hess(rosser())

[Matyas Sustik]

Does octave represent/mark symmetric matrices?  Observe the result of:
octave:1> hess(rosser())
ans =

   6.1100e+02  -4.9693e+02  -6.8294e-15  -6.2369e-15   3.7733e-15   3.1450e-14  -2.2097e-14  -5.0757e-14
  -4.9693e+02   4.0434e+02   1.8737e+01   6.7450e-14   2.9311e-15   4.1902e-14   6.3493e-15   1.4161e-14
   0.0000e+00   1.8737e+01   3.7049e+01   1.0113e+03   2.4212e-14   7.3035e-14  -1.1387e-13   2.2387e-14
   0.0000e+00   0.0000e+00   1.0113e+03  -5.2293e+01  -4.8701e-01  -6.3129e-14   9.4838e-14   9.6069e-15
   0.0000e+00   0.0000e+00   0.0000e+00  -4.8701e-01   1.5399e+01  -1.2437e+02   3.4570e-14  -5.4079e-15
   0.0000e+00   0.0000e+00   0.0000e+00   0.0000e+00  -1.2437e+02   1.0046e+03  -5.0581e-02   2.4911e-13
   0.0000e+00   0.0000e+00   0.0000e+00   0.0000e+00   0.0000e+00  -5.0581e-02   1.0199e+03  -1.4354e-08
   0.0000e+00   0.0000e+00   0.0000e+00   0.0000e+00   0.0000e+00   0.0000e+00  -1.4353e-08   1.0000e+03

The correct result should be symmetric, because rosser() is symmetric.  I implemented myself the orthogonal transformation to upper Hessenberg form by calling the LAPACK dgehrd (general) and dsytrd (symmetric) functions and it seems that the result matches what I get with dgehrd.  Therefore, I suspect that the problem lies with hess() not recognizing (using) the fact that the argument is symmetric.

I do not know how symmetric matrices are handled in Octave.  Is it checked every time?  Is there a mark on the matrix for efficiency? It could also be the case that rosser() does not mark the matrix as symmetric and so hess() uses the wrong algorithm.

-Matyas