1. Scientific Computing Linear Systems – LU Factorization

2. LU Factorization Today we will show that the Gaussian Elimination method can be used to factor (or decompose) a square, invertible, matrix A into two parts, A = LU, where: – L is a lower triangular matrix with 1’s on the diagonal, and – U is an upper triangular matrix

3. LU Factorization A= LU We call this a LU Factorization of A.

4. LU Factorization Example Consider the augmented matrix for Ax=b: Gaussian Elimination (with no row swaps) yields:

5. LU Factorization Example Let’s Consider the first step in the process – that of creating 0’s in first column: This can be achieved by multiplying the original matrix [A|b] by M 1

6. LU Factorization Example Thus, M 1 [A|b] = In the next step of Gaussian Elimination, we pivot on a 22 to get: We can view this as multiplying M 1 [A|b] by

7. LU Factorization Example Thus, M 2 M 1 [A|b] = In the last step of Gaussian Elimination, we pivot on a 33 to get: We can view this as multiplying M 2 M 1 [A|b] by

8. LU Factorization Example Thus, M 3 M 2 M 1 [A|b] = So, the original problem of solving Ax=b can now be thought of as solving M 3 M 2 M 1 Ax = M 3 M 2 M 1 b

9. LU Factorization Example Note that M 3 M 2 M 1 A is upper triangular: Thus, We hope that L is lower triangular

10. LU Factorization Example Recall: In Exercise 3.6 you proved that the inverse to the matrix Is the matrix

11. LU Factorization Example Thus, L = M 1 -1 M 2 -1 M 3 -1

12. LU Factorization Example So, A can be factored as A=LU

13. Solving Ax=b using LU Factorization To solve Ax=b we can do the following: – Factor A = LU – Solve the two equations: Lz = b Ux = z – This is the same as L(Ux) = b or Ax=b.

14. LU Factorization Example Our example was Ax=b: Factor:

15. LU Factorization Example Solve Lz=b using forward substitution: We get:

16. LU Factorization Example Solve Ux=z using back substitution: We get the solution vector for x: (Note typo in Pav, page 36 bottom of page “z” should be “x”)

17. LU Factorization Theorem Theorem 3.3. If A is an nxn matrix, and Gaussian Elimination does not encounter a zero pivot (no row swaps), then the algorithm described in the example above generates a LU factorization of A, where L is a lower triangular matrix (with 1’s on the diagonal), and U is an upper triangular matrix. Proof: (omitted)

18. LU Factorization Matlab Function (1 of 2) function [ L U ] = lu_gauss(A) % This function computes the LU factorization of A % Assumes A is not singular and that Gauss Elimination requires no row swaps [n,m] = size(A); % n = #rows, m = # columns if n ~= m; error('A is not a square matrix'); end for k = 1:n-1 % for each row (except last) if A(k,k) == 0, error('Null diagonal element'); end for i = k+1:n % for row i below row k m = A(i,k)/A(k,k); % m = scalar for row i A(i,k) = m; % Put scalar for row i in (i,k) position in A. % We do this to store L values. Since the lower % triangular part of A gets zeroed out in GE, we % can use it as a storage place for values of L.

19. LU Factorization Matlab Function (1 of 2) for j = k+1:n % Subtract m*row k from row i -> row i % We only need to do this for columns k+1 to n % since the values below A(k,k) will be zero. A(i,j) = A(i,j) - m*A(k,j); end end % At this point, A should be a matrix where the upper triangular % part of A is the matrix U and the rest of A below the diagonal % is L (but missing the 1's on the diagonal). L = eye(n) + tril(A, -1); % eye(n) is a matrix with 1's on diagonal U = triu(A); end

20. LU Factorization Matlab >> A=[2 1 1 3; 4 4 0 7; 6 5 4 17; 2 -1 0 7] A = 2 1 1 3 4 4 0 7 6 5 4 17 2 -1 0 7 >> [l u] = lu_gauss(A);

21. LU Factorization Matlab >> >> l l = 1 0 0 0 2 1 0 0 3 1 1 0 1 -1 -1 1 >> u u = 2 1 1 3 0 2 -2 1 0 0 3 7 0 0 0 12

22. LU Factorization Matlab Solve Function Class Discussion: What must we do to modify the function lu_gauss so that we can compute the solution to Ax=b using the LU factorization? New function: lu_solve on handouts. Discuss and consider Matlab output ->

23. LU Factorization Matlab >>> b = [7 11 31 15]’; >>> [x z] = lu_solve(A, b); >> z z = 7 -3 13 18 >> x x = 1.5417 -1.4167 0.8333 1.5000

24. Linear Algebra Review Now that we have covered some examples of solving linear systems, there are several important questions: – How many numerical operations does Gaussian Elimination take? That is, how fast is it? – Why do we use LU factorization? In what cases does it speed up calculation? – How close is our solution to the exact solution? – Are there faster solution methods?

