Determinants of Sums.
Determinants of Sums.
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总和的决定因素。
DOI：
10.1080/07468342.1990.11973297
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发表时间：
1990
影响因子：
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通讯作者：
M. Marcus
中科院分区：
文献类型：
--
作者：
M. Marcus
在(1)中：A和B是n方阵；r上的外和在整数0，…，n上；对于特定的r，内和是选自1，…，n的所有长度为r的严格递增的整数序列a和8；A[ai，B](方括号)是A在行a和列8中的r平方子矩阵；B(ai，B)是B的(n-r)平方子矩阵，位于与a互补的行和与a互补的列中，8；而S(A)是a中整数的和。当然，当r=0时，被和数被认为是det(B)，当r=n时，它是det(A)。(1)的证明是矩阵每行行列式的线性和标准拉普拉斯展开定理的一个非常简单的结果。以下是争论的细节。WRITE DET(A+B)=DET(A<；1；+B<；1>；，...，A<；n&>；+B<；n&>；)(2)其中A；表示A的第i行。(2)形式上，(2)的右侧的作用就像“二项式”的乘积。因此，对于从0，...，n中选择的每个r，(2)的右侧在形式项的长度r的所有a上提供总和
In (1): A and B are n-square matrices; the outer sum on r is over the integers 0,..., n; for a particular r, the inner sum is over all strictly increasing integer sequences a and, 8 of length r chosen from 1,..., n; A [ai, B](square brackets) is the r-square submatrix of A lying in rows a and columns, 8; B (ai, B) is the (n-r)-square submatrix of B lying in rows complementary to a and columns complementary to, 8; and s (a) is the sum of the integers in a. Of course, when r= 0 the summand is taken to mean det (B) and when r= n, it is det (A). The proof of (1) is a very simple consequence of the linearity of the determinant in each row of the matrix, and the standard Laplace expansion theorem. Here are the details of the argument. Write det (A+ B)= det (A< 1>+ B< 1>,..., A< n>+ B< n>)(2) where A<;> denotes the ith row of A. The right side of (2) formally acts just like a product of the" binomials" A<;>+ B<;>• i= 1,..., n: this is the meaning of det being linear in the rows. Thus for each r chosen from 0,..., n, the right side of (2) contributes a sum over all a of length r of terms of the form