Matrix products under which the determinant behaves multiplicatively - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T13:59:37Z http://mathoverflow.net/feeds/question/39296 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39296/matrix-products-under-which-the-determinant-behaves-multiplicatively Matrix products under which the determinant behaves multiplicatively slimton 2010-09-19T11:29:53Z 2010-09-19T19:47:12Z <p>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. </p> <p>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 $0$.</p> http://mathoverflow.net/questions/39296/matrix-products-under-which-the-determinant-behaves-multiplicatively/39324#39324 Answer by Terry Tao for Matrix products under which the determinant behaves multiplicatively Terry Tao 2010-09-19T18:56:42Z 2010-09-19T19:08:18Z <p>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 $A \oplus B := \begin{pmatrix} A &amp; 0 \\ 0 &amp; B \end{pmatrix}$) also works:</p> <p>$$\det(A \oplus B) = \det(A) \det(B).$$</p> <p>One can debate whether this operation deserves to be called a "matrix product", though (for instance, it is not distributive over addition).</p> <p>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.</p> http://mathoverflow.net/questions/39296/matrix-products-under-which-the-determinant-behaves-multiplicatively/39330#39330 Answer by Andy B for Matrix products under which the determinant behaves multiplicatively Andy B 2010-09-19T19:47:12Z 2010-09-19T19:47:12Z <p>The determinant of the product of two <i>non square</i> 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.</p> <p>I do not know for certain, but this looks like it has to do with some kind of coproduct?</p>