Is there exist a notion of determinant for linear operator between space the same dimension? Let linear operator $T: E \to F$ with $\dim E = \dim F = n$. Is there exist a notion of determinant of $T$,  and/or  eigenvalue of $T$?
Note that doesn´t make sense that: $Tv=\lambda v$ because $v \notin F$. 
Thanks
 A: First consider two vector spaces  of dimension $1$. Can we identify a unique number to an isomorphism between them? (For example consider two lines in the plane).  This requires a choice of non-zero vectors in both the spaces. 
Given a linear map $T\colon V\to W$ for vector spaces of the same dimension, one can talk of the associated map on the exterior powers $T_r\colon \wedge ^r V\to \wedge ^r W$. For $r=\dim V$ this becomes a linear map of 1-dimensional bases. Any choice of bases in $V$ and $V$ would lead to canonical choice of bases in these 1-dimensional bases and so lead to a scalar associated to the map of 1-dimensional spaces. This ois the determinant map associated to $T$.
A: I think you can relabel the elements of $F$ so that the target space is just the initial space $E$.  Because dim(E) = dim(F), there is an isomorphic mapping - a one-to-one correspondence - between the elements of E and the elements of F. 
Without relabeling, then I don't see how this makes sense, in the context of basic linear algebra.
Also, the spectral theorem, the theory of eigenvalues and eigenvectors always (to my knowledge) start with letting some linear operator map some space $v$ -> $v$.  So it is by definition that you are mapping from one space to itself.  Although I understand your question - and have myself asked this question to my linear algebra professor in the past.  (His response was that it doesn't always have to be from $v$ -> $v$, but he never elaborated on any special cases, so I cannot say much more.)
