The Transposition Operator and Notation



Let's say you have a matrix AGLnA\in\mathrm{GL}n. How do we best denote the inverse of its transpose? You would probably write (AT)1(A^T)^{-1} or (A1)T(A^{-1})^T because it is the same. However, today at the office we decided to henceforth write ATA^{-T} instead. It seems abusive at first, but I can make it formal for you, if you care for that kind of stuff. As we all know, the transpose of ATA^T is AA. In other words, the transposition operator ϑ:GLnGLn\vartheta:\mathrm{GL}_n\to\mathrm{GL}_n which maps AA to its transpose satisfies ϑ2=1\vartheta^2=1. So far, the exponentiation map is defined as GLn×ZGLn\mathrm{GL}_n\times\mathbb{Z}\to\mathrm{GL}_n mapping (A,k)Ak(A,k)\mapsto A^k. We instead consider the ring Z[T]:=Z[t]/(t2)\mathbb{Z}[T]:=\mathbb{Z}[t]/(t^2) and extend the domain of the exponentiation map to GLn×Z[T]\mathrm{GL}_n\times\mathbb{Z}[T]\ast, where Z[T]\mathbb{Z}[T]\ast denotes the homogeneous elements of Z[T]\mathbb{Z}[T]. This way, you can write AkTA^{kT} instead of (Ak)T(A^k)^T and you have AkT2=AkA^{kT^2}=A^k as required. Note that by restricting to homogeneous elements in Z[T]\mathbb{Z}[T], we get Ap+q=ApAqA^{p+q}=A^pA^q and Apq=(Ap)qA^{pq}=(A^p)^q for all p,qZ[T]p,q\in\mathbb{Z}[T]\ast.

Leave a Reply

Your email address will not be published. Required fields are marked *