The Solution to Union Problems
Given two integer lattices L(B1) and L(B2) .Compute a basis for the smallest lattice containing both L(B1) and L(B2).
Assume B1 = [a1 a2 … an1] , B2 = [b1,b2…bn2] . Make a new matrix B3 = [a1 a2 … an1 b1,b2…bn2] . Because a1 a2 … an1 b1,b2…bn2 may be linearly dependent , by using elementary matrix transformations , we can find a matrix B4 that its vectors are linearly independent and its vectors can generate all the vectors that can be generated by B3’s vectors.
B3 = ![]()
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The first ,the second and the forth vectors are linearly independent. These three vectors are the new base for the lattice(L(B4)) that contains both L(B1) and L(B2) .
For B4 , If we take one of the three vectors away , then , the L(B4) will not be able to contain L(B1) and L(B2) . For example : if we take (1 1 -2 7 ) away , we will not able to generate (2 4 4 9) by using (2 1 4 3) and (-1 1 -6 6).
So , L(B4) is the smallest lattice containing both L(B1) and L(B2)