Systems of Linear Equations

Transcription

1 Systems of Linear Equations Last time, we found that solving equations such as Poisson s equation or Laplace s equation on a grid is equivalent to solving a system of linear equations. There are many other examples where systems of linear equations appear, such as eigenvalue problems. In this lecture, we look into different approaches to solving systems of linear equations (SLE s). SLE: a 11 x 1 + a 12 x 2 + a 13 x a 1n x n = b 1 a 21 x 1 + a 22 x 2 + a 23 x a 2n x n = b 2 a 31 x 1 + a 32 x 2 + a 33 x a 3n x n = b 3 n unknowns x j m equations a ij and b i are known a m x 1 + a m 2 x 2 + a m 3 x a mn x n = b m Winter Semester 2006/7 Computational Physics I Lecture 8 1

2 SLE s If m<n, there is not a unique solution for the x s (underconstrained). If m>n, the system can be overconstrained and usually the job is to find the best set {x} to represent the set if equations. We will consider here square matrices m=n, with deta0, in which case the equations are linearly independent and there is a unique solution. Matrix representation: Ax = b a 11 a 12 a 1n a 21 a 22 a 2n A = a n1 a n2 a nn b = b 1 b 2 b n Winter Semester 2006/7 Computational Physics I Lecture 8 2

3 SLE s Note that if the matrix is diagonal, then the solutions is easy: a a 21 a 22 0 a n1 a n2 a nn x 1 x 2 x n = b 1 b 2 b n a 11 x 1 = b 1 x 1 = b 1 /a 11 a 21 x 1 + a 22 x 2 = b 2 x 2 = b2 a 21 b 1 a11 Goal of Gaussian elimination method is to bring A into this form. a 22 Winter Semester 2006/7 Computational Physics I Lecture 8 3

4 SLE s Ax = b a 11 a 12 a 1n x 1 a 21 a 22 a 2n x 2 = a n1 a n2 a nn x n b n Note: 1. Interchanging two rows of A and the same rows of b gives the same set of equations. 2. Solution not changed if replace a row with a linear combination including another row, as long as the b s are treated in a similar way. E.g., a ij k i a ij + k i a i j j = 1,,n 3. Interchanging two columns of A gives the same result if simultaneously two rows of vector x interchanged. Winter Semester 2006/7 Computational Physics I Lecture 8 4 b 1 b 2 b i k i b i + k i b i

5 a i1 Gauss Elimination Method define l i1 = a 11 Transform A by subtracting l i1 a t 1 from row i where a t 1 is the transpose of row 1. t t a 1 a 1 t a 2 a t t 2 l 21 a 1 a n t L 1 = a n t l n1 a 1 t 1 l l l n1 1 or A (1) = L 1 A where is the Frobenius matrix Winter Semester 2006/7 Computational Physics I Lecture 8 5

6 Gauss Elimination Method The result is a 11 a 12 a 1n (1) (1) A (1) 0 a 22 a 2n = (1) (1) 0 a n2 a nn now subtract a (1) i2 /a (1) 22 times the second row from rows 3...n A (2) = L 2 A (1) = L 2 L 1 A L 2 = l l n where l i2 = a (1) (1) i2 /a 22 Winter Semester 2006/7 Computational Physics I Lecture 8 6

7 Gauss Elimination Method Now have: a 11 a 12 a 13 a 1n (1) (1) (1) 0 a 22 a 23 a 2n A (2) (2) (2) = 0 0 a 33 a 3n (2) (2) 0 0 a n3 a nn Keep going until have upper triangular matrix (n 1) steps A (n1) = L n1 L n 2 L 1 A = U u 11 u 12 u 13 u 1n 0 u 22 u 23 u 2n U = 0 0 u 33 u 3n u nn with Ux = c and c = L n 1 L n 2 L 1 b Winter Semester 2006/7 Computational Physics I Lecture 8 7

8 Note that: L 1 = Gauss Elimination Method 1 l l l n1 1 1 l L 1 1 = l l n1 1 and similarly for the other L i, so L n 1 L n 2 L 1 A = U implies A = L 1 1 L 1 2 L 1 n 1 U = LU,with L = L 1 1 L L n1 1 l = l 31 l 32 1 l 43 l n1 l n2 l nn1 1 Winter Semester 2006/7 Computational Physics I Lecture 8 8

9 Gauss Elimination Method i.e., the matrix A has been decomposed (LU decomposition) into an upper triangular and a lower triangular matrix. The solution is now easy. A x = LU x = b First solve L y = b, then U x = y 1 = b 1 y 2 = b 2 l 21 y 1 then going from the bottom y so x n = y n u nn x n 1 = y n 1 x n u n 1,n u n 1,n 1 Winter Semester 2006/7 Computational Physics I Lecture 8 9

