NumPy线性代数完整教程 - 矩阵运算与SVD分解

import numpy as np
from numpy import linalg

np.dot(a, b)          # 点积
linalg.inv(a)         # 逆矩阵
linalg.det(a)         # 行列式
linalg.eig(a)         # 特征值和特征向量
linalg.solve(a, b)    # 解线性方程组
linalg.norm(a)        # 范数
linalg.svd(a)         # 奇异值分解

import numpy as np
from numpy import linalg

A = np.array([[1, 2], [3, 4]])
B = np.array([[5, 6], [7, 8]])

# 矩阵乘法
print(f"A @ B:\n{A @ B}")

# 逆矩阵
A_inv = linalg.inv(A)
print(f"\nA的逆矩阵:\n{A_inv}")
print(f"验证 A @ A_inv:\n{np.round(A @ A_inv)}")

# 行列式
print(f"\nA的行列式: {linalg.det(A):.1f}")

# 矩阵秩
print(f"A的秩: {linalg.matrix_rank(A)}")

A @ B:
[[19 22]
 [43 50]]

A的逆矩阵:
[[-2.   1. ]
 [ 1.5 -0.5]]
验证 A @ A_inv:
[[1. 0.]
 [0. 1.]]

A的行列式: -2.0
A的秩: 2

import numpy as np
from numpy import linalg

A = np.array([[4, 2], [1, 3]])

# 特征值和特征向量
eigenvalues, eigenvectors = linalg.eig(A)

print(f"特征值: {eigenvalues}")
print(f"特征向量:\n{eigenvectors}")

# 验证: A @ v = lambda * v
for i in range(len(eigenvalues)):
    v = eigenvectors[:, i]
    lam = eigenvalues[i]
    print(f"\nlambda={lam:.2f}: A@v = {A @ v}, lambda*v = {lam * v}")

特征值: [5. 2.]
特征向量:
[[ 0.89442719 -0.70710678]
 [ 0.4472136   0.70710678]]

lambda=5.00: A@v = [4.47213595 2.23606798], lambda*v = [4.47213595 2.23606798]

lambda=2.00: A@v = [-1.41421356  1.41421356], lambda*v = [-1.41421356  1.41421356]

import numpy as np
from numpy import linalg

# 解线性方程组: 2x + y = 5, x + 3y = 10
A = np.array([[2, 1], [1, 3]])
b = np.array([5, 10])

x = linalg.solve(A, b)
print(f"方程组解: x={x[0]:.2f}, y={x[1]:.2f}")
print(f"验证: A@x = {A @ x}")

# SVD分解
M = np.array([[1, 2], [3, 4], [5, 6]])
U, S, Vt = linalg.svd(M)
print(f"\nSVD分解:")
print(f"U shape: {U.shape}, S: {S}, Vt shape: {Vt.shape}")

# 重构矩阵
S_matrix = np.zeros((3, 2))
S_matrix[:2, :2] = np.diag(S)
reconstructed = U @ S_matrix @ Vt
print(f"重构矩阵:\n{np.round(reconstructed)}")

# 范数计算
v = np.array([3, 4])
print(f"\nL2范数: {linalg.norm(v)}")
print(f"L1范数: {linalg.norm(v, ord=1)}")
print(f"无穷范数: {linalg.norm(v, ord=np.inf)}")

方程组解: x=1.00, y=3.00
验证: A@x = [ 5. 10.]

SVD分解:
U shape: (3, 3), S: [9.52551809 0.51430058], Vt shape: (2, 2)
重构矩阵:
[[1. 2.]
 [3. 4.]
 [5. 6.]]

L2范数: 5.0
L1范数: 7.0
无穷范数: 4.0