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Let $V$ be a vector space over an algebraically closed field $K$ and let $f:V\to V$ be an 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 $\dim V=\infty$?

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  • $\begingroup$ @EricWofsey Thanks, I'll post a new question then. $\endgroup$ – Sora. Apr 25 '16 at 8:00
  • $\begingroup$ @EricWofsey In order to avoid confusion, I've rolled back this question to its original wording. $\endgroup$ – Sora. Apr 25 '16 at 8:02
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The answer is no. Consider the space of infinite real sequences and let $T(x)=(0,x_1,x_2,\dots)$. Suppose $T(x)=\lambda x$.

If $\lambda=0$, then $x$ has to be zero, so $0$ is not an eigenvalue.

If $\lambda \neq 0$, then $0=\lambda x_1$, so $x_1=0$, but then $x_1=0=\lambda x_2$. Continuing by induction we again conclude that $x=0$. So $\lambda$ is not an eigenvalue either.

So $T$ has no eigenvalues.

Note that there is no contradiction with the finite dimensional case, because the finite dimensional analogue of this "forward shift" operator must necessarily "forget" about $x_n$, which means that $(0,0,\dots,1)$ will be an eigenvector with eigenvalue zero.

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No, the shift homomorphism defined on $(e_1,...,e_n,...)$ by $f(e_i)=e_{i+1}$ does not have an eigenvalue.

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