Section Partitioned Matrices and LU Factorization

Transcription

1 Section Partitioned Matrices and LU Factorization Gexin Yu College of William and Mary

2 partition matrices into blocks In real world problems, systems can have huge numbers of equations and un-knowns. Standard computation techniques are inefficient in such cases, so we need to develop techniques which exploit the internal structure of the matrices. In most cases, the matrices of interest have lots of zeros.

3 partition matrices into blocks In real world problems, systems can have huge numbers of equations and un-knowns. Standard computation techniques are inefficient in such cases, so we need to develop techniques which exploit the internal structure of the matrices. In most cases, the matrices of interest have lots of zeros. One approach to simplify the computation is to partition a matrix into blocks.

4 partition matrices into blocks In real world problems, systems can have huge numbers of equations and un-knowns. Standard computation techniques are inefficient in such cases, so we need to develop techniques which exploit the internal structure of the matrices. In most cases, the matrices of interest have lots of zeros. One approach to simplify the computation is to partition a matrix into blocks Ex: A =

5 partition matrices into blocks In real world problems, systems can have huge numbers of equations and un-knowns. Standard computation techniques are inefficient in such cases, so we need to develop techniques which exploit the internal structure of the matrices. In most cases, the matrices of interest have lots of zeros. One approach to simplify the computation is to partition a matrix into blocks Ex: A = This partition can also be written as the following 2 3 block matrix: [ ] A11 A A = 12 A 13 A 21 A 22 A 23

6 partition matrices into blocks In real world problems, systems can have huge numbers of equations and un-knowns. Standard computation techniques are inefficient in such cases, so we need to develop techniques which exploit the internal structure of the matrices. In most cases, the matrices of interest have lots of zeros. One approach to simplify the computation is to partition a matrix into blocks Ex: A = This partition can also be written as the following 2 3 block matrix: [ ] A11 A A = 12 A 13 A 21 A 22 A 23 In the block form, we have blocks A 11 = [ 3 0 ] and so on.

7 Operations on partitioned matrices (Addition and scalar multiplication) IF matrices A and B are the same size and are partitioned in exactly the same way, namely A = (A ij ) and B = (B ij ), then A + B = (A ij + B ij ), and ra = (ra ij )

8 Operations on partitioned matrices (Addition and scalar multiplication) IF matrices A and B are the same size and are partitioned in exactly the same way, namely A = (A ij ) and B = (B ij ), then A + B = (A ij + B ij ), and ra = (ra ij ) (multiplication) If the partitions of A and B are comfortable for block multiplication, namely, the column partition of A matches the row partition of B, then if A = (A ij ) m n and B = (B ij ) n p, then AB = (C ij ) m p, where C ij = A i1 B ij + A i2 B 2j A in B nj

9 Example Find AB, where A = [ A11 A = 12 A 21 A 22 ], B = = 5 2 [ B1 B 2 ]

10 Example Find AB, where A = [ A11 A = 12 A 21 A 22 ], B = = 5 2 [ B1 B 2 ] Soln: [ ] [ ] [ ] A11 A AB = 12 B1 A11 B = 1 + A 12 B 2 A 21 A 22 B 2 A 21 B 1 + A 22 B =

11 Example [ ] [ ] A O C I Compute, where A, B, C and D are n n, O is the I B O D n n zero matrix, and I is the n n identity matrix.

12 Example Soln: [ ] [ ] A O C I Compute, where A, B, C and D are n n, O is the I B O D n n zero matrix, and I is the n n identity matrix. [ ] [ ] [ ] [ ] A O C I AC AI AC A = = I B O D IC II + BD C I + BD

13 Example [ ] [ ] A O C I Compute, where A, B, C and D are n n, O is the I B O D n n zero matrix, and I is the n n identity matrix. [ ] [ ] [ ] [ ] A O C I AC AI AC A Soln: = = I B O D IC II + BD C I + BD If we do direct multiplication: to compute each term uses 2n multiplications and 2n additions, or 4n flops (floating point operations). Since there are 4n 2 terms in the resulting matrix, direct multiplication uses 4n 4n 2 = 16n 3 flops.

