LU Decomposition

04/20/23http://

numericalmethods.eng.usf.edu 1

LU Decomposition

Civil Engineering Majors

Authors: Autar Kaw

http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM

Undergraduates

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LU Decomposition

http://numericalmethods.eng.usf.edu




LU DecompositionLU Decomposition is another method to solve a set of

simultaneous linear equations

Which is better, Gauss Elimination or LU Decomposition?

To answer this, a closer look at LU decomposition is needed.

MethodFor most non-singular matrix [A] that one could conduct Naïve Gauss Elimination forward elimination steps, one can always write it as

[A] = [L][U]

where

[L] = lower triangular matrix

[U] = upper triangular matrix


LU Decomposition


How does LU Decomposition work?


LU DecompositionHow can this be used?

Given [A][X] = [C]

1. Decompose [A] into [L] and [U]

2. Solve [L][Z] = [C] for [Z]

3. Solve [U][X] = [Z] for [X]


When is LU Decomposition better than Gaussian Elimination?

To solve [A][X] = [B]

Table. Time taken by methods

where T = clock cycle time and n = size of the matrix

So both methods are equally efficient.

Gaussian Elimination LU Decomposition

3

412

3

8 23 n

nn

T

3

412

3

8 23 n

nn

T


To find inverse of [A]

Time taken by Gaussian Elimination Time taken by LU Decomposition

3

412

3

8

||2

34 n

nn

T

CTCTn BSFE

3

2012

3

32

|||

23 n

nn

T

CTnCTnCT BSFSLU

n 10 100 1000 10000

CT|inverse GE / CT|inverse LU 3.28 25.83 250.8 2501

Table 1 Comparing computational times of finding inverse of a matrix using LU decomposition and Gaussian elimination.


Method: [A] Decompose to [L] and [U]

33

2322

131211

3231

21

00

0

1

01

001

u

uu

uuu

ULA

[U] is the same as the coefficient matrix at the end of the forward elimination step.

[L] is obtained using the multipliers that were used in the forward elimination process


Finding the [U] matrixUsing the Forward Elimination Procedure of Gauss Elimination

112144

1864

1525

112144

56.18.40

1525

56.212;56.225

64

RowRow

76.48.160

56.18.40

1525

76.513;76.525

144

RowRow

Step 1:


Finding the [U] Matrix

Step 2:

76.48.160

56.18.40

1525

7.000

56.18.40

1525

5.323;5.38.4

8.16

RowRow

7.000

56.18.40

1525

U

Matrix after Step 1:


Finding the [L] matrix

Using the multipliers used during the Forward Elimination Procedure

1

01

001

3231

21

56.225

64

11

2121

a

a

76.525

144

11

3131

a

a

From the first step of forward elimination

112144

1864

1525


Finding the [L] Matrix

15.376.5

0156.2

001

L

From the second step of forward elimination

76.48.160

56.18.40

15255.3

8.4

8.16

22

3232

a

a


Does [L][U] = [A]?

7.000

56.18.40

1525

15.376.5

0156.2

001

UL ?

Example: Cylinder Stresses

To find the maximum stresses in a compound cylinder, the following four simultaneous linear equations need to solved.

0

0070

0

108877

1060573102857400

1538405615384056

1046195102857410461951028574

0010230791028574 3

4

3

2

1

57

5757

57

.

.

c

c

c

c

..

....

....

..


In the compound cylinder, the inner cylinder has an internal radius of a = 5”, and outer radius c = 6.5”, while the outer cylinder has an internal radius of c = 6.5” and outer radius, b=8”. Given E = 30×106 psi, ν = 0.3, and that the hoop

stress in outer cylinder is given by

2432

11

1 rcc

E

find the stress on the inside radius of the outer cylinder. Find the values of c1, c2, c3 and c4 using LU decomposition.


Use Forward Elimination Procedure of Gauss Elimination to find [U]

Step 1

57

5757

57

106057.3102857.400

15384.05.615384.05.6

104619.5102857.4104619.5102857.4

00102307.9102857.4

112;11028574

10285747

7

RowRow.

.

57

575

57

106057.3102857.400

15384.05.615384.05.6

104619.5102857.4107688.30

00102307.9102857.4


777

105167.113;105167.11028574

56RowRow

.

.

57

575

57

106057.3102857.400

15384.05.629384.00

104619.5102857.4107688.30

00102307.9102857.4

Step 1 cont.


014;0102857.4

07

RowRow

57

575

57

106057.3102857.400

15384.05.629384.00

104619.5102857.4107688.30

00102307.9102857.4

Step 1 cont.

This is the matrix after Step 1.


Step 2

775

107966.723;107966.71076883

293840RowRow

.

.

57

575

57

106057.3102857.400

57968.0914.2600

104619.5102857.4107688.30

00102307.9102857.4


024;0107688.3

05

RowRow

57

575

57

106057.3102857.400

57968.0914.2600

104619.5102857.4107688.30

00102307.9102857.4

Step 2 cont.



Step 3

667

105923.134;105294.191426

1028574RowRow

.

.

5

575

57

106250.5000

57968.0914.2600

104619.5102857.4107688.30

00102307.9102857.4



5

575

57

106250.5000

57968.0914.2600

104619.5102857.4107688.30

00102307.9102857.4

][U


Use the multipliers from Forward Elimination to find [L]

From the 1st step of forward elimination

1

01

001

0001

434241

3231

21

0102857.4

0

105167.1102857.4

5.6

1102857.4

102857.4

711

4141

77

11

3131

7

7

11

2121

a

a

a

a

a

a

57

5757

57

106057.3102857.400

15384.05.615384.05.6

104619.5102857.4104619.5102857.4

00102307.9102857.4


From the 2nd step of forward elimination

1

01

001

0001

434241

3231

21

0107688.3

0

107966.7107688.3

29384.0

522

4242

75

22

3232

a

a

a

a

57

575

57

106057.3102857.400

15384.05.629384.00

104619.5102857.4107688.30

00102307.9102857.4


From the 3rd step of forward elimination

1

01

001

0001

434241

3231

21

67

33

4343 105294.1

914.26

102857.4

a

a

57

575

57

106057.3102857.400

57968.0914.2600

104619.5102857.4107688.30

00102307.9102857.4


1105924.100

01107966.7105167.1

0011

0001

6

77L


Does [L][U] = [A]?

