Given a linear map $T : \mathbb{Q}^4 \rightarrow \mathbb{Q}^4$, there exists a nonzero proper subspace $V$ such that $T(V ) \subset V$.

Question

Given a linear transformation $T : \mathbb{Q}^4 \rightarrow \mathbb{Q}^4$, there exists a nonzero proper subspace $V$ of $\mathbb{Q}^4$ such that $T(V ) \subseteq V$.

Is the statement true?

I think the answer is no. But why? I am not sure about it. Is there any theorem which contradicts the above statement?

I also know that if the above condition holds then there exists a $\lambda$ such that $Tv = \lambda v$ for each $v \in \mathbb{Q}^4$. Somehow this may contradicts.

Please help me.

Answer 1

The answer is indeed negative. Take$$T(a,b,c,d)=(-d,a,b,c).$$Then the minimal polynomial of $T$ is $x^4+1$. If there was such a space $V$, the minimal polynomial of $T|_V$ would divide $x^4+1$. But it would also belong to $\Bbb Q[x]$, and $x^4+1$ is irreducible in $\Bbb Q[x]$.