A double Sylvester determinant

Darij Grinberg

Abstract


Given two $\left( n+1\right) \times\left( n+1\right) $-matrices $A$ and $B$ over a commutative ring, and some $k\in\left\{ 0,1,\ldots,n\right\} $, we consider the $\dbinom{n}{k}\times\dbinom{n}{k}$-matrix $W$ whose entries are $\left( k+1\right) \times\left( k+1\right) $-minors of $A$ multiplied by corresponding $\left( k+1\right) \times\left( k+1\right) $-minors of $B$. Here we require the minors to use the last row and the last column (which is why we obtain an $\dbinom{n}{k}\times\dbinom{n}{k}$-matrix, not a $\dbinom{n+1}{k+1}\times\dbinom{n+1}{k+1}$-matrix). We prove that the determinant $\det W$ is a multiple of $\det A$ if the $\left( n+1,n+1\right) $-th entry of $B$ is $0$. Furthermore, if the $\left( n+1,n+1\right) $-th entries of both $A$ and $B$ are $0$, then $\det W$ is a multiple of $\left( \det A\right) \left( \det B\right) $. This extends a previous result of Olver and the author.


Keywords


determinant; compound matrix; Sylvester's determinant; polynomials



ISSN: 1855-3974

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