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as the title, why if $S$ is a linear operator on an odd-dimensional real vector space, then it has a real eigenvalue?

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    $\begingroup$ Please, edit your post. $\endgroup$ – Sigur Dec 3 '12 at 18:26
  • $\begingroup$ How do you compute the eigenvalue of $S$? Is it a root? $\endgroup$ – Sigur Dec 3 '12 at 18:27
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Hint: Consider the degree of the characteristic polynomial of the transformation. What does that imply about the nature of its roots? (Recalling that complex roots come in pairs)

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Not true if your vector space is not real: for example, the map $t\mapsto it$ from $\mathbb C$ to $\mathbb C$ has the only eigenvalue $i$ and no other eigenvalue.

But it is true if the vector space is real. This has to do with the fact that complex roots of a real polynomial appear in pairs.

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