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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.

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  • $\begingroup$ Welcome to MSE. I have two questions: 1) what is $TL(V)$, and 2) what did you try to solve this problem? $\endgroup$ – A.P. Apr 20 '15 at 18:29
  • $\begingroup$ TL(V) means T is a linear operator from V to V. Only thing i could do is thinking about the characteristic polynomial.. $\endgroup$ – Ezgi Gürbüz Apr 20 '15 at 18:50
  • $\begingroup$ Presumably what is meant is $T \in L(V)$. $\endgroup$ – Robert Israel Apr 20 '15 at 18:50
  • $\begingroup$ You must assume your vector space, or at least the subspace, is finite-dimensional: there are examples with infinite-dimensional invariant subspaces. $\endgroup$ – Robert Israel Apr 20 '15 at 18:52
  • $\begingroup$ Yes, it must be finite dimensional but i still can't show it. $\endgroup$ – Ezgi Gürbüz Apr 20 '15 at 18:55
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More precisely (allowing for infinite-dimensional vector spaces): there is no finite-dimensional invariant subspace with odd dimension.

Hint: consider the characteristic polynomial of the restriction of $T$ to an odd-dimensional invariant subspace. What do you know about real roots of polynomials of odd degree?

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Assume to the contrary that there exists a $T$-invariant subspace $W$ that has odd dimension. Consider $T|_W$, the restriction of $T$ on $W$. Let $f(t)$ be the characteristic polynomial of $T|_W$, since $T$ has no real eigenvalues, so is $T|_W$, and by the Fundamental Theroem of Algebra, $f(t)=a(t-c_1)...(t-c_{2k-1})$ where $c_i \in \mathbb C$ for all $i$ (not necessarily distinct) and $a \in \mathbb R$. But this contradicts to $f(t)$ has real coefficients.

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  • $\begingroup$ Or using the fact that real polynomials of odd degrees must have a real root as suggested by Robert $\endgroup$ – Yang May 19 '15 at 14:43

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