25. Operation Count for Gaussian Elimination How many floating point operations (+-*/) are used by the Gaussian Elimination algorithm? Definition: Flop = floating point operation. We will consider a division to be equivalent to a multiplication, and a subtraction equivalent to an addition. Thus, 2/3 = 2*(1/3) will be considered a multiplication. 2-3 = 2 + (-3) will be considered an addition.

26. Operation Count for Gaussian Elimination In Gaussian Elimination we use row operations to reduce to

27. Operation Count for Gaussian Elimination Consider the number of flops needed to zero out the entries below the first pivot a 11.

28. Operation Count for Gaussian Elimination First a multiplier is computed for each row below the first row. This requires (n-1) multiplies. m = A(i,k)/A(k,k); Then in each row below row 1 the algorithm performs n multiplies and n adds. (A(i,j) = A(i,j) - m*A(k,j);) Thus, there is a total of (n-1) + (n-1)*2*n flops for this step of Gaussian Elimination. For k=1 algorithm uses 2n 2 –n -1 flops

29. Operation Count for Gaussian Elimination For k =2, we zero out the column below a 22. There are (n-2) rows below this pivot, so this takes 2(n-1) 2 –(n-1) -1 flops. For k =3, we would have 2(n-2) 2 –(n-2) -1 flops, and so on. To complete Gaussian Elimination, it will take I n flops, where

30. Operation Count for Gaussian Elimination Now, So, I n = (2/6)n(n+1)(2n+1) – (1/2)n(n+1) – n = [(1/3)(2n+1)-(1/2)]*n(n+1) – n = [(2/3)n – (1/6)] * n(n+1) - n = (2/3)n 3 + (lower power terms in n) Thus, the number of flops for Gaussian Elimination is O(n 3 ).

31. Operation Count for LU Factorization In the algorithm for LU Factorization, we only do the calculations described above to compute L and U. This is because we save the multipliers (m) and store them to create L. So, the number of flops to create L and U is O(n 3 ).

32. Operation Count for using LU to solve Ax = b Once we have factored A into LU, we do the following to solve Ax = b: Solve the two equations: Lz = b Ux = z How many flops are needed to do this?

33. Operation Count for using LU to solve Ax = b To solve Lz=b we use forward substitution z = z 1 = b 1, so we use 0 flops to find z 1. z 2 = b 2 – l 21 * z 1, so we use 2 flops to find z 2. z 3 = b 3 – l 31 * z 1 – l 32 * z 2, so we use 4 flops to find z 2, and so on.

34. Operation Count for using LU to solve Ax = b To solve Lz=b we use forward substitution z = Totally, 0+2+4+ … + 2*(n-1)= 2*(1+2+…+(n-1)) = 2*(1/2)*(n-1)(n) = n 2 – n. So, the number of flops for forward substitution is O(n 2 ).

35. Operation Count for using LU to solve Ax = b To solve Ux=z we use backward substitution A similar analysis to that of forward substitution shows that the number of flops for backward substitution is also O(n 2 ). Thus, the number of flops for using LU to solve Ax=b is O(n 2 ).

36. Summary of Two Methods Gaussian Elimination requires O(n 3 ) flops to solve the linear system Ax = b. To factor A = LU requires O(n 3 ) flops Once we have factored A = LU, then, using L and U to solve Ax = b requires O(n 2 ) flops. Suppose we have to solve Ax = b for a given matrix A, but for many different b vectors. What is the most efficient way to do this?

37. Summary of Two Methods Suppose we have to solve Ax = b for a given matrix A, but for many different b vectors. What is the most efficient way to do this? Most efficient is to use LU decomposition and then solve Lz = b Ux = z Computing LU is O(n 3 ), but every time we solve Lz=b, Ux=z we use O(n 2 ) flops!