scraps22

This is the Title PhL 1.11.15 Note that page numbering is turned on in this template and view is 125%, located in phil/roaming/microsoft/templates size about 219K. Recall from Section ** in the V*W* discussion that. (λiλ'j)(v,w) = λi(v)λj'(w) // λiλ'j: VxW → K Now with W* = V* we have instead (λiλj)(v1,v2) = λi(v2)λj(v2) . // λiλj: VxV → K If we evaluate the V*2 functional (λi^ λj) at the V2 location (v1,v2) we find, (λi^ λj)(v1,v2) = (λiλj)(v1,v2) - (λjλi)(v1,v2) = λi(v1)λj(v2) - λj(v1)λi(v2) = det . Notice that : (1) the basis function (λi^ λj)(v1,v2) is linear on both v1 and v2, so it is a "bilinear" function. This follows from the fact that the λk(v) functions are linear as was shown in ***. (2) the basis function (λi^ λj)(v1,v2) is antisymmetric under v1↔ v2, (λi^ λj)(v2,v1) = - (λi^ λj)(v1,v2) This is obvious from the determinant form since we switch two rows. As in Section 3, the wedge product of two vectors in V* can be expressed in terms of certain determinants which are minors of a "tall matrix" whose columns are α and β, α ^ β = Σij αiβj (λi ^ λj) = Σi<j (αiβj- ajβi) (λi ^ λj) = Σi<j det (λi ^ λj) . Because we are now in the dual space of linear functionals, we claim no particular geometric significance of the above wedge product when n = 2. One could of course blithely set V*2 = R2 and repeat the previous geometric discussion, but usually one does not discuss "geometry" in the dual space context. Instead, one notes that, if the linear functional α ^ β is evaluated at (v1,v2) in V2, one gets a statement about functions, (α ^ β)(v1,v2) = Σij αiβj (λi ^ λj)(v1,v2) = Σi<j (αiβj- ajβi) (λi ^ λj)(v1,v2) = Σi<j det (λi ^ λj)(v1,v2) = Σi<j det [ λi(v1)λj(v2) - λj(v1)λi(v2) ] = Σi<j det det . The function (α ^ β)(v1,v2) is bilinear and antisymmetric because (λi ^ λj)(v1,v2) is bilinear and antisymmetric. Evaluating at (ei, ej) we find, using the full double sum Σnm , (α ^ β)(ei,ej) = Σnm αnβm [ λn(ei)λm(ej) - λm(ei)λn(ej) ] = Σnm αnβm ( δi,n δj,m - δi,n δj,m) = αiβj - αjβi = det