?

106250.5000

57968.0914.2600

104619.5102857.4107688.30

00102307.9102857.4

1105924.100

01107966.7105167.1

0011

0001

5

575

57

6

77

UL

0zz105924.1

007.0zz1079662.7z1051667.1

0zz

10887.7z

436

327

17

21

31


Set [L][Z] = [C]

0

007.0

0

10887.7

1105924.100

01107966.7105167.1

0011

0001 3

4

3

2

1

6

77

z

z

z

z

Solve for [Z]


Solve for [Z] 3

1 108877 .z

3

3

12

108877

108877

.

.

zz

2

3737

27

17

3

1019531

108877107966710887710516710070

107966710516710070

.

.....

z.z..z


Solving for [Z] cont.

19034

10195311059241

1059241

26

36

4

..

z.z

19034

101953.1

10887.7

10887.7

z

z

z

z

Z2

3

3

4

3

2

1


Set [U][C] = [Z]

19034

1019531

108877

108877

1062505000

5796809142600

1046195102857410768830

0010230791028574

2

3

3

4

3

2

1

5

575

57

.

.

.

c

c

c

c

.

..

...

..

190341062505

101953157968091426

108877104619510285741076683

1088770010230791028574

45

243

34

53

72

5

3432

51

7

c.

.c.c.

.c.c.c.

.ccc.c.

The four equations become:


Solve for [C]

2

54

1038373

1062505

19034

.

.c

4

22

42

42

3

108469.2

914.26

103837.357968.0101953.1

914.26

57968.0101953.1

91426

5796801019531

c

.

c..c


3

5

25473

54

53

73

2

1026154

1076883

1038373104619510846921028574108877

1076883

10461951028574108877

.

.

.....

.

c.c..c

5

7

353

7432

53

1

1022449

1028574

10261541023079108877

1028574

001023079108877

.

.

...

.

c-cc..c

Solve for [C] cont.


Solution:

The solution vector is

2

4

3

5

4

3

2

1

1038373

1084692

1026154

1022449

.

.

.

.

c

c

c

c

The stress on the inside radius of the outer cylinder is then given by

psi30683

5.6

3.01103837.33.01108469.2

3.01

1030

11

1

224

2

6

2432

rcc

E


Finding the inverse of a square matrix

The inverse [B] of a square matrix [A] is defined as

[A][B] = [I] = [B][A]


Finding the inverse of a square matrixHow can LU Decomposition be used to find the inverse?

Assume the first column of [B] to be [b11 b12 … bn1]T

Using this and the definition of matrix multiplication

First column of [B] Second column of [B]

0

0

1

1

21

11

nb

b

b

A

0

1

0

2

22

12

nb

b

b

A

The remaining columns in [B] can be found in the same manner


Example: Inverse of a MatrixFind the inverse of a square matrix [A]

112144

1864

1525

A

7000

561840

1525

153765

01562

001

.

..

..

.ULA

Using the decomposition procedure, the [L] and [U] matrices are found to be


Example: Inverse of a MatrixSolving for the each column of [B] requires two steps

1)Solve [L] [Z] = [C] for [Z]

2)Solve [U] [X] = [Z] for [X]

Step 1:

0

0

1

15.376.5

0156.2

001

3

2

1

z

z

z

CZL

This generates the equations:

05.376.5

056.2

1

321

21

1

zzz

zz

z


Example: Inverse of a MatrixSolving for [Z]

23

5625317650

537650

562

15620

5620

1

213

12

1

.

...

z.z.z

.

.

z. z

z

23

562

1

3

2

1

.

.

z

z

z

Z


Example: Inverse of a Matrix

Solving [U][X] = [Z] for [X]

3.2

2.56

1

7.000

56.18.40

1525

31

21

11

b

b

b

2.37.0

56.256.18.4

1525

31

3121

312111

b

bb

bbb


Example: Inverse of a MatrixUsing Backward Substitution

04762.0

25

571.49524.05125

51

9524.08.4

571.4560.156.28.4

560.156.2

571.47.0

2.3

312111

3121

31

bbb

bb

b So the first column of the inverse of [A] is:

571.4

9524.0

04762.0

31

21

11

b

b

b


Example: Inverse of a MatrixRepeating for the second and third columns of the inverse

Second Column Third Column

0

1

0

112144

1864

1525

32

22

12

b

b

b

000.5

417.1

08333.0

32

22

12

b

b

b

1

0

0

112144

1864

1525

33

23

13

b

b

b

429.1

4643.0

03571.0

33

23

13

b

b

b


Example: Inverse of a Matrix

The inverse of [A] is

429.1000.5571.4

4643.0417.19524.0

03571.008333.004762.01A

To check your work do the following operation

[A][A]-1 = [I] = [A]-1[A]

Additional ResourcesFor all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit http://numericalmethods.eng.usf.edu/topics/lu_decomposition.html

http://numericalmethods.eng.usf.edu/topics/gaussian_elimination.html

http://numericalmethods.eng.usf.edu/topics/lu_decomposition.html

http://numericalmethods.eng.usf.edu/topics/lu_decomposition.html

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