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Problem for diagonalisation of a complex 6*6 matrix

Jonathan Da Silva
Posted 11 years ago
POSTED BY: Jonathan Da Silva
7 Replies
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Michael Helmle
Michael Helmle
Posted 11 years ago
POSTED BY: Michael Helmle

If you can use exact rational numbers as input, Mathematica can do the calculation exactly (as you saw originally). So, use 11/10 instead of 1.1. If you must use decimal points, get the rational representation using Rationalize[].
POSTED BY: John Doty

Hi thanks, actually there is a direct way to define with precision the parameters with `` as given in http://reference.wolfram.com/language/ref/Accuracy.html. If you also define a parameter called zero for 0 and the others like

a = 1``50;
b = 100``50;
c = 10^10``50;
zero = 0``50;

you get the result I expected, and the execution speed is not so bad.
POSTED BY: Jonathan Da Silva

Mathematica allows you to do arbitrary precision, but it's not always clear how that's implemented.

See: http://mathematica.stackexchange.com/questions/5390/quadruple-precision-128-bit-arithmetic

So yes you can use higher precision numbers if that will help. It will likely affect the execution speed.

POSTED BY: Sean Clarke

Ok thanks, actually I need a better precision especially for the very small eigenvalues, do you know if quadruple precision is supported by mathematica ?
POSTED BY: Jonathan Da Silva

This looks like standard case of floating point error. If we subtract the two results and use Chop we get:

{0, 0, 0, -9.9996*10^-7, 1.*10^-6, 6.60625*10^-10 + 5.08267*10^-9 I}

This is pretty close to a vector of 0s, so the results aren't far apart. I wouldn't say it's completely wrong. Do you believe that the floating point error is too high? If so are you mixing symbolic and numerical computations? Symbolic manipulations on floating point numbers aren't always the most numerical stable way to calculate something.

You can always improve the accuracy/precision of the results by staying with symbolic evaluation or using higher precision numbers.

POSTED BY: Sean Clarke

It appears to be entirely numeric, whether exact or approximate. Given the difference in scales of the inputs, I would not expect a computation at machine precision to do well in such circumstances. The scale factor of 10 and the error of around 10^(-6) are consistent with working at 16 digits of precision (machine double, that is).
POSTED BY: Daniel Lichtblau
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