# What's an example of a linear function $f: \mathbb Q^3 \to \mathbb Q^3$ such that there are no proper invariant subspaces?

Is there a linear automorphism $$f$$ on $$\mathbb Q^3$$ such that the only $$f$$-invariant subspaces of $$\mathbb Q^3$$ are $$0$$ and $$\mathbb Q^3$$, and, if so, then what is an example?

• Try $f(a,b,c)=(2c,a,b)$? Jul 19, 2019 at 15:19

Hint It suffices to find a linear transformation $$\Bbb Q^3 \to \Bbb Q^3$$ whose characteristic polynomial is irreducible over $$\Bbb Q$$. On the other hand, we can build a companion matrix with prescribed characteristic polynomial $$p(t) = t^3 + a_2 t^2 + a_1 t + a_0$$, namely, $$C_p := \pmatrix{ 0 & 0 & -a_0 \\ 1 & 0 & -a_1 \\ 0 & 1 & -a_2 } .$$
So, to find a linear transformation $$\Bbb Q^3 \to \Bbb Q^3$$ we can pick any (monic) cubic polynomial $$t^3 + a_2 t^2 + a_1 t + a_0$$ irreducible over $$\Bbb Q$$ and interpret its companion matrix $$C_p$$ as a linear transformation $$T_p : \Bbb Q^3 \to \Bbb Q^3$$ by fixing any basis of $$\Bbb Q^3$$; taking the standard basis defines the transformation $$T_p (x, y, z) = (y, z, -(a_0 x + a_1 y + a_2 z)) .$$ Applying the Rational Root Theorem to the polynomial $$p$$ with $$a_0 = -2, a_1 = a_2 = 0$$ quickly shows that it is irreducible, giving a concrete example that coincides with Mindlack's example in the comments up to reordering of components.