使用Power方法从Python 3x3矩阵中获取特征值 [英] Getting eigenvalues from 3x3 matrix in Python using Power method

问题描述

我正在尝试使用Python中的Power方法从3x3矩阵中获取所有特征值.但是由于某种原因，我的方法从正确的特征值中返回了不同的特征值.

I'm trying to get all eigenvalues from a 3x3 matrix by using Power Method in Python. However my method returns diffrent eigenvalues from the correct ones for some reason.

我的矩阵:A = [[1、2、3]，[2、4、5]，[3、5，-1]]

My matrix: A = [[1, 2, 3], [2, 4, 5], [3, 5,-1]]

正确的特征值:[8.54851285，-4.57408723，0.02557437]

Correct eigenvalues: [ 8.54851285, -4.57408723, 0.02557437 ]

我的方法返回的特征值:[8.5485128481521926，4.5740872291939381，9.148174458392436]

Eigenvalues returned by my method: [ 8.5485128481521926, 4.5740872291939381, 9.148174458392436 ]

因此，第一个是正确的，第二个是错误的符号，而第三个都是错误的.我不知道自己做错了什么，也看不出哪里出错了.

So the first one is correct, second one has wrong sign and the third one is all wrong. I don't know what I'm doing wrong and I can't see where have I made mistake.

这是我的代码:

import numpy as np
import numpy.linalg as la

eps = 1e-8 # Precision of eigenvalue

def trans(v): # translates vector (v^T)
    v_1 = np.copy(v)
    return v_1.reshape((-1, 1))

def power(A):
    eig = []
    Ac = np.copy(A)
    lamb = 0
    for i in range(3):
        x = np.array([1, 1, 1])
        while True:
            x_1 = Ac.dot(x) # y_n = A*x_(n-1)
            x_norm = la.norm(x_1) 
            x_1 = x_1/x_norm # x_n = y_n/||y_n||
            if(abs(lamb - x_norm) <= eps): # If precision is reached, it returns eigenvalue
                break
            else:
                lamb = x_norm
                x = x_1
        eig.append(lamb)

        # Matrix Deflaction: A - Lambda * norm[V]*norm[V]^T
        v = x_1/la.norm(x_1)
        R = v * trans(v)
        R = eig[i]*R
        Ac = Ac - R

    return eig

def main():
    A = np.array([1, 2, 3, 2, 4, 5, 3, 5, -1]).reshape((3, 3))
    print(power(A))



if __name__ == '__main__':
    main()

PS.有没有一种简单的方法可以从幂方法中获得第二和第三特征值，而不是矩阵偏转?

PS. Is there a simpler way to get the second and third eigenvalue from power method instead of matrix deflaction?

推荐答案

使用

lamb = x_norm

您只能计算特征值的绝对值.更好地将它们计算为

you ever only compute the absolute value of the eigenvalues. Better compute them as

lamb = dot(x,x_1)

其中 x 被认为是标准化的.

where x is assumed to be normalized.

由于您没有删除负特征值-4.57408723，而是有效地添加了负特征值，因此第三阶段的最大特征值是 2 * -4.574 .. = -9.148 .. ，在此您再次进行了计算绝对值.

As you do not remove the negative eigenvalue -4.57408723, but effectively add it instead, the largest eigenvalue in the third stage is 2*-4.574.. = -9.148.. where you again computed the absolute value.

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