列主元消去法和LU矩阵分解以及矩阵迭代求解
目录
上三角矩阵求解
下三角矩阵求解
高斯消去法
LU矩阵分解及其求解
雅克比迭代法
高斯塞得迭代法
松弛法
源码
上三角矩阵求解
为了方便后文求解矩阵，先将上三角矩阵、和下三角矩阵求解打包，这里输入上三角矩阵A和解矩阵b。注意这里的矩阵是A，b都是二维数组。例如：
A =np.array([3.01,6.03,1.99,1.27,4.16,-1.23,0.987,-4.81,9.34],dtype=float).reshape(3,3)
B = np.array([1,1,1], dtype=float).reshape(3, 1)#上三角求解：
def up_solve(A,b):
n = len(A[0])
A[n-1,n-1] = b[n-1][0]/A[n-1,n-1]
for i in range(1,n):
k = n-i-1#取逆序
A[k,k] = (b[k][0] - sum(A[k]*A[k+1]))/A[k,k]
A[k]=np.hstack((A[k,:k+1],A[k+1,k+1:]))#拼接新数组
# print('方程解为',A[0])
return A[0]下三角矩阵求解
同上文一样，这里下三角矩阵求解：
#下三角求解
def down_solve(A,b):
n = len(A[0])
A[0,0] = b[0][0] /A[0,0]
for i in range(1, n):
k = i # 取逆序
A[k, k] = (b[k][0] - sum(A[k] * A[k-1])) / A[k, k]
A[k] = np.hstack((A[k-1, :k], A[k, k:])) # 拼接新数组
# print('方程解为', A[-1])
return A[-1]
高斯消去法
这里主要是用到的列主元消去法，为了更直观的观察消去过程，在代码中增加了消去过程输出。消元后得到上三角矩阵，然后调用上面上三角矩阵求解以及输出了系数矩阵的行列式。
#列主元高斯消去
def gaussian(A,b):
#最大值索引
n = len(A[0])
A = np.hstack([A,b])
change_index=0
for i in range(n):
tem_lie = abs(A[:,i])#列主元最大值
index = np.where(tem_lie==np.max(tem_lie[i:]))[0][0]#注意消去后的新索引
if index !=i:
change_index +=1
print('交换次序%s行与%s行：'%(i,index))
A[i,:],A[index,:]=A[index,:].copy(),A[i,:].copy()
print(A)
#消去部分：
for j in range(i+1,n):
tem = A[i].copy()/A[i,i]
A[j]=tem*A[j,i]-A[j]
print('消去结果：\n',A)
#回带求解
A,b = A[:,:n],A[:,-1]
# print(A,b)
print('高斯消去后行列式|A|', np.linalg.det(A) * ((-1) ** change_index))
x = up_solve(A,b.reshape(n,1))
print('求解x为：',x)
return xLU矩阵分解及其求解
这里借鉴过其他博主的思路，先求得分解的的LU矩阵，再利用下三角求解获取y,再通过下三角求解获取X
#Lu及其求解
def LU_decomposition(A):
n = len(A[0])
L = np.zeros([n, n])
U = np.zeros([n, n])
for i in range(n):
L[i][i] = 1
if i == 0:
U[0][0] = A[0][0]
for j in range(1, n):
U[0][j] = A[0][j]
L[j][0] = A[j][0] / U[0][0]
else:
for j in range(i, n): # U
temp = 0
for k in range(0, i):
temp = temp + L[i][k] * U[k][j]
U[i][j] = A[i][j] - temp
for j in range(i + 1, n): # L
temp = 0
for k in range(0, i):
temp = temp + L[j][k] * U[k][i]
L[j][i] = (A[j][i] - temp) / U[i][i]
return L, U
# 计算线性方程组的解
def answer(matrix_l,matrix_u,matrix_an):#LU以及B矩阵
y = down_solve(matrix_l,matrix_an)
print('y的值为：',y)
x = up_solve(matrix_u,y.reshape(len(y),1))
print('x的解为：',x)
return x
雅克比迭代法
通过对谱半径求解，判断改迭代矩阵是否发散，再利用范数判断是否退出迭代：输出迭代次数，以及迭代的解值，初值全部默认设为0，误差默认为1e-6有需要可以自己设置初值，与上文一样，A,b都是二维数组
#Jacobian雅克比迭代
def jacobian(A,b,x0=None,mistake=1e-6):
if x0 ==None:
x0 = [[0] for i in range(len(A[0]))]
