The determinant of a linear transformation on a finite vector space Given a finite vector space $V$ and a linear transformation $f : V \rightarrow V,$ is it true that for any two ordered bases of $V$, call them $a$ and $b$, the determinant of the matrix of $f$ with respect to $a$ will always equal the determinant of the matrix of $f$ with respect to $b$?
If so, is there an explicit definition of the determinant of $f$ making no reference to a choice of basis?

Yes, the determinant is independent. If $M$ is the matrix of $f$ in basis $a$, then the matrix in basis $b$ will be of the form $AMA^{-1}$ for an invertible matrix $A$. Then $$\det(AMA^{-1}) = \det(A)\det(M)\det(A^{-1}) = \det(M).$$
The only way I know to give a coordinate-free description of the determinant is to use exterior algebra. If $V$ has dimension $n$, then you can form the vector space $\Lambda^n V$ generated by symbols $v_1 \wedge v_2 \wedge \ldots \wedge v_n$, subject to relations that make $\wedge$ bilinear and $v \wedge w = -w \wedge v$. Then $\Lambda^n V$ will be one dimensional, and $f:V \rightarrow V$ induces a map $\Lambda^n f:\Lambda^n V \rightarrow \Lambda^n V$. Since $\Lambda^n V$ is one dimensional, $\Lambda^n f$ must be a scalar, and that scalar is just $\det f$. (This is really just a fancy way of characterizing the determinant as the unique bilinear, anticommutative function on the rows of a matrix.)
