Suppose $V$ is a real vector space and $T \in \mathcal L (V)$ has no real eigenvalues.
Prove that every subspace of $V$ invariant under $T$ has even dimension.
Solution :
Suppose $U$ is a subspace of $V$ that is invariant under $T$. If $\dim U$ were odd, then $T|_{U}$ would have an eigenvalue $\lambda \in \Bbb R$. $\exists v \neq 0, v\in U$ such that $T|_{U} u = \lambda u$.Then $\lambda$ is an eigenvalue of $T$.But $T$ has no eigenvalues, so $\dim U $ must be even.
Why that happened when $T|_{U}$ has odd dimension?
Best Answer
In order to understand why this is only true for odd dimension you should investigate the characteristic polynomial of the restricted linear map $$T\rvert_U : U \rightarrow U .$$
There is something special happening with its degree. Then think of what the relationship between eigenvalues and the characteristic polynomial is.