# $T \in \mathcal L (V)$ has no real eigenvalues. Prove that every subspace of $V$ invariant under $T$ has even dimension.

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?

• Do you mean no real eigenvalues? Is $V$ finite dimensional? – copper.hat Jun 11 '19 at 15:50
• @copper.hat The topic do not mention that, but I think so. – Maggie Jun 11 '19 at 15:53
• Think of something odd. – copper.hat Jun 11 '19 at 16:06
• I have added adjective "real" in front of "eigenvalues". I wish you don't see an objection. – Jean Marie Jun 11 '19 at 16:46

In order to understand why this is only true for odd dimension you should investigate the characteristic polynomial of the restricted linear map $$T\rvert_U : U \rightarrow U .$$