Smith normal form
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.
If and , exchange rows 1 and .
If there is an entry at position such that , then set and choose such that
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 .
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 .
|Title||Smith normal form|
|Date of creation||2013-03-22 13:52:08|
|Last modified on||2013-03-22 13:52:08|
|Last modified by||Mathprof (13753)|