       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

```

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