# why does an automomorphism of a vector space have an eigenvalue

"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

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