Re: Dot product confusion

*To*: mathgroup at smc.vnet.net*Subject*: [mg109955] Re: Dot product confusion*From*: inOr <harderm at onid.orst.edu>*Date*: Wed, 26 May 2010 07:09:28 -0400 (EDT)*References*: <htg90l$k7h$1@smc.vnet.net>

On May 25, 3:32 am, "S. B. Gray" <stev... at ROADRUNNER.COM> wrote: > Given > > ptsa = {{x1, y1, z1}, {x2, y2, z2}, {x3, y3, z3}}; > > I thought the following expressions would be identical: > > {aa, bb, cc}.ptsa (* expression 1 *) > ptsa.{aa, bb, cc} (* expression 2 *) > > but they are not. They evaluate respectively as: > > {aa x1 + bb x2 + cc x3, aa y1 + bb y2 + cc y3, > aa z1 + bb z2 + cc z3} > > {aa x1 + bb y1 + cc z1, aa x2 + bb y2 + cc z2, > aa x3 + bb y3 + cc z3} > > Since ptsa is itself three xyz coordinates, the expressions might be > ambiguous, but I assumed the dot product would always commute. Should > there be a warning? > > The first result is the one I want. > > Steve Gray Steve, The dot product is most definitely NON-commutative. To see this for a case like yours, think of vector-matrix multiplication as the transformation of a vector into a new vector. In real number representation, pre-multiplication of a vector by a matrix creates a new vector that is the sum of the column vectors of the matrix weighted by the elements of the vector. In this case, the matrix multiplication creates a column vector out of another column vector. When the vector is post-multiplied by the matrix, the result is a vector that is the sum of the ROW vectors of the matrix, still weighted by the components of the vector. In this case, a row vector is transformed into another row vector. Disregarding the row- / column- difference, the two results are equal only if the rows and columns of the matrix are identical (symmetric matrix). Mark Harder