14 Example [ ] [ ] A O C I Compute, where A, B, C and D are n n, O is the I B O D n n zero matrix, and I is the n n identity matrix. [ ] [ ] [ ] [ ] A O C I AC AI AC A Soln: = = I B O D IC II + BD C I + BD If we do direct multiplication: to compute each term uses 2n multiplications and 2n additions, or 4n flops (floating point operations). Since there are 4n 2 terms in the resulting matrix, direct multiplication uses 4n 4n 2 = 16n 3 flops. In the above computation, to compute AC takes 2n n 2 flops, to compute BD takes 2n n 2 flops, to add I + BD takes n flops, so partitioned multiplication uses 4n 3 + n? flops.

15 Example [ ] [ ] A O C I Compute, where A, B, C and D are n n, O is the I B O D n n zero matrix, and I is the n n identity matrix. [ ] [ ] [ ] [ ] A O C I AC AI AC A Soln: = = I B O D IC II + BD C I + BD If we do direct multiplication: to compute each term uses 2n multiplications and 2n additions, or 4n flops (floating point operations). Since there are 4n 2 terms in the resulting matrix, direct multiplication uses 4n 4n 2 = 16n 3 flops. In the above computation, to compute AC takes 2n n 2 flops, to compute BD takes 2n n 2 flops, to add I + BD takes n flops, so partitioned multiplication uses 4n 3 + n? flops. Take n = How many steps do we save?

16 Inverse of Partitioned Matrices [ ] A11 A A matrix of the form 12 is said to be block upper triangular. 0 A 22 Assume that A 11 is p p, A 22 is q q, and A is invertible. Find a formula for A 1.

17 Inverse of Partitioned Matrices [ ] A11 A A matrix of the form 12 is said to be block upper triangular. 0 A 22 Assume that A 11 is p p, A 22 is q q, and A is invertible. Find a formula for A 1. Soln: Let A 1 be B and partition B so that [ ] [ ] [ ] A11 A 12 B11 B 12 Ip 0 = 0 A 22 B 21 B 22 0 I q

18 Inverse of Partitioned Matrices [ ] A11 A A matrix of the form 12 is said to be block upper triangular. 0 A 22 Assume that A 11 is p p, A 22 is q q, and A is invertible. Find a formula for A 1. Soln: Let A 1 be B and partition B so that [ ] [ ] [ ] A11 A 12 B11 B 12 Ip 0 = 0 A 22 B 21 B 22 0 I q So we will have A 11 B 11 + A 12 B 21 = I p (1) A 11 B 12 + A 12 B 22 = 0 (2) A 22 B 21 = 0 (3) A 22 B 22 = I q (4)

19 As A is invertible, A 22 must be invertible. So from (??), we get B 22 = A 1 22, and from (??), we have B 21 = 0.

20 As A is invertible, A 22 must be invertible. So from (??), we get B 22 = A 1 22, and from (??), we have B 21 = 0. As A is invertible, A 11 must be invertible. So from (??) and B 21 = 0, we get B 11 = A 1 11.

21 As A is invertible, A 22 must be invertible. So from (??), we get B 22 = A 1 22, and from (??), we have B 21 = 0. As A is invertible, A 11 must be invertible. So from (??) and B 21 = 0, we get B 11 = A From (??), we get A 11 B 12 = A 12 B 22 = A 12 A So B 12 = A 1 11 A 12A 1 22.

22 As A is invertible, A 22 must be invertible. So from (??), we get B 22 = A 1 22, and from (??), we have B 21 = 0. As A is invertible, A 11 must be invertible. So from (??) and B 21 = 0, we get B 11 = A From (??), we get A 11 B 12 = A 12 B 22 = A 12 A So B 12 = A 1 11 A 12A So [ ] 1 A 1 A11 A = 12 = 0 A 22 [ A 1 11 A 1 11 A 12A 1 ] 22 0 A 1 22

23 As A is invertible, A 22 must be invertible. So from (??), we get B 22 = A 1 22, and from (??), we have B 21 = 0. As A is invertible, A 11 must be invertible. So from (??) and B 21 = 0, we get B 11 = A From (??), we get A 11 B 12 = A 12 B 22 = A 12 A So B 12 = A 1 11 A 12A So [ ] 1 A 1 A11 A = 12 = 0 A 22 [ A 1 11 A 1 11 A 12A 1 ] 22 0 A 1 22 A block diagonal matrix is a partitioned matrix with zero blocks off the main diagonal (of blocks). From the above example, such a matrix is invertible if and only if each block on the diagonal is invertible.

24 Example: find the inverse of the following matrix

25 LU Factorization An LU-factorization of an m n matrix A is an equation that expresses A as a product of an m m lower triangular matrix and an m n upper triangular matrix.

