I've started studied eigenvalues and eigenvectors.
If there is a transformation T: V->V
I can find out a matrix of T with fixed basis and characteristic polynomial of T.
With this characteristic polynomial of T, I can find out eigenvalues and eigenvectors
Does this mean that this transformation T: V->V has eigenvalues and eigenvectors?
And,
If there is a transformation T: S->V, S is a subspace of V.
then Matrix of T is not square, and therefore I can not find out characteristic polynomial. Does this imply that there exists no eigenvalues and eigenvectors under Transformation from subspace into Space ?
To answer your first question: yes. If you can find the characteristic polynomial of a transformation, then its eigenvectors are the zeros of that polynomial. Note that these eigenvalues will be the same, regardless of your choice of basis.
As for your second question: eigenvalues (like the determinant) are only defined on transformations that take a space to itself (or to a space of the same dimension). So, $T:S \to S$ has eigenvalues, as does $T:V \to V$, but not $T:S \to V$ (even if $S$ is a subspace of $V$).
If $S$ is a $T$-invariant subspace of $V$, then we could take the restriction of the map $T: V \to V$ to get the "smaller" transformation $T:S \to S$. Also, if $S \subset V$ and $T:V \to S$, then we can think of $T$ as a (non-surjective) map from $V$ to $V$ and thereby find its eigenvalues.