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"Let f be an automorphism of a complex space $V \ne O$. Then f has an eigenvalue." Could you explain why this is true? Thanks. edit: V has a finite dimension

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    $\begingroup$ I feel like you need the assumption that $V$ is finite for this to be true. $\endgroup$ – Cameron Williams Mar 5 '16 at 21:56
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    $\begingroup$ What do you know about polynomials over $\mathbb{C}$? $\endgroup$ – carmichael561 Mar 5 '16 at 22:00
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    $\begingroup$ An eigenvalue is a root of the characteristic polynomial of $f$. But a complex polynomial always has a root. $\endgroup$ – Crostul Mar 5 '16 at 22:09
  • $\begingroup$ carmichael561 That they always have a root? $\endgroup$ – Kryštof Mar 5 '16 at 22:10
  • $\begingroup$ @Kryštof - yes. That is the fundamental theorem of algebra: Every polynomial over the complex numbers has a root. $\endgroup$ – Paul Sinclair Mar 6 '16 at 0:31

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