矩阵的乘法
numpy.dot(a, b[, out])
#计算两个矩阵的乘积,如果是一维数组则是它们的内积。
import numpy as np
x = np.array([1, 2, 3, 4, 5])
y = np.array([2, 3, 4, 5, 6])
z = np.dot(x, y)
print(z) # 70
x = np.array([[1, 2, 3], [3, 4, 5], [6, 7, 8]])
print(x)
# [[1 2 3]
# [3 4 5]
# [6 7 8]]
y = np.array([[5, 4, 2], [1, 7, 9], [0, 4, 5]])
print(y)
# [[5 4 2]
# [1 7 9]
# [0 4 5]]
z = np.dot(x, y)
print(z)
# [[ 7 30 35]
# [ 19 60 67]
# [ 37 105 115]]
z = np.dot(y, x)
print(z)
# [[ 29 40 51]
# [ 76 93 110]
# [ 42 51 60]]
在线性代数里面讲的维数和数组的维数不同,如线代中提到的n维行向量在 Numpy 中是一维数组,而线性代数中的n维列向量在 Numpy 中是一个shape为(n, 1)的二维数组。
特征向量
numpy.linalg.eig(a) 计算方阵的特征值和特征向量。
特征值
numpy.linalg.eigvals(a) 计算方阵的特征值。
import numpy as np
# 创建一个对角矩阵!
x = np.diag((1, 2, 3))
print(x)
# [[1 0 0]
# [0 2 0]
# [0 0 3]]
print(np.linalg.eigvals(x))
# [1. 2. 3.]
a, b = np.linalg.eig(x)
# 特征值保存在a中,特征向量保存在b中
print(a)
# [1. 2. 3.]
print(b)
# [[1. 0. 0.]
# [0. 1. 0.]
# [0. 0. 1.]]
# 检验特征值与特征向量是否正确
for i in range(3):
if np.allclose(a[i] * b[:, i], np.dot(x, b[:, i])):
print('Right')
else:
print('Error')
# Right
# Right
# Right
判断对称阵是否为正定阵(特征值是否全部为正)
import numpy as np
A = np.arange(16).reshape(4, 4)
print(A)
# [[ 0 1 2 3]
# [ 4 5 6 7]
# [ 8 9 10 11]
# [12 13 14 15]]
A = A + A.T # 将方阵转换成对称阵
print(A)
# [[ 0 5 10 15]
# [ 5 10 15 20]
# [10 15 20 25]
# [15 20 25 30]]
B = np.linalg.eigvals(A) # 求A的特征值
print(B)
# [ 6.74165739e+01 -7.41657387e+00 1.82694656e-15 -1.72637110e-15]
# 判断是不是所有的特征值都大于0,用到了all函数,显然对称阵A不是正定的
if np.all(B > 0):
print('Yes')
else:
print('No')
# No
奇异值分解
u, s, v = numpy.linalg.svd(a, full_matrices=True, compute_uv=True, hermitian=False)#奇异值分解
#a 是一个形如(M,N)矩阵
#full_matrices的取值是为False或者True,默认值为True,这时u的大小为(M,M),v的大小为(N,N)。否则u的大小为(M,K),v的大小为(K,N) ,K=min(M,N)。
#compute_uv的取值是为False或者True,默认值为True,表示计算u,s,v。为False的时候只计算s。
#总共有三个返回值u,s,v,u大小为(M,M),s大小为(M,N),v大小为(N,N),a = u*s*v。
#其中s是对矩阵a的奇异值分解。s除了对角元素不为0,其他元素都为0,并且对角元素从大到小排列。s中有n个奇异值,一般排在后面的比较接近0,所以仅保留比较大的r个奇异值。
Numpy中返回的v是通常所谓奇异值分解a=usv’中v的转置。**
import numpy as np
A = np.array([[4, 11, 14], [8, 7, -2]])
print(A)
# [[ 4 11 14]
# [ 8 7 -2]]
u, s, vh = np.linalg.svd(A, full_matrices=False)
print(u.shape) # (2, 2)
print(u)
# [[-0.9486833 -0.31622777]
# [-0.31622777 0.9486833 ]]
print(s.shape) # (2,)
print(np.diag(s))
# [[18.97366596 0. ]
# [ 0. 9.48683298]]
print(vh.shape) # (2, 3)
print(vh)
# [[-0.33333333 -0.66666667 -0.66666667]
# [ 0.66666667 0.33333333 -0.66666667]]
a = np.dot(u, np.diag(s))
a = np.dot(a, vh)
print(a)
