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 $$v_1R+\cdots+v_nR=R.$$ 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 $$1=\operatorname{det}(U)=v_1r_1+\cdots+v_nr_n.$$