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How to evaluate the success of LDLT.solve using isApprox?

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zhanxw
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Hello,

I am trying to use solve() in the LDLT class.
The documentation (http://eigen.tuxfamily.org/dox/classEigen_1_1LDLT.html) suggests this code:

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bool a_solution_exists = (A*result).isApprox(b, precision);


In practice, what is the suggested value of choosing precision?

I basically use MatrixXf and found 1e-3 is practically good.
But I am looking forward to other ideas or suggestions, ideally with more justice.

Thank you.


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