# [[ 4. 11. 14.]
# [ 8. 7. -2.]]
QR分解
q,r = numpy.linalg.qr(a, mode='reduced')
#计算矩阵a的QR分解。
#a是一个(M, N)的待分解矩阵。
#mode = reduced:返回(M, N)的列向量两两正交的矩阵q,和(N, N)的三角阵r(Reduced QR分解)。
#mode = complete:返回(M, M)的正交矩阵q,和(M, N)的三角阵r(Full QR分解)。
import numpy as np
A = np.array([[2, -2, 3], [1, 1, 1], [1, 3, -1]])
print(A)
# [[ 2 -2 3]
# [ 1 1 1]
# [ 1 3 -1]]
q, r = np.linalg.qr(A)
print(q.shape) # (3, 3)
print(q)
# [[-0.81649658 0.53452248 0.21821789]
# [-0.40824829 -0.26726124 -0.87287156]
# [-0.40824829 -0.80178373 0.43643578]]
print(r.shape) # (3, 3)
print(r)
# [[-2.44948974 0. -2.44948974]
# [ 0. -3.74165739 2.13808994]
# [ 0. 0. -0.65465367]]
print(np.dot(q, r))
# [[ 2. -2. 3.]
# [ 1. 1. 1.]
# [ 1. 3. -1.]]
a = np.allclose(np.dot(q.T, q), np.eye(3))
print(a) # True
mode=complete
import numpy as np
A = np.array([[1, 1], [1, -2], [2, 1]])
print(A)
# [[ 1 1]
# [ 1 -2]
# [ 2 1]]
q, r = np.linalg.qr(A, mode='complete')
print(q.shape) # (3, 3)
print(q)
# [[-0.40824829 0.34503278 -0.84515425]
# [-0.40824829 -0.89708523 -0.16903085]
# [-0.81649658 0.27602622 0.50709255]]
print(r.shape) # (3, 2)
print(r)
# [[-2.44948974 -0.40824829]
# [ 0. 2.41522946]
# [ 0. 0. ]]
print(np.dot(q, r))
# [[ 1. 1.]
# [ 1. -2.]
# [ 2. 1.]]
a = np.allclose(np.dot(q, q.T), np.eye(3))
print(a) # True
Cholesky分解
L = numpy.linalg.cholesky(a)
#返回正定矩阵a的 Cholesky 分解a = L*L.T,其中L是下三角。
import numpy as np
A = np.array([[1, 1, 1, 1], [1, 3, 3, 3],
[1, 3, 5, 5], [1, 3, 5, 7]])
print(A)
# [[1 1 1 1]
# [1 3 3 3]
# [1 3 5 5]
# [1 3 5 7]]
print(np.linalg.eigvals(A))
# [13.13707118 1.6199144 0.51978306 0.72323135]
L = np.linalg.cholesky(A)
print(L)
# [[1. 0. 0. 0. ]
# [1. 1.41421356 0. 0. ]
# [1. 1.41421356 1.41421356 0. ]
# [1. 1.41421356 1.41421356 1.41421356]]
print(np.dot(L, L.T))
# [[1. 1. 1. 1.]
# [1. 3. 3. 3.]
# [1. 3. 5. 5.]
# [1. 3. 5. 7.]]
范数
numpy.linalg.norm(x, ord=None, axis=None, keepdims=False)
#计算向量或者矩阵的范数。
import numpy as np
x = np.array([1, 2, 3, 4])
print(np.linalg.norm(x, ord=1))
# 10.0
print(np.sum(np.abs(x)))
# 10
print(np.linalg.norm(x, ord=2))
# 5.477225575051661
print(np.sum(np.abs(x) ** 2) ** 0.5)
# 5.477225575051661
print(np.linalg.norm(x, ord=-np.inf))
# 1.0
print(np.min(np.abs(x)))
# 1
print(np.linalg.norm(x, ord=np.inf))
# 4.0
print(np.max(np.abs(x)))
# 4
矩阵的范数
import numpy as np
A = np.array([[1, 2, 3, 4], [2, 3, 5, 8],
[1, 3, 5, 7], [3, 4, 7, 11]])
print(A)
# [[ 1 2 3 4]
# [ 2 3 5 8]
# [ 1 3 5 7]
# [ 3 4 7 11]]
print(np.linalg.norm(A, ord=1)) # 30.0
print(np.max(np.sum(A, axis=0))) # 30
print(np.linalg.norm(A, ord=2))
# 20.24345358700576
print(np.max(np.linalg.svd(A, compute_uv=False)))
# 20.24345358700576
print(np.linalg.norm(A, ord=np.inf)) # 25.0
print(np.max(np.sum(A, axis=1))) # 25
print(np.linalg.norm(A, ord='fro'))
# 20.273134932713294
print(np.sqrt(np.trace(np.dot(A.T, A))))
# 20.273134932713294
方阵的行列式
numpy.linalg.det(a)
#计算行列式。
import numpy as np
x = np.array([[1, 2], [3, 4]])
print(x)
# [[1 2]
# [3 4]]
print(np.linalg.det(x))
# -2.0000000000000004
矩阵的秩
numpy.linalg.matrix_rank(M, tol=None, hermitian=False)
#返回矩阵的秩。
import numpy as np
I = np.eye(3) # 先创建一个单位阵
print(I)
