determinant of an operator

A determinant usually attaches to a matrix, but the deeper object is the determinant of a linear OPERATOR — a map T from a finite-dimensional space V to itself, with no basis mentioned. Pick any basis, write T as a matrix, take that matrix's determinant: the number you get does not depend on the basis you chose. So det(T) is a genuine invariant of the operator itself.

Why it is well defined: changing basis replaces the matrix [T] by P^-1 [T] P for some invertible P. By multiplicativity, det(P^-1 [T] P) = det(P^-1) det([T]) det(P) = det([T]), because det(P^-1) det(P) = 1. The conjugating factors cancel exactly, so every basis yields the same number. This similarity-invariance is the entire reason det(T) makes sense.

Read coordinate-free, det(T) is the factor by which T scales oriented volume on V — independent of how you set up coordinates, because ratios of volumes are. The slick high-level definition uses the top exterior power: T induces a linear map on the one-dimensional space of top alternating forms, and multiplying a one-dimensional space is just scaling by a number; that number is det(T), no basis required at all.

Caveat: det(T) needs a finite-dimensional domain equal to its codomain (an operator, V -> V), and it needs the dimension finite — there is no determinant for a general operator on an infinite-dimensional space (special 'determinant-class' operators on Hilbert space carry a refined Fredholm determinant, a separate story). For matrices that are not square, det simply is not defined.