Existence of a non-zero proper subspace True/ False:
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) $\subset$ V.
This type of question is kind of new to me and how much I can think of it is that I assume there is a play between the invariant subspace and the role to the Jordan form. But I'm thinking how to apply them to reach the solution out. Also, am I on the right track?
Thanks!
 A: Let $V$ be such a subspace and let $p(t)$ denote the characteristic polynomial of $T$. Show the following:

*

*$T|_V$ is a linear transformation from $V$ to $V$.

*Let $q(t)$ be the characteristic polynomial of above. Show that $q(t) \mid p(t)$.

*Show that there exists an irreducible fourth-degree polynomial (e.g., $t^4 - 2$).

*Show that given any polynomial, there exists an operator having that as its characteristic polynomial. Can you conclude?

A: If we pick a basis of $V$ and extend it to a basis of $\Bbb Q^4$, then relative to this basis, $T$ is given by a block triangular matrix
$$ \pmatrix{A&B\\0&C}$$
where $A$ may be $1\times 1$, $2\times 2$, or $3\times 3$, and $C$ accordingly $3\times 3$, $2\times 2$, or $1\times 1$. Then $\det T=\det A\det C$ as well as $\det (T-\lambda)=\det (A-\lambda)\det (C-\lambda)$. In short: If the characteristic polynomial of $T$ is irreducible over the rationals, then no such $V$ can exist. Try to find such a polynomial and a $T$ with that as characteristic polynomial.
