基础方法没啥好叙述的，书上随处可见，给出了具体的代码实现以及注释，错误之处请留言。谢谢

Mathematical expression of gaussian elimination

elimination-step and get a upper triangular matrix

for k=0 to n-1

mki=a(k)ika(k)kk(i=k+1,…,n−1)

m_i^k= \frac {a_{ik}^{(k)}} {a_{kk}^{(k)}} (i=k+1,\dots,n-1)

ak+1ij=aij(k)−mk∗a(k)kj(i,j=k+1,…,n−1)

a_{ij}^{k+1}=a_{ij}{(k)}-m_k*a_{kj}^{(k)} (i,j=k+1,\dots,n-1)

bk+1i=bi(k)−mik∗b(k)k(i=k+1,…,n−1)

b_{i}^{k+1}=b_{i}{(k)}-m_{ik}*b_k^{(k)} (i=k+1,\dots,n-1)

back-substitution-step

xn−1=b(n−1)n−1an−1(n−1)(n−1)

x_{n-1}=\frac{b_{n-1}^{(n-1)}}{a_{(n-1)(n-1)}^{n-1}}
for i=n-2 to 0

xi=b(n)i−∑n−1j=i+1a(n)ijxjanii

x_i=\frac{b_i^{(n)}-\sum_{j=i+1}^{n-1}a_{ij}^{(n)}x_j} {a_{ii}^{n}}

Mathematical expression of LU decomposition

1.implement A=LU

u1i=a1i(i=1,2,…,n)

u_{1i}=a_{1i} (i=1,2,\dots,n)

li1=ai1u11(i=1,2,…,n)

l_{i1}=\frac{a_{i1}}{u_{11}} (i=1,2,\dots,n)
for k=1 to n-1

uki=aki−∑j=0k−1lkjuji(i=k,k+1,…,n−1)

u_{ki}=a_{ki}-\sum_{j=0}^{k-1}l_{kj}u_{ji} (i=k,k+1,\dots,n-1)

lik=aik−∑k−1j=0lijujkukk(i=k+1,…,n−1,k!=n)

l_{ik}=\frac{a_{ik}-\sum_{j=0}^{k-1}l_{ij}u_{jk}}{u_kk} (i=k+1,\dots,n-1,k!=n)

2.计算LY=b

y1=b1

y1=b1
for k=1 to n-1

yk=bk−∑j=0k−1lkjyj(k=2,3,…,n−1)

y_k=b_k- \sum_{j=0}^{k-1}l_{kj}y_j (k=2,3,\dots,n-1)

计算UX=Y

xn−1=bn−1u(n−1)(n−1)

x_{n-1}=\frac{b_{n-1}}{u_{(n-1)(n-1)}}
for k= n-2 to 0

xk=yk−∑n−1j=k+1ukjxjukk

x_k=\frac{y_k-\sum_{j=k+1}^{n-1}u_{kj}x_j} {u_{kk}}

import numpy as np  #a package to calculate
from scipy.sparse import identity# to get a identity matrix
import time#calculate time of diffient method

create a random matrix M with dimension 100*100 ,generate A by adding an identify matrix

np.random.seed(1)   #set seed to make s unchanged
s=np.random.rand(100,100)
A=s+identity(100).toarray()
#add a identity to make sure has a inverse matrix
print A
"""[[  1.41702200e+00   7.20324493e-01   1.14374817e-04 ...,   5.73679487e-012.87032703e-03   6.17144914e-01][  3.26644902e-01   1.52705810e+00   8.85942099e-01 ...,   2.34362086e-016.16778357e-01   9.49016321e-01][  9.50176119e-01   5.56653188e-01   1.91560635e+00 ...,   7.96260777e-029.82817114e-01   1.81612851e-01]..., [  2.47870273e-01   1.11841381e-01   5.13540776e-01 ...,   1.83527618e+002.39285522e-01   7.30797255e-02][  9.52966404e-01   1.12326974e-01   8.02396496e-01 ...,   4.95854311e-011.50837092e+00   8.47333803e-02][  4.37268153e-01   8.63201246e-01   6.80236396e-01 ...,   5.95336949e-021.08043656e-01   1.76279378e+00]]"""

gengerate a vector x=(1,2,3,+⋯+100)Tx=(1,2,3,+ \cdots + 100)^T

x=np.array(range(1,101)).T

generate a vector b as b=AXb=AX

b=np.dot(A,x)
#get result by calling function ,also you can calculate by yourself
#for i in range(100):
#    b[i]=0
#    for j in range(100):
#        b[i]+=A[i][j]*x[j][0]

