求助 用mathematica编一个求任意整数矩阵的smith标准形

firlimin

跪求用mathematica求任意整数矩阵A的smith标准形

算法为(1)找出A的元素中的最大公约数
(2)将它调到(1,1)位置上去
(3)使第一行和第一列的所有其他元素都为零
这可用基本行或列来变换:
交换两行(列)
一行(列)减去另一行(列)的整数倍
这样得到矩阵B,置S11=b11,对(m-1)x(n-1)的矩阵[bij]重复以上算法,重复r次就可得到smith标准形.
This topic gives a version of the Gauss elimination algorithm for a commutative principal ideal domain which is usually described only for a field.

Let be a -matrix with entries from a commutative principal ideal domain . For denotes the number of prime factors of . Start with and choose to be the smallest column index of with a non-zero entry.

(I)
If and , exchange rows 1 and .
(II)
If there is an entry at position such that , then set and choose such that
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By left-multiplication with an appropriate matrix it can be achieved that row 1 of the matrix product is the sum of row 1 multiplied by and row multiplied by . Then we get at position , where . Repeating these steps one obtains a matrix having an entry at position that divides all entries in column .
(III)
Finally, adding appropriate multiples of row , it can be achieved that all entries in column except for that at position are zero. This can be achieved by left-multiplication with an appropriate matrix.
Applying the steps described above to the remaining non-zero columns of the resulting matrix (if any), we get an -matrix with column indices where , each of which satisfies the following:
the entry at position is non-zero;
all entries below and above position as well as entries left of are zero.
Furthermore, all rows below the -th row are zero.
Now we can re-order the columns of this matrix so that elements on positions for are nonzero and for ; and all columns right of the -th column (if present) are zero. For short set for the element at position . has non-negative integer values; so is equivalent to being a unit of . can either happen if and differ by a unit factor, or if they are relatively prime. In the latter case one can add column to column (which doesn't change and then apply appropriate row manipulations to get . And for and one can apply step (II) after adding column to column . This diminishes the minimum -values for non-zero entries of the matrix, and by reordering columns etc. we end up with a matrix whose diagonal elements satisfy .

Since all row and column manipulations involved in the process are invertible, this shows that there exist invertible and -matrices so that is



This is the Smith normal form of the matrix. The elements are unique up to associates and are called elementary divisors

有没有谁能够根据上述算法,将程序编出来.............
谢谢!!!!!!!

toniby1978

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