Linear Algebra Section 2.6 : LU Decomposition Section 2.7 : Permutations and transposes Wednesday, February 13th Math 301 Week #4

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1 Linear Algebra Section. : LU Decomposition Section. : Permutations and transposes Wednesday, February 1th Math 01 Week # 1

2 The LU Decomposition We learned last time that we can factor a invertible matrix as A = LU where L is a lower triangular matrix recording the multipliers `ij we used in elimination, and U is the upper triangular matrix produced by the elimination process. L = ` `1 ` 1 0 `1 ` ` 1 U = X X X X 0 X X X 0 0 X X X The pivots appear on the diagonal of U. Because the matrix is invertible, the pivots are all non-zero

3 The LU Decomposition If the elimination completes with n non-zero pivots, then we have shown that the original matrix A is invertible. Question : How can you determine whether a matrix has an inverse or not? (Answer : Try to finds its LU decomposition. We have also greatly simplified the task of solving a linear system involving A.

4 Solving systems using LU How can we solve a system using the LU factorization? Ax = b Step 0 : Factor A into LU Row-reduction LUx = b Step 1 : Solve Ly = b Forward substitution Step : Solve Ux = y Back substitution

5 Cost of elimination The cost of eliminating entries below the 1 st pivot : To eliminate (or zero out ) the column below the first pivot, we must do (n 1) = () multiplications and subtractions.

6 Cost of elimination The cost of eliminating entries below the nd pivot : To eliminate (or zero out ) the column below the second pivot, we must do (n ) = () multiplications and subtractions.

7 Cost of elimination The cost of eliminating entries below the nd pivot : To eliminate (or zero out ) the column below the second pivot, we must do (n ) = () multiplications and subtractions.

8 Cost of elimination The cost of eliminating entries below the rd pivot : 0 0 X X X 0 0 X X X 0 0 X X X To eliminate (or zero out ) the column below the third pivot, we must do (n ) = () multiplications and subtractions. 8

9 Cost of elimination The cost of eliminating entries below the rd pivot : 0 0 X X X X X X X To eliminate (or zero out ) the column below the fourth pivot, we must do (n ) = (1) multiplications and subtractions. 9

10 Cost of elimination The total number of multiplications is then : n 1 X k=1 k = (n 1)(n)(n 1) 1 n The number of subtractions is the same. We say that elimination is an n process. This is consider expensive for a linear solve. 10

11 Cost of the back solve Step k in the back solve requires k multiplications and k additions. So the total work for a back solve is : nx k=1 k = n(n + 1) 1 n Considerably cheaper than the elimination procedure. 11

12 The LU decomposition If we have more than one right hand side (as is often the case) Ax = b i, i =1,,...,M We can actually store the work involved carrying out the elimination by storing the multipliers used to carry out the row operations. 1

13 Solving systems using LU How can we solve a system using the LU factorization? Ax = b LUx = b Step 0 : Factor A into LU Row-reduction Step 1 : Solve Ly = b Forward substitution Step : Solve Ux = y Back substitution 1

14 Row exchanges What if we start with a system that looks like : A = All we need to do is exchange the rows of A, and do the decomposition on LU = PA where P is a permutation matrix, i.e P =

15 Partial pivoting We can also do row exchanges not just to avoid a zero pivot, but also to make the pivot as large as possible. This is called partial pivoting. Find the largest pivot in the entire column, and do a row exchange. One can also do full pivoting by looking for the largest pivot in the entire matrix. But this is rarely done. 1

16 Permutation matrices A permutation matrix is one which permutes (without changing the numerical values) the rows or columns of a matrix. Permute the rows by multiplying on the left by P : = Permute the columns by multiplying on the right by P : =

17 Permutation matrices A permutation matrix has exactly one 1 in each row and column, and zeros every where else. Alternatively, it has the rows of the identity matrix in any order. The inverse of P is its transpose = T = To get the matrix transpose, flip rows and columns. 1

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