unimodular matrixUnimodularMatrix2005-01-26 16:37:522006-09-09 23:51:24Definitionunimodular linear transformationunimodular rowunimodular columnunimodular groupunimodular vectorunimodularunimodularity% this is the default PlanetMath preamble. as your knowledge
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An $n\times n$ square matrix over a field is \emph{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 \emph{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 \emph{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 \emph{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 \emph{unimodular row}, or a \emph{unimodular vector}. A $n\times 1$ \emph{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 $$1=\operatorname{det}(U)=v_1r_1+\cdots+v_nr_n.$$