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?

  • 1
    $\begingroup$ Do you mean no real eigenvalues? Is $V$ finite dimensional? $\endgroup$
    – copper.hat
    Commented Jun 11, 2019 at 15:50
  • 1
    $\begingroup$ @copper.hat The topic do not mention that, but I think so. $\endgroup$
    – Maggie
    Commented Jun 11, 2019 at 15:53
  • $\begingroup$ Think of something odd. $\endgroup$
    – copper.hat
    Commented Jun 11, 2019 at 16:06
  • $\begingroup$ I have added adjective "real" in front of "eigenvalues". I wish you don't see an objection. $\endgroup$
    – Jean Marie
    Commented Jun 11, 2019 at 16:46

1 Answer 1


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.


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