numpy arbitrary precision linear algebra

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SymPy can calculate arbitrary precision:

from sympy import exp, N, S
from sympy.matrices import Matrix

data = [[S("-800.21"),S("-600.00")],[S("-600.00"),S("-1000.48")]]
m = Matrix(data)
ex = m.applyfunc(exp).applyfunc(lambda x:N(x, 100))
vecs = ex.eigenvects()
print vecs[0][0] # eigen value
print vecs[1][0] # eigen value
print vecs[0][2] # eigen vect
print vecs[1][2] # eigen vect

output:

-2.650396553004310816338679447269582701529092549943247237903254759946483528035516341807463648841185335e-261
2.650396553004310816338679447269582701529092549943247237903254759946483528035516341807466621962539464e-261
[[-0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999994391176386872]
[                                                                                                      1]]
[[1.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000560882361313]
[                                                                                                    1]]

you can change 100 in N(x, 100) to other precision, but, as I tried 1000, the calculation of eigen vect failed.

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