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This article says that the elementary operations required are unimodular transformations. My questions are the following:

  • Why does it have to be unimodular? What is so special about it?

  • Why are the elementary operations called "unimodular transformations"?

Since unimodular matrix is defined to be a square integer matrix having determinant +1 or -1, then would unimodular transformation mean that it transforms some matrx $M$ to have unimodular properties? (Does this answer my 2nd question?)

  • What is the significance of a unimodular transformation and hermite normal form? In particular, why would someone want to transform their matrix in hermite normal form? Like most of us know gaussian elimination computes for solutions, and reduced row echelon form help us determine if the vectors are linearly independent.

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A unimodular matrix of dimension $n$ with coefficients in a (commutative) ring $R$ is simply an element of the group $ \operatorname{GL}(n,R)$, i.e. an invertible matrix,such that its inverse has coefficients in $R$.

The condition is this $$U\in \operatorname{GL}(n,R)\iff \det U\in R^\times \enspace (\text{the group of units in }R).$$ So here, with matrices with coefficients in $\mathbf Z$, it means that $\det U=\pm 1$.

When you transform a matrix $A$ by a unimodular transformation $U$, so that $H=UA$, you can do the backwards transformation: $A=U^{-1}H$, since $U$ is invertible and $U^{-1}$ is unimodular.

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  • $\begingroup$ group of units in $R$. According to wikipedia this has something to do with Rings. I haven't read up on any ring theory yet. $\endgroup$ – TheLast Cipher Feb 16 '18 at 12:00
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    $\begingroup$ You just have to know what a ring is , to understand this: the group of units is the set of elements in $R$ which have a multiplicative inverse (if we have a field, it's the non-zero elements, by definition). These form a multiplicative group. $\endgroup$ – Bernard Feb 16 '18 at 12:04
  • $\begingroup$ What does matrix $A$ got to do in the condition you've given? $\endgroup$ – TheLast Cipher Feb 16 '18 at 12:13
  • $\begingroup$ Sorry, a bad copy-paste. I'll fix it at once. $\endgroup$ – Bernard Feb 16 '18 at 12:14
  • $\begingroup$ Okay. But I think I am missing the point on why the facts you've given makes unimodular transformation important. If I already have matrix $A$, then why would I like to construct the unimodular matrix $U$ to obtain $H$? $\endgroup$ – TheLast Cipher Feb 16 '18 at 12:26

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