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?

  • 2
    $\begingroup$ Try $f(a,b,c)=(2c,a,b)$? $\endgroup$
    – Aphelli
    Jul 19, 2019 at 15:19

1 Answer 1


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.


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