A double Sylvester determinant. (arXiv:1901.11109v1 [math.RA])

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 ( arXiv:1802.02900 ).查看全文

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 你的名字 留空匿名提交 你的Email或网站 用户可以联系你 标题 简单描述 内容 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 ( arXiv:1802.02900 ).
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