第十一周作业

最新推荐文章于 2022-11-29 17:40:36 发布

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本文通过Python的NumPy和SciPy库进行了一系列矩阵运算实验，包括矩阵加法、乘法、求解线性方程组、计算范数及奇异值等，并分析了矩阵的Frobenius范数、无穷范数、最小和最大奇异值。此外，还探讨了矩阵大小N与最大奇异值之间的关系，以及概率p与最大奇异值的关系。

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import numpy as np
import scipy.linalg
import matplotlib.pyplot as plt
 
#initialize
n, m = 200, 500
A = np.mat(np.random.normal(size = (200, 500)))
B = np.mat(scipy.linalg.toeplitz([np.random.normal(0, 1) for i in range(m)]))
 
def f1(Lambda):
	return A * (B - Lambda * np.eye(m))
 
#Excercise 9.1
AaddA = A + A
AAT = A * A.T
ATA = A.T * A
AB = A * B
f1(1)
 
#Exercise 9.2
b = [1 for i in range(m)]
scipy.linalg.solve(B, b)
 
#Exercise 9.3
FrobeniusNormOfA = scipy.linalg.norm(A, 'fro')
inﬁnityNormOfB = scipy.linalg.norm(B, np.inf)
SmallestingularValueOfB = min(scipy.linalg.svdvals(B))
LargestingularValueOfB = max(scipy.linalg.svdvals(B))
 
#Exercise 9.4
 
Z = np.random.normal(size = (500, 500))
bk = np.random.normal(size = (500, ))
i = 0
while 1:
	bk1 = np.dot(Z, bk)
	bk1 /= scipy.linalg.norm(bk1)
	i += 1
	if abs(scipy.linalg.norm(bk, np.inf) - scipy.linalg.norm(bk1, np.inf)) < 10e-6: break
	bk = bk1
print(i)
 
#Exercise 9.5
#Analyse the relationship between the N and the largest singular value
p = 0.5
nValues, LargestingularValues = [], []
for i in range(10):
	N = (i + 1) * 50
	nValues.append(N)
	C = [[1 if np.random.random() > p else 0 for k in range(N)] for j in range(N)]	
	U, sigma, VT=scipy.linalg.svd(C)
	LargestingularValues.append(max(sigma))
plt.scatter(nValues, LargestingularValues, s = 100) 
plt.xlabel("N", fontsize=14)  
plt.ylabel("largest singular value", fontsize=14)
plt.show()
 
#Analyse the relationship between the p and the largest singular value
N = 100
pValues, LargestingularValues = [], []
for i in range(10):
	p = (i + 1) * 0.1
	pValues.append(p)
	C = [[1 if np.random.random() > p else 0 for k in range(N)] for j in range(N)]	
	U, sigma, VT=scipy.linalg.svd(C)
	LargestingularValues.append(max(sigma))
plt.scatter(pValues, LargestingularValues, s = 100) 
plt.xlabel("p", fontsize=14)  
plt.ylabel("largest singular value", fontsize=14)
plt.show()
 
#Exercise 9.6
def f2(z, A):
	return A[np.argmin([abs(A[i] - z) for i in range(len(A))])]
 
f2(0.5, [np.random.normal(0, 1) for i in range(500)])