Eigenvalues existence

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?

The answer is "yes" if $V$ is allowed to be infinite-dimensional, and "no" if $V$ is finite-dimensional.

For the example in infinite dimensions, simply consider $\mathbb C^\omega:=\{(x_1,x_2,\ldots):x_i\in\mathbb C\}$ and the linear function $f\colon\mathbb C^\omega\to \mathbb C^\omega$ defined by $f(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$. There is nothing special about $\mathbb C$ here; the same example works over any field

On the other hand, if $V$ is a finite-dimensional dimensional vector space over $\mathbb C$, suppose $\dim V=n$ and consider $L(V,V)$, the space of linear functions from $V$ to itself, which is also a vector space over $\mathbb C$ under pointwise addition and scalar multiplication and has dimension $n^2$.

For $f\in L(V,V)$, and a polynomial $p(t)=a_0+a_1t+\dots+a_kt^k\in\mathbb C[t]$, we can define $p(f)=a_0\text{id}_V+a_1f+a_2f^{\circ 2}+\dots+a_kf^{\circ k}$ where $\text{id}_V$ is the identity map on $V$ and $f^{\circ k}=f\circ f\circ\cdots\circ f$, $k$ times. Note that $p(f)\in L(V,V)$ as well, and that we can factor $p(f)$ (using $\circ$ as multiplication) in the same way as we would factor $p(t)\in\mathbb C[t]$.

Now, consider the set $\{\text{id}_V,f,f^{\circ 2},\dots,f^{\circ n^2}\}$; this is a set of $n^2+1$ vectors in $L(V,V)$, so there must be a non-trivial linear dependence $c_0\text{id}_V+c_1f+\dots+c_{n^2}f^{\circ n^2}=O_V$ where $O_V\in L(V,V)$ is the map which sends everything to $0$. In other words, if $p(t)=c_0+c_1t+c_2t^2+\dots+c_{n^2}t^{n^2}$, then $p$ is a non-zero polynomial in $\mathbb C[t]$ with $p(f)=O_V$. Suppose that $k$ is the largest non-zero index with $c_k\neq 0$, so by dividing by $c_k$, we can suppose that $p(t)=c_0+c_1t+\dots+c_{k-1}t^{k-1}+t^k$.

Now, using the fundamental theorem of algebra, we can factor $p(t)=\prod_{i=1}^k(t-\lambda_i)$ where $\lambda_i\in\mathbb C$ and thus $O_V=p(f)=(f-\lambda_1\text{id}_V)\circ\cdots\circ(f-\lambda_k\text{id}_V)$. Now, $\ker p(f)=V\neq\{0\}$, but if it were the case that $\ker(f-\lambda_i\text{id}_V)=\{0\}$ for all $i$, then this could not happen, so there must be some $i\in[k]$ for which $\ker(f-\lambda_i\text{id}_V)\neq\{0\}$. Hence, find $v\in\ker(f-\lambda_i\text{id}_V)\setminus\{0\}$, so $f(v)-\lambda_i\text{id}_V(v)=0\implies f(v)=\lambda_i v$, i.e. $v$ is an eigenvector for $f$ with eigenvalue $\lambda_i$.