[Scilab-Dev] ./ Operator Question

[Serge Steer]

Le 27/09/2014 01:19, Santiago Chialvo a écrit :
> Hi, first of all, sorry for my english, it's not good enough. I'm 
> using Scilab 5.5.0 Version, and I found a problem that I don't 
> understand how it works.
>
> First of all, when I do the operation
>
> R = A/B;
>
> With A = Scalar and B = Vector, the result that gives me is a R vector 
> that satisfy:
>
> R*B = A;
>
> But, my question is, how does it works? I mean, we have
>
> R(1)*B(1) + R(2)*B(2) .... + R(N)*B(N) = A;
>
> Where we know the value of the components of B, and the value of A, 
> but it's only 1 equation with N unknowns!
>
> How does it works, to have the values of R? Thanks,
As the problem is underdetermined  the equation as an infinity of 
solutions as follow. I will describe the process with the equivalent 
equation B'*R'=A that can be written suing standard notations as X*A=B, 
where  A is a column vector X the row vector of the  unknowns
A can be factored as the product Q*R where Q is an othonormal square 
matrix and R a column vector whose all elements but the first one are zeros
So the equation X*A=B can be rewritten (X*Q)*R=B or Y*R=B, all Y of the 
form [B/R(1),y2,....yn] are solutions, and consequently all X= 
[B/R(1),y2,....yn]*Q' are solutions of the initial equation

Example
-->A=(1:5)';B=4.5;
-->[Q,R]=qr(A);R
-->Y=[B/R(1),rand(1,4)];
-->Y*R-B
  ans  =
     0.
  -->X=Y*Q'
-->X*A-B
  ans  =
   - 8.882D-16

The particular solution computed by B/A serach for y2...yn that maximize 
the number of 0 in the solution X.

Serge
> Santiago.
>
>
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