python第11周作业
1.Generate matrices A, with random Gaussian entries, B, a Toeplitz matrix, where A ∈ Rn×m and B ∈ Rm×m,
Calculate A + A, AA>, A>A and AB. Write a function that computes A(B − λI) for any λ.
Generate a vector b with m entries and solve Bx = b.
smallest singular values of B.
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.
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.
Hint: Use the built-in functionality of Numpy rather than writing code to find this value manually. In
particular, use brackets and argmin.
for n = 200, m = 500.
import numpy as np
import time
import scipy
import scipy.linalg
np.random.seed(int(time.time()))
A = np.random.normal(0, 1, (200, 500))
B = scipy.linalg.toeplitz(list(range(1,501)))
Calculate A + A, AA>, A>A and AB. Write a function that computes A(B − λI) for any λ.
np.random.seed(int(time.time()))
A = np.random.normal(0, 1, (200, 500))
B = scipy.linalg.toeplitz(list(range(1,501)))
AaA = A + A
AAT = np.matmul(A, A.T)
ATA = np.matmul(A.T, A)
AB = np.matmul(A, B)
def func(c):
T = B - c * np.identity(500)
return np.matmul(A, T)结果：
[[-2.43481625 0.30609404 -1.04089433 ..., 2.06474909 -1.40572007
1.59820044]
[-2.145266 -1.37180988 0.47677339 ..., 3.13621931 -2.34917689
-2.86347192]
[ 1.39579581 2.3582495 0.30430538 ..., 0.36531283 1.11695448
0.81979878]
...,
[-0.36408384 4.20189382 1.64300896 ..., -3.91156688 -2.00446271
-0.37220382]
[ 0.6513696 -0.69248735 -0.06958598 ..., 0.36882953 -0.71120829
0.75370296]
[-3.21804384 1.67379258 -3.94242806 ..., 0.23953906 2.15602927
1.08936674]]
[[ 4.88042142e+02 3.24708429e-01 -1.23663493e+01 ..., -3.84843748e+01
2.81886255e+01 4.09658717e+01]
[ 3.24708429e-01 4.71415031e+02 -2.02349093e+01 ..., -6.15256393e+00
-1.44221999e+00 3.95308830e+01]
[ -1.23663493e+01 -2.02349093e+01 5.11639534e+02 ..., -4.26799895e+01
-1.27156660e+01 -7.32146132e+00]
...,
[ -3.84843748e+01 -6.15256393e+00 -4.26799895e+01 ..., 5.14871637e+02
-1.37220534e+01 -4.47321696e+01]
[ 2.81886255e+01 -1.44221999e+00 -1.27156660e+01 ..., -1.37220534e+01
4.93422434e+02 -4.62980419e+01]
[ 4.09658717e+01 3.95308830e+01 -7.32146132e+00 ..., -4.47321696e+01
-4.62980419e+01 4.97840981e+02]]
[[ 185.10360245 11.87797242 -0.79199176 ..., 6.89548094 -25.2321327
-16.28313649]
[ 11.87797242 189.94158862 0.51663773 ..., 13.813414 4.53860253
-15.29495292]
[ -0.79199176 0.51663773 173.34197838 ..., 5.91059159
8.83135864 1.4515256 ]
...,
[ 6.89548094 13.813414 5.91059159 ..., 221.45655496
9.81504408 -1.26396403]
[ -25.2321327 4.53860253 8.83135864 ..., 9.81504408 201.9162066
-7.10646931]
[ -16.28313649 -15.29495292 1.4515256 ..., -1.26396403
-7.10646931 198.86807285]]
[[ 2751.75395236 2750.13073873 2748.81361915 ..., -3154.9529813
-3155.9570643 -3158.36686737]
[ 3823.04583129 3811.89239909 3799.36715702 ..., 663.95298093
678.17379594 690.04543406]
[ 2658.88399701 2664.56583641 2672.60592531 ..., -4794.86319598
-4801.08599283 -4806.19183519]
...,
[ 4124.91097027 4112.01367127 4103.3182661 ..., 2126.4145235
2141.32440519 2154.22982417]
[ 11423.95393771 11400.39055979 11376.13469451 ..., 660.00127418
684.17352703 707.63457158]
[ -6648.26444508 -6629.581224 -6609.22421034 ..., -4276.13198888
-4301.27864982 -4324.26928148]]
[[ 2754.18876861 2749.82464469 2749.85451348 ..., -3157.01773039
-3154.55134423 -3159.96506781]
[ 3825.19109729 3813.26420897 3798.89038363 ..., 660.81676162
680.52297283 692.90890598]
[ 2657.4882012 2662.20758691 2672.30161993 ..., -4795.22850881
-4802.20294731 -4807.01163397]
...,
[ 4125.27505411 4107.81177745 4101.67525713 ..., 2130.32609038
2143.3288679 2154.60202799]
[ 11423.30256811 11401.08304714 11376.2042805 ..., 659.63244465
684.88473532 706.88086862]
[ -6645.04640123 -6631.25501658 -6605.28178228 ..., -4276.37152794
-4303.43467908 -4325.35864823]]
[Finished in 0.8s]Generate a vector b with m entries and solve Bx = b.