# [[1. 0. 0.]
# [0. 1. 0.]
# [0. 0. 1.]]
r = np.linalg.matrix_rank(I)
print(r) # 3
I[1, 1] = 0 # 将该元素置为0
print(I)
# [[1. 0. 0.]
# [0. 0. 0.]
# [0. 0. 1.]]
r = np.linalg.matrix_rank(I) # 此时秩变成2
print(r) # 2
矩阵的迹
numpy.trace(a, offset=0, axis1=0, axis2=1, dtype=None, out=None)
#方阵的迹就是主对角元素之和。
import numpy as np
x = np.array([[1, 2, 3], [3, 4, 5], [6, 7, 8]])
print(x)
# [[1 2 3]
# [3 4 5]
# [6 7 8]]
y = np.array([[5, 4, 2], [1, 7, 9], [0, 4, 5]])
print(y)
# [[5 4 2]
# [1 7 9]
# [0 4 5]]
print(np.trace(x)) # A的迹等于A.T的迹
# 13
print(np.trace(np.transpose(x)))
# 13
print(np.trace(x + y)) # 和的迹 等于 迹的和
# 30
print(np.trace(x) + np.trace(y))
# 30
逆矩阵(inverse matrix)
numpy.linalg.inv(a)
#计算矩阵a的逆矩阵(矩阵可逆的充要条件:det(a) != 0,或者a满秩)。
import numpy as np
A = np.array([[1, -2, 1], [0, 2, -1], [1, 1, -2]])
print(A)
# [[ 1 -2 1]
# [ 0 2 -1]
# [ 1 1 -2]]
# 求A的行列式,不为零则存在逆矩阵
A_det = np.linalg.det(A)
print(A_det)
# -2.9999999999999996
A_inverse = np.linalg.inv(A) # 求A的逆矩阵
print(A_inverse)
# [[ 1.00000000e+00 1.00000000e+00 -1.11022302e-16]
# [ 3.33333333e-01 1.00000000e+00 -3.33333333e-01]
# [ 6.66666667e-01 1.00000000e+00 -6.66666667e-01]]
x = np.allclose(np.dot(A, A_inverse), np.eye(3))
print(x) # True
x = np.allclose(np.dot(A_inverse, A), np.eye(3))
print(x) # True
A_companion = A_inverse * A_det # 求A的伴随矩阵
print(A_companion)
# [[-3.00000000e+00 -3.00000000e+00 3.33066907e-16]
# [-1.00000000e+00 -3.00000000e+00 1.00000000e+00]
# [-2.00000000e+00 -3.00000000e+00 2.00000000e+00]]
求解线性方程组
numpy.linalg.solve(a, b)
#求解线性方程组或矩阵方程。
# x + 2y + z = 7
# 2x - y + 3z = 7
# 3x + y + 2z =18
import numpy as np
A = np.array([[1, 2, 1], [2, -1, 3], [3, 1, 2]])
b = np.array([7, 7, 18])
x = np.linalg.solve(A, b)
print(x) # [ 7. 1. -2.]
x = np.linalg.inv(A).dot(b)
print(x) # [ 7. 1. -2.]
y = np.allclose(np.dot(A, x), b)
print(y) # True
参考博文
team-learning-program/IntroductionToNumpy/task09 线性代数/13. 线性代数…ipynb
numpy判断两个向量是否相近-numpy.allclose
Cholesky分解
Cholesky分解及一个例子
Cholesky分解