A linear transformation has a 3D Invariant subspace

Question

Suppose $V$ is vector space on $\mathbb{Q}$; $\mathscr{A}$ is a non-zero linear transformation on $V$ satisfying $\mathscr{A}^4=4\mathscr{A}^2-2\mathscr{A}$. We have easily that $V=\mathscr{A} V\oplus \mathscr{A}^{-1}(0)$. But how can I deduce that $\mathscr{A}$ has a three dimensional invariant subspace? Thank you.

Answer 1

If for all $v\in V$, we have $\mathscr{A}v=0$, then $\mathscr{A}\equiv 0$. Then this is true when $\mathrm{dim}\,V\geqslant 3$.

Otherwise, let $v\in V$ s.t. $\mathscr{A}v\ne 0$. Denote $\mu=\mathscr{A}v$, then$$(\mathscr{A}^3-2\mathscr{A}+2I)\mu=\mathscr{A}(\mathscr{A}^3-2\mathscr{A}+2I)v=0.$$Therefore if we can prove $\mu,\mathscr{A}\mu,\mathscr{A}^2\mu$ are linearly independent on $\mathbb{Q}$, then $\mathrm{span}\{\mu,\mathscr{A}\mu,\mathscr{A}^2\mu\}$ would be a three dimensional invariant subspace.This is obivious.

First, $\mu,\mathscr{A}\mu$ are linearly independent on $\mathbb{Q}$, otherwise let $a\mu+\mathscr{A}\mu=0$,where $a\in\mathbb{Q}$. Then $a$ must satisfy $a^3-4a+2=0$, which is impossible for $a\in\mathbb{Q}$.

Furthermore, if $\mu,\mathscr{A}\mu,\mathscr{A}^2\mu$ are linearly dependent on $\mathbb{Q}$, let $$a\mu+b\mathscr{A}\mu+\mathscr{A}^2\mu=0,$$where $a,b\in\mathbb{Q}$. Then we must have $a+bx+x^2|x^3-4x+2$, which is impossible since $f(x)=x^3-4x+2$ is irreducible on $\mathbb{Q}$.