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I know there are linear maps $f:\Bbb R^{n} \rightarrow \Bbb R^{n}$ where $\Bbb R^{n}$ is on $\Bbb R$ that do not have eigenvalues.

Can we find a linear map $f:V \rightarrow V$ where $V$ is a vector space on $\Bbb C$ such that $f$ does not have eigenvalues?

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    No, every such map has eigenvalues, and this statement is more or less equivalent to the fundamental theorem of algebra. – 2017-02-27
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    It's exactly equivalent to the fundamental theorem of algebra because of characteristic polynomials and [companion matrices](https://en.wikipedia.org/wiki/Companion_matrix) – 2017-02-27

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we can find a map $T\colon v\mapsto v$ where $v$ is a vector space over $c$ such that $$T(f(x))=xf(x)$$ Then $T$ has no Eigenvalue