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# unimodular matrix

An $n\times n$ square matrix over a field is *unimodular* if its determinant is 1. The set of all $n\times n$ unimodular matrices forms a group under the usual matrix multiplication. This group is known as the special linear group. Any of its subgroup is simply called a *unimodular group*. Furthermore, unimodularity is preserved under similarity transformations: if $S$ any $n\times n$ invertible matrix and $U$ is unimodular, then $S^{{-1}}US$ is unimodular. In view of the last statement, the special linear group is a normal subgroup of the group of all invertible matrices, known as the general linear group.

A linear transformation $T$ over an $n$-dimensional vector space $V$ (over a field $F$) is *unimodular* if it can be represented by a unimodular matrix.

The concept of the unimodularity of a square matrix over a field can be readily extended to that of a square matrix over a commutative ring. Unimodularity in square matrices can even be extended to arbitrary finite-dimensional matrices: suppose $R$ is a commutative ring with 1, and $M$ is an $m\times n$ matrix over $R$ (entries are elements of $R$) with $m\leq n$. Then $M$ is said to be *unimodular* if it can be “completed” to a $n\times n$ square unimodular matrix $N$ over $R$. By completion of $M$ to $N$ we mean that $m$ of the $n$ rows in $N$ are exactly the rows of $M$. Of course, the operation of completion from a matrix to a square matrix can be done via columns too.

Let $M$ is an $m\times n$ matrix and $v$ is any row of $M$. If $M$ is unimodular, then $v$ is unimodular viewed as a $1\times n$ matrix. A $1\times n$ unimodular matrix is called a *unimodular row*, or a *unimodular vector*. A $n\times 1$ *unimodular column* can be defined via a similar procedure. Let $v=(v_{1},\ldots,v_{n})$ be a $1\times n$ matrix over $R$. Then the unimodularity of $v$ means that

$v_{1}R+\cdots+v_{n}R=R.$ |

To see this, let $U$ be a completion of $v$ with $\operatorname{det}(U)=1$. Since $\operatorname{det}$ is a multilinear operator over the rows (or columns) of $U$, we see that

$1=\operatorname{det}(U)=v_{1}r_{1}+\cdots+v_{n}r_{n}.$ |

## Mathematics Subject Classification

20H05*no label found*15A04

*no label found*15A09

*no label found*

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