Matlab:
m1 = [ 333.33333333 83.33333333 0. ; 83.33333333 333.33333333 83.33333333 ; 0. 3.33333333 166.66666667]
k1 = [ 800. -400. 0.; -400. 800. -400.; 0. -400. 400.]
[vec, val] = eig(m1, k1)
vec =
-0.0106 -0.0289 0.0394
0.0183 -0.0000 0.0683
-0.0211 0.0289 0.0789
python:
import scipy
import numpy as np
m1 = np.array[[333.33333333, 83.33333333, 0.], [ 83.33333333, 333.33333333, 83.33333333], [0., 83.33333333, 166.66666667]]
k1 = np.array[[800., -400., 0.], [-400., 800., -400.], [0., -400., 400.]]
val, vec = scipy.linalg.eig(m1, k1)
vec =
[[ 3.53553391e-01, -7.07106781e-01, 3.53553391e-01],
[ 6.12372436e-01, 1.88119544e-16, -6.12372436e-01],
[ 7.07106781e-01, 7.07106781e-01, 7.07106781e-01]]
So matlab and python eigenvectors vec are not matching. Matlab document says
[V,D] = eig(A,B) returns diagonal matrix D of generalized eigenvalues and full matrix V whose columns are the corresponding right eigenvectors, so that
A*V = B*V*D.
and m1 * vec = k1 * vec * val satisfies for matlab output but not for python output. How can I get similar eigenvectors as given by matlab using python, numpy?
Eigenvectors are not unique; if x is an eigenvector, then a * x is still an eigenvector for any nonzero scalar a.
Look at the two results from python and matlab; the first column of vec from matlab looks like a scaled version of the third column of vec from python. The second column from matlab seems to be a scaled version of the second column from python (recall 1.88119544e-16 is almost zero). The third column corresponds to the first column. So my guess is that val from matlab is a reversed version of val from python; I can't verify this because I don't have matlab right now, but you can verify this.
Also in python, remember * is element-wise multiplication (i.e., .* in matlab). You can use @ for matrix multiplication.
import scipy.linalg
import numpy as np
m1 = np.array([[333.33333333, 83.33333333, 0.], [ 83.33333333, 333.33333333, 83.33333333], [0., 83.33333333, 166.66666667]])
k1 = np.array([[800., -400., 0.], [-400., 800., -400.], [0., -400., 400.]])
val, vec = scipy.linalg.eig(m1, k1)
print(m1 @ vec)
print(k1 @ vec * val)
As you can see in the following output, we have the same output (the imaginary part is 0.j, so you can ignore that).
Output:
[[ 1.68882167e+02 -2.35702260e+02 6.68200939e+01]
[ 2.92512493e+02 -3.14270210e-09 -1.15735798e+02]
[ 1.68882167e+02 1.17851130e+02 6.68200939e+01]]
[[ 1.68882167e+02+0.j -2.35702260e+02+0.j 6.68200939e+01+0.j]
[ 2.92512493e+02+0.j -3.14263578e-09+0.j -1.15735798e+02+0.j]
[ 1.68882167e+02+0.j 1.17851130e+02+0.j 6.68200939e+01+0.j]]