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.

When you say $\subset$, do you mean $\subseteq$ or $\subsetneq$? (This confusion is why people universally need to stop using $\subset$ when the sets are allowed to be equal. The subset notation clearly reflects $<$ and $\leq$, and using it any other way is just not right.) — Arthur, Nov 05 '21 at 08:21

score 2 · Accepted Answer · answered Nov 05 '21 at 08:28

2

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 $\Bbb Q[x]$.

answered Nov 05 '21 at 08:28

José Carlos Santos

440,053

Hello Jose sir why you didn't take $T(a,b,c,d)=(a,b,c,d)$? – wasiu Jul 14 '22 at 02:29
1

@wasiu Because then there are lots of subspaces $V$ of $\Bbb Q^4$ with $T(V)\subset V$. In fact, every subspace of $\Bbb Q^4$ has that property. – José Carlos Santos Jul 14 '22 at 02:32
okay thanks you@Jose Carlos sir – wasiu Jul 14 '22 at 02:35

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

1 Answers1