multiplicativity of det

The product rule says the determinant of a product is the product of the determinants: det(AB) = det(A) det(B). It is the most useful single fact about determinants, and from the volume viewpoint it is almost obvious — apply map B (scaling volume by det B), then map A (scaling by det A), and the total volume scaling is the product. Composition of scalings multiplies.

The clean proof uses the uniqueness characterization. Fix A and define D(B) = det(AB) as a function of B's columns. Because AB just applies A to each column of B, D is alternating and multilinear in those columns, so by uniqueness D(B) = c * det(B) for some constant c. Evaluate at B = I to find c = det(A). Hence det(AB) = det(A) det(B) for all B, with not one matrix entry computed.

The consequences cascade. det(A^-1) = 1/det(A), since det(A) det(A^-1) = det(I) = 1, which also re-proves that invertible forces det != 0. Determinant is similarity-invariant: det(P^-1 A P) = det(A), since the det(P) and det(P^-1) cancel — and that invariance is exactly what lets us define the determinant of an operator with no reference to a basis. It also gives det(A^T) = det(A) and det of block-triangular matrices as a product of block determinants.

Caveat: the rule is about products, not sums. In general det(A + B) is NOT det(A) + det(B) — there is no simple additive law, because volume scaling does not behave linearly under matrix addition. Also det(cA) = c^n det(A) for an n-by-n matrix, not c det(A); the scalar gets raised to the dimension because it scales all n edges of the box at once.