高级编程技术，第十一周

Generate matrices A , with random Gaussian entries, B , a Toeplitz matrix, where A 2 R n × m and B 2 R m × m ,

for n = 200, m = 500

import numpy as np
from scipy import linalg

n = 200
m = 500
A = np.random.normal(size = (n,m) )
B = linalg.toeplitz(np.random.normal(size = (1,m)), np.random.normal(size = (1,m)))

9.1

Calculate A + A, AAT; ATA and AB. Write a function that computes A(B - λI) for any λ

print("A+A=",A+A)
print("AAT=",np.dot(A,A.T))
print("ATA=",np.dot(A.T,A))
print("AB=",np.dot(A,B))

def func(x):
	return np.dot(A,B - x * np.eye(m))

9.2

Generate a vector b with m entries and solve Bx = b

b = np.random.random(m)
x = np.linalg.solve(B,b)
print("Bx = ",np.dot(B,x))
print("b = ",b)

9.3

Compute the Frobenius norm of A: kAkF and the infinity norm of B: kBk1. Also find the largest and
smallest singular values of B

print("the fro norm of A = ",linalg.norm(A))
print("the infinity norm of B = ",linalg.norm(B, np.inf))
u,sigma,vt = linalg.svd(B)
print("the largest singular values of B = ",np.max(sigma))
print("the smallest singular values of B = ",np.min(sigma))

9.4

Generate a matrix Z, n × n, with Gaussian entries, and use the power iteration to find the largest
eigenvalue and corresponding eigenvector of Z. How many iterations are needed till convergence?
Optional: use the time.clock() method to compare computation time when varying n

import numpy as np
from scipy import linalg
import time


def eig(n):
	Z = np.random.normal(size = (n,n))
	v = np.ones(n)
	a = linalg.norm(v, np.inf)
	temp = 0
	x = 0 
	t = time.clock()
	while True:
		temp = a
		y = np.dot(Z, v)
		a = linalg.norm(y, np.inf)
		v = y / a
		x = x + 1
		if abs(temp - a) < 0.00000100:
			break
	t = time.clock() - t
	print("times:",x)
	print("time:",t)
	print("eigenvector:",v)
	print("eigenvalue:",a)

9.5

Generate an n × n matrix, denoted by C, where each entry is 1 with probability p and 0 otherwise. Use
the linear algebra library of Scipy to compute the singular values of C. What can you say about the
relationship between n, p and the largest singular value?

import numpy as np
from scipy import linalg

def Singular(n,p):
	C = np.random.binomial(1,p,[n,n])
	u,sigma,vt = linalg.svd(C)
	print("p = ", p, "n = ",n, "MAX SINGULAR = ",np.max(sigma))


Singular(50,0.3)
Singular(50,0.6)
Singular(80,0.2)
Singular(100,0.9)
Singular(200,0.5)

执行效果如下：

结论是：N足够大时，此时最大的奇异值收敛于np

9.6

Write a function that takes a value z and an array A and finds the element in A that is closest to z. The
function should return the closest value, not index

import numpy as np

def find(A,z):
	a = np.abs(A - z)
	return A[np.argmin(a)]