L = (-1)*np.tril(A,-1)
D =np.diag(np.diag(A))
U = (-1)*np.triu(A,1)
B = np.dot(np.linalg.inv(D),(L+U))
b = np.dot(np.linalg.inv(D),b)
t_a,t_b = np.linalg.eig(B)
if max(abs(t_a)) >1:
print('改矩阵发散')
return
k = 1
while True:
x1 = np.dot(B,x0)+b
print('第%s次迭代'%k,x1)
norm = max(abs(x1-x0))
if norm 高斯塞得迭代法
与上文类似，更改迭代矩阵，如下：
#高斯迭代
def gauss_seidel(A,b,x0=None,mistake=1e-6):
if x0 == None:
x0 = [[0] for i in range(len(A[0]))]
L = (-1) * np.tril(A, -1)
D = np.diag(np.diag(A))
U = (-1) * np.triu(A, 1)
B = np.dot(np.linalg.inv(D-L),U)
b = np.dot(np.linalg.inv(D-L), b)
t_a, t_b = np.linalg.eig(B)
if max(abs(t_a)) > 1:
print('改矩阵发散')
return
k = 1
while True:
x1 = np.dot(B, x0) + b
print('第%s次迭代' % k, x1)
norm = max(abs(x1 - x0))
# print(norm)#这个莫可以用来判断下降速度
if norm < mistake:
break
x0 = x1
k += 1松弛法
这里多出了一个松弛因子w,w=1时就是高斯塞得迭代，这里w为必须参数
def sor(A,b,w,x0=None,mistake=1e-6):
if x0 == None:
x0 = [[0] for i in range(len(A[0]))]
L = (-1) * np.tril(A, -1)
D = np.diag(np.diag(A))
U = (-1) * np.triu(A, 1)
B = np.dot(np.linalg.inv(D-w*L),(1-w)*D+w*U)
b = np.dot(np.linalg.inv(D-w*L), b)*w
t_a, t_b = np.linalg.eig(B)
if max(abs(t_a)) > 1:
print('改矩阵发散')
return
k = 1
while True:
x1 = np.dot(B, x0) + b
print('第%s次迭代' % k, x1)
norm = max(abs(x1 - x0))
# print(norm)
if norm < mistake:
break
x0 = x1
k += 1源码
这里给出上面全部源码以及使用案列
import numpy as np
#上三角求解：
def up_solve(A,b):
n = len(A[0])
A[n-1,n-1] = b[n-1][0]/A[n-1,n-1]
for i in range(1,n):
k = n-i-1#取逆序
A[k,k] = (b[k][0] - sum(A[k]*A[k+1]))/A[k,k]
A[k]=np.hstack((A[k,:k+1],A[k+1,k+1:]))#拼接新数组
# print('方程解为',A[0])
return A[0]
#下三角求解
def down_solve(A,b):
n = len(A[0])
A[0,0] = b[0][0] /A[0,0]
for i in range(1, n):
k = i # 取逆序
A[k, k] = (b[k][0] - sum(A[k] * A[k-1])) / A[k, k]
A[k] = np.hstack((A[k-1, :k], A[k, k:])) # 拼接新数组
# print('方程解为', A[-1])
return A[-1]
#列主元高斯消去
def gaussian(A,b):
#最大值索引
n = len(A[0])
A = np.hstack([A,b])
change_index=0
for i in range(n):
tem_lie = abs(A[:,i])#列主元最大值
index = np.where(tem_lie==np.max(tem_lie[i:]))[0][0]#注意消去后的新索引
if index !=i:
change_index +=1
print('交换次序%s行与%s行：'%(i,index))
A[i,:],A[index,:]=A[index,:].copy(),A[i,:].copy()
print(A)
#消去部分：
for j in range(i+1,n):
tem = A[i].copy()/A[i,i]
A[j]=tem*A[j,i]-A[j]
print('消去结果：\n',A)
#回带求解
A,b = A[:,:n],A[:,-1]
# print(A,b)
print('高斯消去后行列式|A|', np.linalg.det(A) * ((-1) ** change_index))
x = up_solve(A,b.reshape(n,1))
print('求解x为：',x)
