Linear transformation from 2D to 3D or vice versa. Will there be Eigenvectors? I am studying Eigenvectors and I encountered this observation and couldn't find an answer to it. Let's say a transformation applied on a vector, if a vector knocked off its span during the transformation, it won't count as eigenvector. 
Thus, my question is, what if a 2D vector transformed to 3D vector and it kept the criteria about being a vector stretched or squashed during the transformation without being knocked off its 2D span. Would it still be eigenvector? what about the opposite way?
 A: An eigenvector must satisfy $Av = \lambda v$. If $ A: \mathbb{R}^m \mapsto \mathbb{R}^n$ with $m \neq n$, then the vectors on either side won't even be the same type, let alone proportional. So eigenvectors only make sense for transforms between vector spaces of the same dimension.
However, your question raises two interesting questions: 


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*For a transform $ A: \mathbb{R}^2 \mapsto \mathbb{R}^3$, does there exist a 2-dimensional subspace (namely, a plane spanned by axes) such that a vector v in the subspace satisfies $Av = v'$, where the first two components of v' are proportional (by the same $\lambda$) to the corresponding elements of v, and the third component of v' is zero? 


The answer to this question is yes. In fact, all linear transforms  $ A: \mathbb{R}^2 \mapsto \mathbb{R}^3$ have a 2-dimensional image (subspace). Whenever this subspace is precisely a plane spanned by axes, we can talk about eigenvectors. Note that this is solely because such a transform is the exact same as a transform $ A': \mathbb{R}^2 \mapsto \mathbb{R}^2$, which of course has eigenvectors (real or complex).  Even if such an image space is not spanned by axes, then the restriction of the matrix transform to its 2-dimensional image subspace can have real or complex eigenvectors, but these will again be the eigenvectors of a corresponding 2 x 2 matrix transform with respect to some other basis.


*For a transform $ A: \mathbb{R}^2 \mapsto \mathbb{R}^3$, define a "quasi-eigenvector" as a vector satisfying $v = \lambda v'$, where $v'$ is the "shadow" (projection) of the vector in 3-D space onto some 2-D space (i.e. a plane spanned by axes). Even then, the combined transform of first taking a 2-D vector and turning it into a 3-D vector, then projecting it into 2-space, is the same (i.e. isomorphic) as a transform $ A: \mathbb{R}^2 \mapsto \mathbb{R}^2$ (i.e. a square matrix). Eigenvectors are thus well defined, but no different from taking eigenvectors between spaces of same dimension.


My answer might be a bit unsatisfying, but it is a demonstration of why eigenvectors aren't well defined for linear mappings between spaces of different dimension. And if they ever make sense, it is because of a choice of restriction that makes the domain and codomain of equal dimension.