26 LU Factorization An LU-factorization of an m n matrix A is an equation that expresses A as a product of an m m lower triangular matrix and an m n upper triangular matrix. In particular, if A can be row reduced to echelon form without row interchanges, then we could make the main diagonal of L to be 1s. Or u 11 u 12 u u 1n l u 22 u u 2n A = LU = l 31 l l m1 l m2 l m u u 3n u nm

27 Motivation for LU Factorization If we want to solve the matrix equation Ax = b and we have LU-factorization of A = LU, then we can the following: Ax = LUx = L(Ux) = b so let y = Ux, we have Ly = b. Solve Ly = b to get y. Sove Ux = y to get x.

28 An LU factorization algorithm Suppose that A can be reduced to an echelon form U using only row replacements that add a multiple of one row to another row below it.

29 An LU factorization algorithm Suppose that A can be reduced to an echelon form U using only row replacements that add a multiple of one row to another row below it. So there are unit lower triangular elementary matrices E 1, E 2,..., E p such that E p... E 1 A = U

30 An LU factorization algorithm Suppose that A can be reduced to an echelon form U using only row replacements that add a multiple of one row to another row below it. So there are unit lower triangular elementary matrices E 1, E 2,..., E p such that E p... E 1 A = U Then A = (E p... E 1 ) 1 U = LU, with L = (E p... E 1 ) 1 = E Ep 1

31 An LU factorization algorithm Suppose that A can be reduced to an echelon form U using only row replacements that add a multiple of one row to another row below it. So there are unit lower triangular elementary matrices E 1, E 2,..., E p such that E p... E 1 A = U Then A = (E p... E 1 ) 1 U = LU, with L = (E p... E 1 ) 1 = E Ep 1 The above argument suggests the following algorithm to find an LU factorization of A:

32 An LU factorization algorithm Suppose that A can be reduced to an echelon form U using only row replacements that add a multiple of one row to another row below it. So there are unit lower triangular elementary matrices E 1, E 2,..., E p such that E p... E 1 A = U Then A = (E p... E 1 ) 1 U = LU, with L = (E p... E 1 ) 1 = E Ep 1 The above argument suggests the following algorithm to find an LU factorization of A: 1 Use row operations to reduce A to an upper triangles, and record the elementary matrices E 1, E 2,..., E p.

33 An LU factorization algorithm Suppose that A can be reduced to an echelon form U using only row replacements that add a multiple of one row to another row below it. So there are unit lower triangular elementary matrices E 1, E 2,..., E p such that E p... E 1 A = U Then A = (E p... E 1 ) 1 U = LU, with L = (E p... E 1 ) 1 = E Ep 1 The above argument suggests the following algorithm to find an LU factorization of A: 1 Use row operations to reduce A to an upper triangles, and record the elementary matrices E 1, E 2,..., E p. 2 Then L = E Ep 1. Or I = E 1... E p L.

34 Example Ex. Find an LU factorization of A =

35 Example Ex. Find an LU factorization of A = We first reduce A to an upper triangular form by row operations, and record each step with an elementary matrix A = A 1 (5) = U (6)

36 In the first step, we used row operations R2 + 2R1, R3 R1, R4 + 3R1, which corresponds to elementary matrices E 1 = , E 2 = , E 3 =

37 In the first step, we used row operations R2 + 2R1, R3 R1, R4 + 3R1, which corresponds to elementary matrices E 1 = , E 2 = , E 3 = In the second step, we used row operations R3 + 3R2, R4 4R2, which corresponds to elementary matrices E 4 = , E 5 =

38 In the first step, we used row operations R2 + 2R1, R3 R1, R4 + 3R1, which corresponds to elementary matrices E 1 = , E 2 = , E 3 = In the second step, we used row operations R3 + 3R2, R4 4R2, which corresponds to elementary matrices E 4 = , E 5 = In the last step, the row operation is R4 2R3, and the corresponding elementary matrix is E 6 = ,

39 One important observation (the position of d could be anywhere): if E = d , then E 1 = d

40 One important observation (the position of d could be anywhere): if E = d , then E 1 = d So (pay attend to the result) L 1 = E1 1 E 2 1 E 3 1 I =

41 One important observation (the position of d could be anywhere): if E = d , then E 1 = d So (pay attend to the result) L 1 = E1 1 E 2 1 E 3 1 I = Thus L = L 1 E4 1 E 5 1 E 6 1 =