MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The determinant behaves multiplicatively with respect to the usual matrix product $$ \det(AB) = \det(A)\det(B), $$ and also with respect to the Kronecker (or tensor) product of square matrices $$ \det(A\otimes B) = \det(A)^q \det(B)^p, $$ when $A$ and $B$ are $p\times p$ and $q \times q$ matrices, respectively.

Are there other natural types of matrix products under which the determinant behaves multiplicatively? To be completely precise, the property I need is that the determinant of the product is $0$ if and only if the determinant of at least one of its factors is $0$.

share|cite|improve this question
First of all, what do you mean by a "product"? Do you mean an algebra structure on $\oplus_{n=0}^{\infty}Mat(n,\Bbbk)$? – Qfwfq Sep 19 '10 at 15:20
Thinking about the 'natural' part of the question, would this be a good formulation? Fix a group $G$ and $G$-modules $U$ and $V$ (over a ground field $K$, say). Classify the $G$-modules $W$ and bilinear $G$-module morphisms $\mu: End(U)\times End(V) \to End(W)$ with the property that $\det_W(\mu(A,B)) = c\ \det_U(A)^q\det_V(B)^p$ for nonzero $c$ and $p,q>0$? The special case of $G=GL(n,\mathbb{R})$ and $U=V=\mathbb{R}^n$, would encompass regular matrix multiplication, reversed matrix multiplication, and the tensor product, though there are other examples. – Robert Bryant Jul 21 '11 at 19:44

Direct summation (taking a $p \times p$ matrix $A$ and a $q \times q$ matrix $B$ and returning a block-diagonal $(p+q) \times (p+q)$ matrix $A \oplus B := \begin{pmatrix} A & 0 \\\ 0 & B \end{pmatrix}$) also works:

$$\det(A \oplus B) = \det(A) \det(B).$$

One can debate whether this operation deserves to be called a "matrix product", though (for instance, it is not distributive over addition).

EDIT: Another (somewhat trivial) example is the reversed multiplication operation $(A, B) \mapsto BA$. More generally, if there was a linear automorphism $T$ on $Mat_n$ that preserved the singular variety $\{ A \in Mat_n: \det A = 0 \}$, one could conjugate the usual matrix multiplication operation by $T$. In the above example, $T$ is the transpose operation $T: A \mapsto A^t$. As another example, one could let $T$ be a left multiplication operator $A \mapsto SA$ for some invertible $S$, in which case the matrix multiplication operation becomes $(A, B) \mapsto ASB$, which also seems to work. One can combine the two and obtain another operation $(A, B) \mapsto BSA$. I'm not sure if these are the only examples that can be constructed by this method.

share|cite|improve this answer

The determinant of the product of two non square matrices is nicely expressed by the Binet-Cauchy formula: $$ \det(AB) = \sum_I \det A_I \det B_I $$ Here $A$ is $n \times m$ and $B$ is $m \times n$ and the sum ranges over $n$-subsets $I$ of the numbers $\{1,2,...,m\}$. $A_I$ means "select columns of $A$ indexed by $I$" and $B_I$ means "select the rows of $B$ indexed by $I$". If either $A$ or $B$ has rank less than $n$ than the determinant of $AB$ is, thus, zero.

I do not know for certain, but this looks like it has to do with some kind of coproduct?

share|cite|improve this answer
That definitely has nothing to do with a coproduct. That's just what you get if you compute the determinant of the square matrix resulting from multiplying matrices in the way noted above. If I had to guess how to begin proving it, it would be something like: start from Cramer's rule and look at the minors. The reason that the formula is so convoluted is that both matrix multiplication and the determinant have nasty explicit descriptions for matrices (although their formal properties arise from canonical constructions). – Harry Gindi Sep 19 '10 at 22:25
The proof is via the functoriality of the exterior algebra, along with the fact that the adjoint of the map induced on exterior algebras is the map on exterior algebras induced by the adjoint. It might have to do with a coproduct, as it resembles the coproduct on the ring of matrix coefficients. However, a proof along those lines would be circuitous. – Charlie Frohman Sep 20 '10 at 11:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.