python vs matlab中的特征值和特征向量

问题描述

我注意到，matlab计算矩阵的特征值和特征向量的方式有所不同，其中matlab返回实值，而numpy返回复杂值的特征值和向量.例如:

I have noticed there is a difference between how matlab calculates the eigenvalue and eigenvector of a matrix, where matlab returns the real valued while numpy's return the complex valued eigen valus and vector. For example:

对于矩阵:

A=
     1    -3     3
     3    -5     3
     6    -6     4

Numpy:
w, v = np.linalg.eig(A)

w
array([ 4. +0.00000000e+00j, -2. +1.10465796e-15j, -2. -1.10465796e-15j])


v
array([[-0.40824829+0.j        ,  0.24400118-0.40702229j,
             0.24400118+0.40702229j],
           [-0.40824829+0.j        , -0.41621909-0.40702229j,
            -0.41621909+0.40702229j],
           [-0.81649658+0.j        , -0.66022027+0.j        , -0.66022027-0.j        ]])

Matlab:
[E, D] = eig(A)
E

   -0.4082   -0.8103    0.1933
   -0.4082   -0.3185   -0.5904
   -0.8165    0.4918   -0.7836
D

    4.0000         0         0
         0   -2.0000         0
         0         0   -2.0000

有没有一种方法可以像在Matlab中那样获取python中的真实特征值?

Is there a way of getting the real eigen values in python as it is in matlab?

推荐答案

要使NumPy在复杂部分较小时返回真实特征值的对角线数组，可以使用

To get NumPy to return a diagonal array of real eigenvalues when the complex part is small, you could use

In [116]: np.real_if_close(np.diag(w))
Out[116]:
array([[ 4.,  0.,  0.],
       [ 0., -2.,  0.],
       [ 0.,  0., -2.]])

---

根据 Matlab文档，[E, D] = eig(A)返回满足A*E = E*D的E和D:我没有Matlab，所以我将使用Octave来检查您发布的结果:

---

According to the Matlab docs,[E, D] = eig(A) returns E and D which satisfy A*E = E*D:I don't have Matlab, so I'll use Octave to check the result you posted:

octave:1> A = [[1, -3, 3],
               [3, -5, 3],
               [6, -6, 4]]

octave:6> E = [[ -0.4082, -0.8103, 0.1933],
               [ -0.4082, -0.3185, -0.5904],
               [ -0.8165, 0.4918, -0.7836]]

octave:25> D = [[4.0000, 0, 0],
               [0, -2.0000, 0],
               [0, 0, -2.0000]]

octave:29> abs(A*E - E*D)
ans =

   3.0000e-04   0.0000e+00   3.0000e-04
   3.0000e-04   2.2204e-16   3.0000e-04
   0.0000e+00   4.4409e-16   6.0000e-04

误差的幅度主要是由于Matlab报告的值是显示的精度低于Matlab在内存中保存的实际值.

The magnitude of the errors is mainly due to the values reported by Matlab beingdisplayed to a lower precision than the actual values Matlab holds in memory.

在NumPy中，w, v = np.linalg.eig(A)返回满足以下条件的w和vnp.dot(A, v) = np.dot(v, np.diag(w)):

In NumPy, w, v = np.linalg.eig(A) returns w and v which satisfynp.dot(A, v) = np.dot(v, np.diag(w)):

In [113]: w, v = np.linalg.eig(A)

In [135]: np.set_printoptions(formatter={'complex_kind': '{:+15.5f}'.format})

In [136]: v
Out[136]:
array([[-0.40825+0.00000j, +0.24400-0.40702j, +0.24400+0.40702j],
       [-0.40825+0.00000j, -0.41622-0.40702j, -0.41622+0.40702j],
       [-0.81650+0.00000j, -0.66022+0.00000j, -0.66022-0.00000j]])

In [116]: np.real_if_close(np.diag(w))
Out[116]:
array([[ 4.,  0.,  0.],
       [ 0., -2.,  0.],
       [ 0.,  0., -2.]])

In [112]: np.abs((np.dot(A, v) - np.dot(v, np.diag(w))))
Out[112]:
array([[4.44089210e-16, 3.72380123e-16, 3.72380123e-16],
       [2.22044605e-16, 4.00296604e-16, 4.00296604e-16],
       [8.88178420e-16, 1.36245817e-15, 1.36245817e-15]])

In [162]: np.abs((np.dot(A, v) - np.dot(v, np.diag(w)))).max()
Out[162]: 1.3624581677742195e-15

In [109]: np.isclose(np.dot(A, v), np.dot(v, np.diag(w))).all()
Out[109]: True

这篇关于python vs matlab中的特征值和特征向量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！