Hi Suat
Thanks for the info. The problem is that with current tip, it is
enabled for all Symmetric matrices by template magic in the
SelfAdjoint solver which obviously is a problem.
Cheers
Benjamin
On 31.10.2011 22:10, Suat Gedikli wrote:
Hi,
thank you for the report.
This method (even if the name is misleading) is for symmetric
positive semidefinite matrices _only_ => nonnegative
eigenvalues.
If in your case the largest eingenvalue is zero, then this would
mean that for a positivesemidefinite matrix the smallest one is
also zero => the first if would apply in line 195:
if((evals(2)evals(0))<=Eigen::NumTraits<Scalar>::epsilon())
{
// all three equal
evecs.setIdentity();
}
so, the matrix you're using can not produce correct results, since
its negative definite. If it should be a covariance matrix (as
used for the normal estimation) check the code where you're
calculating the matrix. If not, just use the SelfAdjoint Solver
from Eigen which should give you correct results, but is also much
slower.
Cheers,
Suat
On 10/31/2011 11:14 AM, Radu B. Rusu wrote:
Could be interesting for you :)
Cheers,
Radu.
Begin forwarded message:
Hi
I just stumbled over a problem in the analytic 3x3
eigenvalue/eigenvector for symmetric matrices.
I have a matrix, where all row vectors are parallel to
each other and
there is a zero eigenvalue as the largest eigenvalue.
What happens is that:
Vector cross01 = tmp.row (0).cross (tmp.row (1));
Vector cross02 = tmp.row (0).cross (tmp.row (2));
Scalar n01 = cross01.squaredNorm();
Scalar n02 = cross02.squaredNorm();
here, both n01 and n02 become zero and as a result,
the entire
eigenvector matrix becomes nan's.
Reproducing it is easy. Its a different case, but
shows the same behavior:
Matrix3f m = Matrix3f::Zero();
m.diagonal() = Vector3f(2000, 1, 0);
Matrix3f evecs;
Vector3f eivs;
eigenSymm33(m, evecs, eivs);
std::cout << evecs << std::endl
<< eivs.transpose() << std::endl;
Output is:
nan nan nan
nan nan nan
nan nan nan
0.0110269 0.98902 2000
Now, one might think that 2000,1,0 is exploiting
numerics. But in fact
its not. Test 2, 1, 1 as the diagonal. The routine
will give out the
correct result, but if you look closely on what
happens, both n01 and
n02 will be extremely small, like 1e15, bot not zero.
But analytically,
zero would be the correct value! So nan's are simply
prevented because
of numerical errors.
Cheers
Benjamin