b = np.random.random(500) * 500
x = np.linalg.solve(B, b)结果：
[ 2.91863964e-01 1.10752802e+02 -2.64336406e+02 2.37546766e+02
-1.09032341e+02 9.16481966e+01 -1.79691165e+02 2.55673896e+02
-1.73717600e+02 -3.70264353e+00 4.96868138e+01 -6.08914025e+01
-4.53682636e+01 6.03112957e+01 2.22032512e+02 -3.27236563e+02
7.98886999e+01 1.22327097e+02 -1.21550891e+02 9.71128273e+01
2.27727003e+01 -4.17147694e+01 -1.44261875e+02 2.13355012e+02
1.29749380e+01 -2.45278907e+02 1.68072690e+02 3.62124701e+01
-1.16325155e+02 1.61347938e+02 -2.44412686e+02 1.67543448e+02
8.06111200e+01 -1.66895771e+02 -4.27944144e+01 1.46689230e+02
-1.64667279e+02 1.13228575e+02 4.16131875e+01 2.80574337e+01 ...]Exercise 9.3: Norms
Compute the Frobenius norm of A: kAkF and the infinity norm of B: kBk∞. Also find the largest andsmallest singular values of B.
norma = np.linalg.norm(A, 'fro')
normb = np.linalg.norm(B, np.inf)
s = np.linalg.svd(B, compute_uv=0)
print(norma)
print(normb)
print(max(s))
print(min(s))结果：
314.732306846
125250.0
87334.5204564
0.500004934835Generate 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.
from time import clock
def power_iteration(Z, stop):
u = np.random.normal(size=[len(Z)]).transpose()
u = u/np.max(u)
v = u.copy()
eigen_change = np.inf
eigen_val = np.max(v)
iteration = 0
s_time = time.clock()
while abs(eigen_change) > stop:
v = np.matmul(Z, u)
max_v = np.max(v)
eigen_change = eigen_val - max_v
eigen_val = max_v
u = v/max_v
iteration += 1
print("The result is {}".format(np.allclose(np.matmul(Z, u), eigen_val*u)))
print("Total iteration: {}".format(iteration))
print("Total time: {}s".format(time.clock() - s_time))
return [eigen_val, u, iteration, time.clock() - s_time]
Z = np.random.normal(0, 1, (200, 200))
power_iteration(Z, 0.0000001)Exercise 9.5: Singular values
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?
p_ = [0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9]
for p in p_:
C = np.random.random(size=[n, n])
for i in range(n):
for j in range(n):
C[i][j] = 1 if C[i][j] > p else 0
s= np.linalg.svd(C, compute_uv=False)
m = np.max(s) 关系： n*(1-p) ≈ max_singular
Exercise 9.6: Nearest neighborWrite 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.
Hint: Use the built-in functionality of Numpy rather than writing code to find this value manually. In
particular, use brackets and argmin.
def func(A, z):
index = np.argmin(np.abs(A - z))
return A[index]