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Let $V$ be an $\mathbb{R}$-vector space with odd dimension, and let $\varphi$ be an endomorphism on $V$. Show $V$ has a one-dimensional $\varphi$-invariant subspace.

I already know that $\ker f(\varphi)$ is a $\varphi$-invariant subspace for any polynomial $f$ with coefficients in $\mathbb{R}$. Can I somehow use this to find the desired one-dimensional subspace of $V$?

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Hint: The characteristic polynomial of $\varphi$ has a real root because it has odd degree.

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  • $\begingroup$ This is what I was thinking, but why are we guaranteed that the characteristic polynomial will have odd degree? $\endgroup$
    – morrowmh
    Jun 3, 2020 at 23:04
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    $\begingroup$ Nevermind, if the matrix representation of $\varphi$ is an $n\times n$ matrix, then $n=\dim V$ and $n$ is the degree of the characteristic polynomial. $\endgroup$
    – morrowmh
    Jun 3, 2020 at 23:17
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    $\begingroup$ Then we just take the span of one of the eigenvectors. $\endgroup$
    – morrowmh
    Jun 3, 2020 at 23:30
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    $\begingroup$ Exactly. Well done. $\endgroup$
    – lhf
    Jun 4, 2020 at 0:05

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