Let $A, B \in L(V)$ (or equivalently $n \times n$ matrix), where $V$ is a $n$-dimensional vector space. There are a multiple of proofs why $\det(A)\det(B) = \det(AB)$, but I couldn't find a satisfactory that is intuition-appealing.
First, the most imminent motivation for determinant is volume: $\det(A)$ is oriented volume of parallelepiped consisting $n$-column vectors of $A$. Given three matrices $A, B, AB$, we have three distinct parallelepiped each of which has oriented volume $\det(A)$, $\det(B)$, and $\det(AB)$, respectively.
Second, the given $A, B \in L(V)$ as in linear transformation, $AB$ is simply a composition of linear transformation.
I am trying to relate these two concepts, but it feels that the gap between two concepts are too broad. Is there a nice interpretation that fills the gap between these two concepts ?
Best Answer
Here's a slightly modified interpretation of the $\det(A)$ as a volume. $\det(A)$ is more generally the ratio between volumes in the input and output space.
Indeed, in linear transformations, volumes scale uniformly, and the scale factor from domain to codomain of any set transformed by a linear map (not just the fundamental parallelepiped) gives the same ratio, which we know is the determinant.
Thus, take an arbitrary set in the vector space. The determinant of the composition mapping $AB$ is the ratio of the volumes of the given set under the transformation. $B$ first scales the set's volume by $\det(B)$, and then $A$ does by $\det(A)$, so the total scale factor, which is $\det(AB)$ is just the product of these intermediate scale factors. Hence $$\det(AB) = \det(A) \det(B)$$