A Fibonacci identity and $f(Av)=\det(A)f(v)$ Given a diagonalizable linear transformation $A:V\to V$ over an algebraically closed field $k$, and an eigenbasis $v_1,\ldots v_n$ with $Av_i=\lambda_i v_i$, consider the following procedure to define a function $f_A$.
Given a vector $v\in V$, express it as as $v=\sum a_i v_i$.  Let $f_A(v)=\prod a_i$.  This satisfies the nice property that $f_A(Av)=\det(A)f_A(v)$.  
I came across this construction when trying to find invariants for generalized Fibonacci sequences.  In particular, define $G_0=a, G_1=b, G_{n+1}=G_n+G_{n-1}$ for $n>1$.  Applying this construction to the equation
$$ \pmatrix{1 & 1\\ 1 & 0}^n\pmatrix{b \\ a} =\pmatrix{G_n \\ G_{n-1}}$$
yields that $G_{n}G_{n+2}-G_{n+1}^2=(-1)^n(G_0G_2-G_1^2)=(-1)^n(a(a+b)-b^2)$.  In particular, $|G_{n}G_{n+2}-G_{n+1}^2|$ is independent of $n$, so if one wanted to determine whether $(c,d)=(G_n,G_{n+1})$ for some $n$, then a necessary condition is that $c(c+d)-d^2=(-1)^n(a(a+b)-b^2)$.  
This identity is similar to but more general than the classical identity $F_{n-1}F_{n+1}-F_n^2=(-1)^n$
My question: Has this construction been studied before? Is there a cleaner approach to it?  Does it generalize?  Are there other, better ways to generate functions that satisfy $f(Av)=\det(A)f(v)$?

Update: The construction can be rephrased (up to a constant) as $\det(M(v))$ where $M(v)$ is the matrix whose columns are the $A$-eignevectors which sum to $v$.  The property $f(Av)=\det(A)f(v)$ comes from the property that $M(Av)=AM(v)$.  Therefore, this question could be answered by a classification of the maps $V\to \hom(V,V)$ which commute with $A$. I feel like this kind of tensor construction has a better chance of having been studied.
 A: This is an answer to your motivational question, not your asked question: in an earlier edit you claimed that you did not know how to prove the identity for $G_n$ using only determinants. Actually it is straightforward to do so: take determinants of both sides of the identity
$$\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]^n \left[ \begin{array}{cc} b & a + b \\ a & b \end{array} \right] = \left[ \begin{array}{cc} G_n & G_{n+1} \\ G_{n-1} & G_n \end{array} \right].$$
A: There is a nice alternative to the construction which yields the same function (up to a constant), works if $A$ is not diagonalizable (although is zero if $A$ is diagonalizable with repeated eigenvalues), and always has coefficients in the base field/ring.  We define $M_{A,v}$ to be the matrix whose $j$th column is $A^{j-1}v$, i.e., 
$$M_{A,v}=\pmatrix{v & Av & A^2v & \cdots & A^{n-1}}.$$
It is straight forward that $AM_{A,v}=M_{A,Av}$, and so if we define $f_A(v)=\det(M_{A,v})$, then by the observation in the question update, $f_A(Av)=\det(A)f_A(v)$.
Moreover, we can directly compare this construction to the one in the question. Assume $A$ is diagonalizable, with eigenbasis and eigenvalue in the question.  Let $P$ be the matrix whose columns are the eigenvectors.  If $v=\sum a_i v_i$, then $P^{-1}v$ is the vector whose $j$th component is $a_j$.  $P^{-1}M_{A,v}$ will have $ij$ entry $a_i \lambda_i^{j-1}$. Taking the determinant, factoring out each of the $a_i$'s from the $i$th row and noting that we are left with a Vandermonde matrix, we have 
$$\det(M_{A,v})=\det(P)\left( \prod_i  a_i \right)\prod_{i<j}(\lambda_j-\lambda_i).$$
Thus, up to a constant depending on our choice of eigenbasis and our eigenvalues, this construction agrees with the one in the question.
Unfortunately, this construction is automatically zero if we have repeated eigenvalues (which is not an issue with the original construction), so it is unclear if there is an equally nice construction for those cases.  
