The general linear group

Last update: 13 August 2013

The general linear group

• An $n×n$ matrix $A$ is invertible if there is a matrix ${A}^{-1}$ such that $A{A}^{-1}={A}^{-1}A={I}_{n}\text{.}$
• ${GL}_{n}\left(R\right)=\left\{n×n \text{invertible matrices with entries in} R\right\}\text{.}$

${GL}_{n}\left(R\right)$ is a group with identity ${I}_{n}\text{.}$

Let $R$ be a ring and let ${R}^{*}$ be the group of units in $R\text{.}$

• The matrices ${e}_{ij}$ which contain a 1 in the ${i}^{\text{th}}$ row and the ${j}^{\text{th}}$ column and zeros in all other entries are called matrix units.
• The elementary matrices are $xij(t)= ( 1 ⋱t ⋱ ⋱ 1 ) for 1≤i,j≤n, i≠j,t∈R, sij= ( 10 ⋱ 1 01 1 ⋱ 1 10 1 ⋱ 01 ) , 1≤i,j≤n, i≠j, hi(t)= ( 1 ⋱ 1 t 1 ⋱ 1 ) 1≤i≤n,t∈R*.$

Let $A$ be an $n×n$ matrix.

$xij(t)A is the same as A except that t·(jth row of A) is added to the ith row of A. Axij(t) is the same as A except that t·(jth column of A) is added to the ith column of A. sijA is the same as A except that the jth row of A and the ith row of A are switched. Asij is the same as A except that the jth column of A and the ith column of A are switched. hi(t)A is the same as A except that the ith row of A is multiplied by t. Ahi(t) is the same as A except that the ith column of A is multiplied by t.$

(Smith normal form) Let $R$ be a PID. Any matrix $A\in {M}_{m×n}\left(R\right)$ can be written in the form

$A=PDQ,where D=diag(d1,…,dr,0,…,0),$

where $P\in {GL}_{m}\left(R\right),$ $Q\in {GL}_{n}\left(R\right)$ and ${d}_{1},\dots ,{d}_{r}\in R$ with ${d}_{i}|{d}_{i+1}$ for $1\le i\le r-1\text{.}$