left hand side eigenvector in python?

Your result is correct to my understanding.

However, you might be misinterpreting it. The numpy docs are a bit clearer on what the left eigenvectors should be.

Finally, it is emphasized that v consists of the right (as in right-hand side) eigenvectors of a. A vector y satisfying dot(y.T, a) = z * y.T for some number z is called a left eigenvector of a, and, in general, the left and right eigenvectors of a matrix are not necessarily the (perhaps conjugate) transposes of each other.

I.e. you need to transpose the vectors in vl. vl[:,i].T is the i-th left eigenvector.If I test this, I get, that the results are correct.

>>> import numpy as np>>> from scipy.linalg import eig>>> np.set_printoptions(precision=4)>>> T = np.mat("0.2 0.4 0.4;0.8 0.2 0.0;0.8 0.0 0.2")>>> print "T\n", TT[[ 0.2  0.4  0.4] [ 0.8  0.2  0. ] [ 0.8  0.   0.2]]>>> w, vl, vr = eig(T, left=True)>>> vlarray([[ 0.8165,  0.8165,  0.    ],       [ 0.4082, -0.4082, -0.7071],       [ 0.4082, -0.4082,  0.7071]])>>> [ np.allclose(np.dot(vl[:,i].T, T), w[i]*vl[:,i].T) for i in range(3) ][True, True, True]