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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?

Jean Marie
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Maggie
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1 Answers1

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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.

Paultje
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