Eigenvector mystery

[Albert Chan]

All eigenvectors are correct. The difference is only a scaling factor.

XCas> m := [[-4,-17], [2,2]]
XCas> eigenvalues(m)      → (-1-5*i,-1+5*i)

Solving both vectors at the same time, λ = -1±5i

m * [x,y] = λ * [x,y]
(m - λ) * [x,y] = 0

m - λ = \(\left(\begin{array}{cc}
-3∓5i & -17 \\
2 & 3∓5i
\end{array}\right) \)

If you pick 2nd row, you get results of Wolfram Alpha

2 x + (3∓5i) y = 0
2 x = (-3±5i) y       → [x, y] = [-3±5i, 2] * t, where t is the parameter

If you pick 1st row, you get [x,y] = [-17, 3±5i] * t'
Confirming numpy's result (which is probably what Julia use)

>>> import numpy
>>> k, v = numpy.linalg.eig([[-4,-17],[2,2]])
>>> print k
[-1.+5.j -1.-5.j]
>>> print v
[[ 0.94590530+0.j           0.94590530+0.j        ]
 [-0.16692447-0.27820744j  -0.16692447+0.27820744j]]
>>> print v * (-17/v[0][0])
[[-17.+0.j  -17.+0.j]
 [  3.+5.j    3.-5.j]]

Update: numpy returns normalized eigenvectors (length = 1)
>>> sum(abs(v)**2) ** 0.5       # numpy eigenvectors are column vectors
array([ 1., 1.])