If you want elementwise division you have to use the ./ operator. Then each
element gets divided by the corresponding element in the other vector.
On Wednesday, February 5, 2014 8:16:42 AM UTC+1, Andreas Noack Jensen wrote:
>
> I different framing of Kevin's answer could be that it solves your systems
> of equations: three systems of one equation in three variables
>
> julia> x=a/b
> 3x3 Array{Float64,2}:
> 0.0519481 0.0649351 0.0779221
> 0.103896 0.12987 0.155844
> 0.155844 0.194805 0.233766
>
>
> julia> x*b
> 3-element Array{Float64,1}:
> 1.0
> 2.0
> 3.0
>
> The solutions are not unique and Julia chooses those with minimum norm.
>
> 2014-02-05 Kevin Squire <kevin....@gmail.com <javascript:>>:
> >
> > Hello Fil,
> >
> > Generally, division is like multiplying by the inverse of the divisor.
> For vectors, the inverse is actually a pseudoinverse or generalized
> inverse. In this case,
> >
> > a / b = a * pinv(b)
> >
> > where
> >
> > julia> pinv(b)
> > 1x3 Array{Float64,2}:
> > 0.0519481 0.0649351 0.0779221
> >
> > julia> pinv(b) * b
> > 1-element Array{Float64,1}:
> > 1.0
> >
> > julia> a * pinv(b) # the outer product of a and pinv(b)
> > 3x3 Array{Float64,2}:
> > 0.0519481 0.0649351 0.0779221
> > 0.103896 0.12987 0.155844
> > 0.155844 0.194805 0.233766
> >
> > Cheers!
> > Kevin
> >
> >
> > On Tue, Feb 4, 2014 at 9:19 PM, Fil Mackay
> > <f...@vertigotechnology.com<javascript:>>
> wrote:
> >>
> >> Can anyone explain what's going on here? :)
> >>
> >> julia> a = [1,2,3]
> >> 3-element Array{Int64,1}:
> >> 1
> >> 2
> >> 3
> >>
> >> julia> b = [4,5,6]
> >> 3-element Array{Int64,1}:
> >> 4
> >> 5
> >> 6
> >>
> >> julia> a / b
> >> 3x3 Array{Float64,2}:
> >> 0.0519481 0.0649351 0.0779221
> >> 0.103896 0.12987 0.155844
> >> 0.155844 0.194805 0.233766
> >>
> >> I get the principle of how two vectors (3x) lead to a matrix (3x3)
> result - but the actual values are a mystery?
> >>
> >
>
>
>
> --
> Med venlig hilsen
>
> Andreas Noack Jensen
>
