python 矩阵特征值分解_讲一下numpy的矩阵特征值分解与奇异值分解

1、特征值分解

主要还是调包:

from numpy.linalg import eig

特征值分解:  A = P*B*PT 当然也可以写成 A = QT*B*Q 其中B为对角元为A的特征值的对角矩阵,P=QT,

首先A得对称正定,然后才能在实数域上分解,

>>> A = np.random.randint(-10,10,(4,4))>>>A

array([[6, 9, -10, -1],

[5, 9, 5, -5],

[-8, 7, -4, 4],

[-1, -9, 0, 6]])>>> C =np.dot(A.T, A)>>>C

array([[126, 52, -3, -69],

[52, 292, -73, -80],

[-3, -73, 141, -31],

[-69, -80, -31, 78]])>>> vals, vecs =eig(C)>>>vals

array([357.33597086, 174.10172008, 8.84429957, 96.71800949])>>>vecs

array([[-0.28738314, -0.51589436, -0.38221983, -0.71075449],

[-0.87487263, 0.12873861, -0.24968051, 0.39456798],

[0.2572149 , -0.69304313, -0.33950158, 0.58161018],

[0.29300052, 0.48679627, -0.82237845, -0.02955945]])

故使用时应先将特征值转换为矩阵:

>>> Lambda =np.diag(vals)>>>Lambda

array([[357.33597086, 0. , 0. , 0. ],

[ 0. ,174.10172008, 0. , 0. ],

[ 0. , 0. ,8.84429957, 0. ],

[ 0. , 0. , 0. ,96.71800949]])>>> np.dot(np.dot(vecs, Lambda), vecs.T) #与C=A.T*A相等

array([[126., 52., -3., -69.],

[52., 292., -73., -80.],

[-3., -73., 141., -31.],

[-69., -80., -31., 78.]])>>>np.dot(np.dot(vecs.T, Lambda), vecs)

array([[171.65817919, 45.58778569, 53.20435074, 13.37512137],

[45.58778569, 125.15670964, 28.22684299, 134.91290105],

[53.20435074, 28.22684299, 129.48789571, 80.5284382],

[13.37512137, 134.91290105, 80.5284382 , 210.69721545]])

故验证了使用np中的eig分解为A=P*B*PT而不是A=QT*B*Q,其中P=vecs,

即 C = vecs * np.diag(vals) * vecs.T #这里简写*为矩阵乘法

然后再来看使用np中的eig分解出来的vec中行向量是特征向量还是列向量是特征向量,只需验证:A*vecs[0] = vals[0]*vecs[0]

>>>np.dot(C, vecs[0])

array([-12.84806258, -80.82266859, 6.66283128, 17.51094927])>>> vals[0]*vecs[0]

array([-102.69233303, -184.34761071, -136.58089252, -253.97814676])>>>np.dot(C, vecs[:,0])

array([-102.69233303, -312.62346098, 91.91213634, 104.69962583])>>> vals[0]*vecs[:, 0]

array([-102.69233303, -312.62346098, 91.91213634, 104.69962583])

后者两个是相等的,故使用np中的eig分解出的vecs的列向量是特征向量。

然后我们可以验证P是单位正交矩阵:

>>>np.dot(vecs.T, vecs)

array([[1.00000000e+00, -7.13175042e-17, -2.45525952e-18,2.75965773e-16],

[-7.13175042e-17, 1.00000000e+00, 2.49530948e-17,-5.58839097e-16],

[-2.45525952e-18, 2.49530948e-17, 1.00000000e+00,-7.85564967e-17],

[2.75965773e-16, -5.58839097e-16, -7.85564967e-17,1.00000000e+00]])>>>np.dot(vecs, vecs.T)

array([[1.00000000e+00, 2.97888865e-16, -2.68317972e-16,1.69020590e-16],

[2.97888865e-16, 1.00000000e+00, -4.40952204e-18,-6.24188690e-17],

[-2.68317972e-16, -4.40952204e-18, 1.00000000e+00,-1.13726775e-17],

[1.69020590e-16, -6.24188690e-17, -1.13726775e-17,1.00000000e+00]])#可以看到除对角元外其他都是非常小的数

即PT*P = P*PT = E , PT=P-1。事实上,在求解P的过程中就使用了施密特正交化过程。

另一方面,我们从数学角度来看:

首先补充一些数学知识:可以看我另一篇文章:矩阵知识

A = P*B*P-1  ,其中B为对角元素为A的特征值的对角阵,P的列向量为特征值对应的特征向量(因为B每行乘以P每列)

2、奇异值分解

还是调包:

from numpy.linalg import svd

设任意矩阵A是m*n矩阵

奇异值分解:A = U*Σ*VT , 其中U为满足UTU=E的m阶(m*m)酉矩阵,Σ为对角线上为奇异值σi 其他元素为0的广义m*n对角阵,V为满足VTV=E的n阶(n*n)酉矩阵

a = np.random.randint(-10,10,(4, 3)).astype(float)'''array([[ -9., 3., -7.],

[ 4., -8., -1.],

[ -1., 6., -9.],

[ -4., -10., 2.]])'''In [53]: u, s, vh = np.linalg.svd(a) #这里vh为V的转置

In [55]: u.shape, s.shape, vh.shape

Out[55]: ((4, 4), (3,), (3, 3))'''In [63]: u

Out[63]:

array([[-0.53815289, 0.67354057, -0.13816841, -0.48748749],

[ 0.40133556, 0.1687729 , 0.78900752, -0.43348888],

[-0.59291924, 0.04174708, 0.59180987, 0.54448603],

[ 0.44471115, 0.71841213, -0.09020922, 0.52723647]])

In [64]: s

Out[64]: array([16.86106528, 11.07993065, 7.13719934])

In [65]: vh

Out[65]:

array([[ 0.31212695, -0.760911 , 0.56885078],

[-0.74929793, -0.56527432, -0.3449892 ],

[ 0.58406282, -0.31855829, -0.74658639]])'''In [56]: np.allclose(a, np.dot(u[:, :3] *s, vh))

Out[56]: True#将s转化为奇异值矩阵

In [60]: smat[:3, :3] =np.diag(s)

In [61]: smat

Out[61]:

array([[16.86106528, 0. , 0. ],

[ 0. ,11.07993065, 0. ],

[ 0. , 0. ,7.13719934],

[ 0. , 0. , 0. ]])#验证分解的正确性

In [62]: np.allclose(a, np.dot(u, np.dot(smat, vh)))

Out[62]: True

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