Can a linear transformation going from one vector space to another have eigenvalues? Usually eigenvalues are defined as $T:V \rightarrow V$ from one vector space to itself. But if we consider a vector space in 2D space $V = span\{(1, 0)\}$ and another vector space $W = V^{\perp}$, we get a new linear transformation $T:V \rightarrow W$ which is a linear transformation from one vector space to another. Now I consider the $90^{\circ}$ 2D rotation matrix as the matrix of the linear transformation, $\left( \begin{matrix}
0 & -1 \\
1 & 0 \\
\end{matrix} \right)$, so that I have a linear transformation from one vector space to a different vector space with eigenvalues.
My question is, is what I did correct? And is my statement correct? Can linear transformations into a different vector space have eigenvalues?
 A: Eigenvectors of a linear transformation T are those vectors v, for which T(v) is a scalar multiple of v.  As scalar multiples of v are in the same vector space as V, it does not make sense to talk of eigenvectors when dealing with a linear transformation from one vector space to a different one.   And so eigenvalues are also undefinable in that case.
A: The whole point of the eigenvector/value problem is to find the bases which diagonalize (or quasi-diagonalize) matrix representations of linear operators. The result of this makes iterative numerical procedures more efficient. To iterate of course, one would need the domain and codomain vector spaces to be the same.
That is not to say that $det(A-\lambda I_n)$ can't be calculated for two different vector spaces of the same dimension in some chosen bases. We would just need an interpretation that doesn't really align with the notion of the action of a linear operator on a vector space.
In your example, the rotation matrix actually effects the whole of $\mathbb{R}^2$ and the fixed direction is orthogonal to this plane. In other words $T:V\to V$ where $V = \mathbb{R}^2$. This operator, when viewed in $V = \mathbb{R}^3$ with the standard basis, looks like:
$$A = [T]_{\beta_{std.}} = \begin{bmatrix}0&& -1&& 0\\1&& 0&& 0\\0&&0 && 1\end{bmatrix},$$
from which it is more apparent (by the block form) that the hidden subspace is $T$-invariant.
