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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1$\begingroup$ Please, edit your post. $\endgroup$– SigurDec 3, 2012 at 18:26
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$\begingroup$ How do you compute the eigenvalue of $S$? Is it a root? $\endgroup$– SigurDec 3, 2012 at 18:27
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2 Answers
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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)
answered Dec 3, 2012 at 18:27
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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.
answered Dec 3, 2012 at 18:29