Comment #7 on issue 3189 by pr...@goodok.ru: Calculate eigenvectors
numericly if it impossible calculate symbolical
http://code.google.com/p/sympy/issues/detail?id=3189
But if "a" is real, then:
In [39]: a = Symbol('a', real=True)
In [40]: vals = roots(x**2 - (1 + a)*x + a, x)
In [41]: v1 = vals.keys()[0]
In [42]: v2 = vals.keys()[1]
In [43]: v1
Out[43]:
a │a - 1│ 1
─ - ─────── + ─
2 2 2
In [44]: v1.expand(complex=True)
Out[44]:
a │a - 1│ 1
─ - ─────── + ─
2 2 2
In [45]: v2
Out[45]:
a │a - 1│ 1
─ + ─────── + ─
2 2 2
About this example we can think how to deal with this sets of roots
If a > 1 the the list of roots is: (1/2, a + 1/2), otherwise (a+1/2, 1/2).
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