return x
#Lu及其求解
def LU_decomposition(A):
n = len(A[0])
L = np.zeros([n, n])
U = np.zeros([n, n])
for i in range(n):
L[i][i] = 1
if i == 0:
U[0][0] = A[0][0]
for j in range(1, n):
U[0][j] = A[0][j]
L[j][0] = A[j][0] / U[0][0]
else:
for j in range(i, n): # U
temp = 0
for k in range(0, i):
temp = temp + L[i][k] * U[k][j]
U[i][j] = A[i][j] - temp
for j in range(i + 1, n): # L
temp = 0
for k in range(0, i):
temp = temp + L[j][k] * U[k][i]
L[j][i] = (A[j][i] - temp) / U[i][i]
return L, U
# 计算线性方程组的解
def answer(matrix_l,matrix_u,matrix_an):#LU以及B矩阵
y = down_solve(matrix_l,matrix_an)
print('y的值为：',y)
x = up_solve(matrix_u,y.reshape(len(y),1))
print('x的解为：',x)
return x
#Jacobian雅克比迭代
def jacobian(A,b,x0=None,mistake=1e-6):
if x0 ==None:
x0 = [[0] for i in range(len(A[0]))]
L = (-1)*np.tril(A,-1)
D =np.diag(np.diag(A))
U = (-1)*np.triu(A,1)
B = np.dot(np.linalg.inv(D),(L+U))
b = np.dot(np.linalg.inv(D),b)
t_a,t_b = np.linalg.eig(B)
if max(abs(t_a)) >1:
print('改矩阵发散')
return
k = 1
while True:
x1 = np.dot(B,x0)+b
print('第%s次迭代'%k,x1)
norm = max(abs(x1-x0))
if norm 1:
print('改矩阵发散')
return
k = 1
while True:
x1 = np.dot(B, x0) + b
print('第%s次迭代' % k, x1)
norm = max(abs(x1 - x0))
# print(norm)#这个莫可以用来判断下降速度
if norm < mistake:
break
x0 = x1
k += 1
def sor(A,b,w,x0=None,mistake=1e-6):
if x0 == None:
x0 = [[0] for i in range(len(A[0]))]
L = (-1) * np.tril(A, -1)
D = np.diag(np.diag(A))
U = (-1) * np.triu(A, 1)
B = np.dot(np.linalg.inv(D-w*L),(1-w)*D+w*U)
b = np.dot(np.linalg.inv(D-w*L), b)*w
t_a, t_b = np.linalg.eig(B)
if max(abs(t_a)) > 1:
print('改矩阵发散')
return
k = 1
while True:
x1 = np.dot(B, x0) + b
print('第%s次迭代' % k, x1)
norm = max(abs(x1 - x0))
# print(norm)
if norm < mistake:
break
x0 = x1
k += 1
if __name__ == '__main__':
#系数矩阵
print('实验一(1)\n')
matrix_a = np.array([3.01,6.03,1.99,1.27,4.16,-1.23,0.987,-4.81,9.34],dtype=float).reshape(3,3)
matrix_b = np.array([1,1,1], dtype=float).reshape(3, 1)
print('原增广矩阵为:')
print(np.hstack([matrix_a,matrix_b]))
A=matrix_a
L, U = LU_decomposition(A)
print('矩阵L为：\n',L, '\n矩阵U为\n', U)
x = answer(L,U,matrix_b)
print('\n高斯求解：')
gaussian(matrix_a,matrix_b)
#这里为矩阵迭代
#教材案列
matrix_a = np.array([10,-2,-1,-2,10,-1,-1,-2,5],dtype=float).reshape(3,3)
matrix_b = np.array([3,15,10], dtype=float).reshape(3, 1)
jacobian(matrix_a,matrix_b)#雅克比迭代
gauss_seidel(matrix_a,matrix_b)#高斯迭代
sor(matrix_a,matrix_b,w=0.9)#松弛法迭代w=0.9