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Let $V$ be a vector space over an algebraically closed field $K$ and let $f:V\to V$ be an automorphism, i.e. a bijective endomorphism. If $V$ is finite-dimensional, we know that the characteristic polynomial $\chi_f$ has a zero $\lambda$, which is an eigenvalue of $f$. However, does a similar argument hold if $\dim V=\infty$?

Remark: In my previous question, I asked the same without requiring $f$ to be bijective. The counterexamples from the answers weren't bijective, so now I'm wondering if there's a bijective counterexample as well or if there's a way to prove the statement for the bijective case.

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Here's a counterexample. Let $V$ be the space of functions $\mathbb{Z}\to K$ which have finite support. Define $T:V\to V$ by $(T\varphi)(n)=\varphi(n+1)$ (i.e., thinking of an element of $V$ as a "bi-infinite sequence", shift the terms in the sequence by $1$). Then $T$ is an isomorphism, but if $T\varphi=\lambda\varphi$, then $\varphi(n)=(T^n\varphi)(0)=\lambda^n\varphi(0)$ for all $n\in\mathbb{Z}$, so $\varphi$ cannot have finite support unless it is always $0$. That is, $T\varphi=\lambda\varphi$ implies $\varphi=0$, so $T$ has no eigenvalues.

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