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A.Kaw矩阵代数初步学习笔记 7. LU Decomposition

“矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授。
PDF格式学习笔记下载(Academia.edu)
第7章课程讲义下载(PDF)

Summary

  • For a nonsingular matrix [A] on which one can always write it as [A]=[L][U] where [L] is a lower triangular matrix, [U] is a upper triangular matrix.
  • Note that not all matrices have LU decomposition, such as [0220]. [0220]=[10a1][bc0d]{b=0ab=2 which is contradiction.
  • If one is solving a set of equations [A][X]=[B] then LUX=B L1LUX=L1B UX=L1B=Y then we have {LY=BUX=Y So we can solve the first equation for [Y]by using forward substitution and then use the second equation to calculate the solution vector [X] by back substitution.
  • For instance, solve the following set of equations: [123214152][xyz]=[14817] Applying LU decomposition on the coefficient matrix,
    • Firstly write down an identity matrix (the same size as the coefficient matrix) on the left and the coefficient matrix on the right. L[100010001][123214152]U
    • Then applying elementary row operation on the right while simultaneously updating successive columns of the matrix on the left. For example, if we are doing R1+mR2 on the right then we will do C2mC1 on the left. That is, we will keep the equivalent of the product. [100010001][123214152] {R22R1C1+2C2[100210001][1230310152] {R3R1C1+C3[100210101][1230310031] {R3+R2C2C3[100210111][12303100011] Thus far, the right matrix is an upper triangular matrix (i.e. U) and the left one is a lower triangular matrix (i.e. L).
    • Solving [L][Y]=[B], that is [100210111]Y=[14817]Y=[143633]
    • Solving [U][X]=[Y], that is [12303100011][xyz]=[143633] {x=1y=2z=3

Selected Problems

1. Find the [L] and [U] matrices of the following matrix [25547571612.51222]

Solution: [100010001][25547571612.51222] {R23R1R312R1C1+3C2C1+12C3[1003101201][255408409.520] {R3+1916R2C21916C3[1003101219161][255408400994] That is, L=[1003101219161], U=[255408400994].

2. Using LU decomposition to solve: {4x1+x2x3=25x1+x2+2x3=46x1+x2+x3=6

Solution: [100010001][411512611] {R254R1R332R1C1+54C2C1+32C3[10054103201][41101413401252] {R32R2C2+2C3[10054103221][411014134004] That is, L=[10054103221], U=[411014134004]. Then we solve [L][Y]=[B], [10054103221]Y=[246]Y=[21324] Finally, we solve [U][X]=[Y], [411014134004]X=[21324]X=[3131] Thus the solution is {x1=3x2=13x3=1

3. Find the inverse of [A]=[341271815]

Solution:

To find the inverse of a matrix, actually it is to solve a set of equations: {AX1=[1,0,0]TAX2=[0,1,0]TAX3=[0,0,1]T Firstly, we will find the [L] and [U]. [100010001][341271815] {R223R1R383R1C1+23C2C1+83C3[10023108301][341029353029373] {R3R2C2+C3[10023108311][341029353004] That is, L=[10023108311], U=[341029353004]. Then we solve [L][Y]=[I], note that there are three columns of [Y]: LY1=[10023108311]Y1=[100]Y1=[1,23,2]T LY2=[10023108311]Y2=[010]Y2=[0,1,1]T LY3=[10023108311]Y3=[001]Y3=[0,0,1]T Finally we can solve [X] by [U][X]=[Y]: UX1=Y1[341029353004]X1=[1232]X1=[1758,958,12]T UX2=Y2[341029353004]X2=[011]X2=[19116,7116,14]T UX3=Y3[341029353004]X3=[001]X3=[3116,5116,14]T Thus the inverse of the original matrix is [A]1=[175819116311695871165116121414]

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