calculate x

#Gauss-Jordan elimination
def Gauss(n,A,b):#elimination-step and get a upper triangular matrixfor k in range(0,n-1):for i in range(k+1,n):m=A[i][k]/A[k][k]#calculate multiplierfor j in range(k,n):A[i][j]=A[i][j]-m*A[k][j] #elimination of ith rowb[i]=b[i]-m*b[k]"""back-substitution-step,easy to implement according to arithmetic expression"""newx=np.zeros((n,1))newx[n-1]=(b[n-1]/A[n-1][n-1])i=n-2while(i>=0):sum2=0for j in range(i+1,n):sum2+=A[i][j]*newx[j]newx[i]=(b[i]-sum2)/A[i][i]i-=1return newx"""this method is based on Gauss,just exchange kth row with maxkth
row before calculate m to avoid some mistakes caused by float"""
def Gauss_exchange(n,A,b):for k in range(0,n-1):maxk=kmaxA=A[k][k]for t in range(k+1,n):if(A[t][k]>maxA):maxk=tmaxA=A[t][k]for j in range(k,n):tmp=A[k][j]A[k][j]=A[maxk][j]A[maxk][j]=tmptmp=b[k]    b[k]=b[maxk]b[maxk]=tmp #exchange maxk with kfor i in range(k+1,n):m=A[i][k]/A[k][k]for j in range(k,n):A[i][j]=A[i][j]-m*A[k][j]b[i]=b[i]-m*b[k]#same as Gaussnewx=np.zeros((n,1))newx[n-1]=(b[n-1]/A[n-1][n-1])i=n-2while(i>=0):sum1=0for j in range(i+1,n):sum1+=A[i][j]*newx[j]newx[i]=(b[i]-sum1)/A[i][i]i-=1return  newx
def dolittle(n,A,b):L=np.zeros(A.shape,dtype="float")+identity(100).toarray()U=np.zeros(A.shape,dtype="float")#implement A=LUfor i in range(0,n):U[0][i]=A[0][i]for i in range(0,n):L[i][0]=A[i][0]/U[0][0]for k in range(1,n):for i in range(k,n):sum3=0for j in range(0,k):sum3+=L[k][j]*U[j][i]U[k][i]=A[k][i]-sum3for i in range(k+1,n):sum3=0for j in range(0,k):sum3=sum3+L[i][j]*U[j][k]L[i][k]=(A[i][k]-sum3)/U[k][k]#calculate LY=bY=np.zeros((n,1))Y[0]=b[0]for k in range(1,n):sum3=0for j in range(0,k):sum3+=L[k][j]*Y[j]Y[k]=b[k]-sum3#calculate UX=Ynewx=np.zeros((n,1))newx[n-1]=(Y[n-1]/U[n-1][n-1])i=n-2while(i>=0):sum3=0for j in range(i+1,n):sum3+=U[i][j]*newx[j]newx[i]=(Y[i]-sum3)/U[i][i]i-=1return newx"""copy() function is to make sure  do not change  A
and pass correct parameters in follow functions"""
t0=time.time()  #start time
newx2=Gauss(100,A.copy(),b.copy())
print "the time of Gaussian elimination is ",time.time()-t0
t0=time.time()
newx1=Gauss_exchange(100,A.copy(),b.copy())
print "the time of column principal element elimination is ",time.time()-t0
t0=time.time()
newx3=dolittle(100,A.copy(),b.copy())
print "the time of LU decomposition method is ",time.time()-t0
print "first ",newx1.T, " second",newx2.T,"\n third",newx3.Tthe time of Gaussian elimination is  0.805999994278
the time of column principal element elimination is  0.884999990463
the time of LU decomposition method is  0.444000005722""" first  [[  1.   2.   3.   4.   5.   6.   7.   8.   9.  10.  11.  12.  13.  14.15.  16.  17.  18.  19.  20.  21.  22.  23.  24.  25.  26.  27.  28.29.  30.  31.  32.  33.  34.  35.  36.  37.  38.  39.  40.  41.  42.43.  44.  45.  46.  47.  48.  49.  50.  51.  52.  53.  54.  55.  56.57.  58.  59.  60.  61.  62.  63.  64.  65.  66.  67.  68.  69.  70.71.  72.  73.  74.  75.  76.  77.  78.  79.  80.  81.  82.  83.  84.85.  86.  87.  88.  89.  90.  91.  92.  93.  94.  95.  96.  97.  98.99. 100.]] second [[  1.   2.   3.   4.   5.   6.   7.   8.   9.  10.  11.  12.  13.  14.15.  16.  17.  18.  19.  20.  21.  22.  23.  24.  25.  26.  27.  28.29.  30.  31.  32.  33.  34.  35.  36.  37.  38.  39.  40.  41.  42.43.  44.  45.  46.  47.  48.  49.  50.  51.  52.  53.  54.  55.  56.57.  58.  59.  60.  61.  62.  63.  64.  65.  66.  67.  68.  69.  70.71.  72.  73.  74.  75.  76.  77.  78.  79.  80.  81.  82.  83.  84.85.  86.  87.  88.  89.  90.  91.  92.  93.  94.  95.  96.  97.  98.99. 100.]] third [[  1.   2.   3.   4.   5.   6.   7.   8.   9.  10.  11.  12.  13.  14.15.  16.  17.  18.  19.  20.  21.  22.  23.  24.  25.  26.  27.  28.29.  30.  31.  32.  33.  34.  35.  36.  37.  38.  39.  40.  41.  42.43.  44.  45.  46.  47.  48.  49.  50.  51.  52.  53.  54.  55.  56.57.  58.  59.  60.  61.  62.  63.  64.  65.  66.  67.  68.  69.  70.71.  72.  73.  74.  75.  76.  77.  78.  79.  80.  81.  82.  83.  84.85.  86.  87.  88.  89.  90.  91.  92.  93.  94.  95.  96.  97.  98.99. 100.]]

conclusion:the data is small ,It’s hard to compare performance

                                                    author: cbf17.10.3

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