{X} \cdot \overline{h(X)}$ is a $\mathbb{Q}$-linear transformation.
To say that $R$ is a field is the same thing as saying that $R$ has no nontrivial proper ideals. But you can check that an ideal of $R$ is the same thing as a $T$-invariant subspace of $R$.  
Now $\overline{1}, \overline{X}, \overline{X^2}$ is a basis for $R$.  Identifying these with the standard basis $e_1, e_2, e_3$ for $\mathbb{Q}^3$, and using the fact that $\overline{X^3} = -\overline{2}$, $T$ is really the linear transformation defined on basis elements by
$$T(e_1) = e_2, T(e_2) = e_3, T(e_3) = -2e_1$$
or as a matrix,
$$T = \begin{pmatrix} 0 & 0 & -2 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$$
