[Fwd: Re: correction to full size Q output from a QR factorization]

[Ron Boisvert <boisvert@nist.gov>]

Sender: Sione <sionep@xtra.co.nz>
Subject: Re: correction to full size Q output from a QR factorization



Hello Andreas,

I  have the following example using your code:
==============================

double[][] aa = {{0.5774,    0.5774,    0.5774}};
Matrix A = new Matrix(aa);
A = A.transpose();
   
QRDecomposition QRdecomp = new QRDecomposition(A);
Matrix Q = QRdecomp.getQ(false);
System.out.println("---------- Q ----------");
Q.print(8,4);
   
Matrix R = QRdecomp.getR();
System.out.println("---------- R ----------");
R.print(8,4);

===============================


The output is show below:

---------- Q ----------

  -0.5774   -0.5774   -0.5774
  -0.5774    0.7887   -0.2113
  -0.5774   -0.2113    0.7887

---------- R ----------

  -1.0001



I know that  R is suppose to be a column vector as shown below:

---------- R ----------

-1.0001
        0
        0


Is there any way to make or modify the  method call :   "getR( )"  to 
output the correct column vector as shown above?   I do use your code as 
it is the same as that of Matlab  "qr"  function.

Cheers,
Sione.




Andreas Kyrmegalos wrote:
> Hello again,
> in the QRDecomposition code I have neglected to include the case when 
> rank= n-1 and no householder vector has been calculated. I have an 
> implementation of the QR algorithm that allows pivoting but I m 
> skeptical about releasing it. This slight omission is a good example 
> of why I refrain from its release.
>
> Anyway here's the corrected code (the only change is the addition of 
> the else clause):
>
>     ....
>     if (!economy) {
>        if (QR[n-1][n-1] != 0.0) {
>           for(int k = m-1; k >= n-1;k--) {
>              Q[k][k] = 1.0;
>              double s = 0.0;
>               for (int i = n-1; i <=k; i++) {
>                 s += QR[i][n-1]*Q[i][k];
>              }
>              s = -s/QR[n-1][n-1];
>              Q[k][k] += s*QR[k][n-1];
>              for (int i = m-1; i >k; i--) {
>                 Q[i][k] += s*QR[i][n-1];
>                 Q[k][i] = Q[i][k];
>              }
>           }
>        }
>        else {
>           for(int k = m-1; k >= n-1;k--)
>              Q[k][k] = 1.0;
>        }
>     }      ....
>
>  I apologize to anyone who may have used the code but as I've pointed 
> out before, please do test the code before using it.
>
> Cheers,
> Andreas
>
>
>
>
>