Existence of eigenvalue when domain and codomain are different

I came across this statement: Let V be a complex vector space. If $T:V\rightarrow V$ is a linear transformation, then $T$ has a (complex) eigenvalue. I believe this is a direct consequence of fundamental theorem of algebra. But I was wondering whether $V \rightarrow V$ is necessary (i.e. is the statement true if $T: V\rightarrow S$, where S is not a subspace of $V$?)?
I think it's not true in general because the basis is different. My counterexample in mind is a 90-degree rotation of the x-axis. Then with respect to the standard basis, the eigenvector seems to be 1. But surely the vector changed after the transformation.
I was wondering if this is a valid counterexample? And in general, when we talk about eigenvalue/eigenvector of a linear transformation, do we assume $T:V\rightarrow V$?
Thank you!

Solution 1:

In linear algebra, one usually defines eigenalues and eigenvectors only for linear transformations from a space to itself. However, in functional analysis one considers unbounded linear operators that are only defined on a subspace of a vector space, and there one does consider eigenvalues and eigenvectors.