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HermiteDecomposit of the lattices generated by column or row matrices

Posted 3 years ago
POSTED BY: Hongyi Zhao
4 Replies
Posted 3 years ago

It seems that:

2nd == 3rd.

1st is not equivalent to the above two.

POSTED BY: Hongyi Zhao

You tried three different operations, getting three different results. And wondered why you got three different results? Me, I'd wonder if the results were the same.

POSTED BY: Daniel Lichtblau
Posted 3 years ago

See the following example. According to your suggestion, I got three different results:

In[111]:= mat=tbasSGAK227Left1
%//HermiteDecomposition
Map[Transpose, Reverse[HermiteDecomposition[Transpose[mat]]]]
mat//Transpose//HermiteDecomposition

Out[111]= {{1/2, 0, 0}, {0, 1/2, -(1/2)}, {1/2, 1/2, 1/2}}

Out[112]= {{{1, 0, 0}, {-1, 0, 1}, {-1, -1, 1}}, {{1/2, 0, 0}, {0, 1/
   2, 1/2}, {0, 0, 1}}}

Out[113]= {{{1/2, 0, 0}, {0, 1/2, 0}, {1/2, 1/2, 1}}, {{1, 0, 0}, {0, 
   1, 1}, {0, 0, 1}}}

Out[114]= {{{1, 0, 0}, {0, 1, 0}, {0, 1, 1}}, {{1/2, 0, 1/2}, {0, 1/2,
    1/2}, {0, 0, 1}}}

However, I still haven't seen any clue.

Regards, Zhao

POSTED BY: Hongyi Zhao

This is not a question people here can answer. It depends on context: which lattices you obtain are column-based and which are row-based. This is something you would need to determine. For the former case, something like Map[Transpose, Reverse[HermiteDecomposition[Transpose[mat]]]]` might be what you want. Again, having explicit examples would help for testing proposed answers.

POSTED BY: Daniel Lichtblau
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