10 Gauss Elimination Method Take a concrete example: x x 2 = x 3 Here is some code: x * Loop over the rows. The index i is the row which is not touched in this step Do i=1,n-1 * * For step i, we modify all rows j>i Do j=i+1,n * * Loop over the column elements in this row. Perform linear transformation * on matrix A. Keep the upper diagonal elements in A as we go. Also build * up the lower diagonal matrix at the same time. Lji=A(j,i)/A(i,i) Do k=1,n A(j,k)=A(j,k)-Lji*A(i,k) Enddo L(j,i)=Lji Enddo Enddo Winter Semester 2006/7 Computational Physics I Lecture 8 10

11 Gauss Elimination Method A (1) = L (1) = A (2) = L (2) = A (3) = L (3) = Winter Semester 2006/7 Computational Physics I Lecture 8 11

12 Gauss Elimination Method * Now build up the solution matrix * First the vector y Do i=1,n y(i)=b(i) Do j=1,i-1 y(i)=y(i)-l(i,j)*y(j) Enddo Enddo * * and now x * Do i=n,1,-1 x(i)=y(i) Do j=i+1,n x(i)=x(i)-a(i,j)*x(j) Enddo x(i)=x(i)/a(i,i) Enddo y= x= It s good practice to check result. Is Ax=b? Is A=LU? Winter Semester 2006/7 Computational Physics I Lecture 8 12

13 Gauss-Jordan Elimination The Gauss-Jordan method changes the matrix A into the diagonal matrix in one pass. First try, use 2. from P.4 to recast equations: a 1 j a 1 j = a 1 j j = 1,2,3, 4 a 11 Then take the following linear combination: a a ij = a ij + k i a 1 j = a ij + k 1 j i i = 2,3,4 a 11 and choose k i such that a i1 = 0 = a i1 + k i a 11 = a i1 + k i i = 2,3,4 or k i = a i1 Winter Semester 2006/7 Computational Physics I Lecture 8 13

14 Gauss-Jordan Elimination a 11 a 12 a 13 a 14 Work with a 4x4 matrix to be a 21 a 22 a 23 a 24 concrete: a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 After the first step, our matrix looks like this x 1 x 2 x 3 x 4 = b 1 b 2 b 3 b 4a11 a 1 12 a 13 a 14 a11 a11 a11 a 0 a 22 a 12 a 21 a11 a 23 a 13 a 21 a11 a 24 a a11 a 0 a 32 a 12 a 31 a11 a 33 a 13 a 31 a11 a 34 a a11 a 0 a 42 a 12 a 41 a11 a 43 a 13 a 41 a11 a 44 a a11 x 1 x 2 x 3 x 4 = b 1 a11 b b 2 a 1 21 a11 b b 3 a 1 31 a11 b b 4 a 1 41 a11 Now we move on and make the second column look like the unit matrix, and so on. Winter Semester 2006/7 Computational Physics I Lecture 8 14

15 1. a 2 j = 2. a ij = 3. b i = a 2 j Gauss-Jordan Elimination a 22 a ij a i2 b i a i2 b 2 a 2 j a i = 1, 3, 4 22 a i = 1, 3, 4 22 Note that first column is not affected: a 21 = a i1 = 0 a 22 = 0 a i1 a i2 a 21 a = 22 a i1 a i2 0 a = 22 a i1 i = 1, 3, 4 Winter Semester 2006/7 Computational Physics I Lecture 8 15

16 Gauss-Jordan Elimination After these two sets of transformations, we have 1 0 a a a a 43 a 14 a 24 a 34 a 44 x 1 x 2 x 3 x 4 = b 1 b 2 b 3 b 4 Keep going until we have x 1 x 2 x 3 x 4 = b 1 (4) b 2 (4) b 3 (4) b 4 (4) x 1 x 2 x 3 x 4 = b 1 (4) b 2 (4) b 3 (4) b 4 (4) Winter Semester 2006/7 Computational Physics I Lecture 8 16

17 Gauss-Jordan Elimination Algorithm looks like this: 1. Loop over the n columns. We want to turn the columns one at a time into the unit matrix 2. For column k, we find the linear transformation on the rows which gives the desired result (1 in row k, 0 in other rows) a) Loop over the n rows and make the diagonal element a 1. For i=k, do this by dividing every element in A(i,j) by A(k,k), where j is the column index and i is the row index. Also need to divide b(i) by A(k,k) b) For ik, make the element in column k a 0. We do this by subtracting A(i,k)*A(k,j)/A(k,k) from A(i,j). For b(i), we subtract A(i,k)*b(k)/A(k,k) Test matrices: A 1 = b 1 = A 2 = b 2 = Winter Semester 2006/7 Computational Physics I Lecture 8 17

18 Gauss-Jordan Elimination Start with A 1 : b First iteration Second iteration Third iteration Fourth iteration Winter Semester 2006/7 Computational Physics I Lecture 8 18

19 Gauss-Jordan Elimination We can use the same transformations to get the inverse of A: AA 1 = E now suppose the transformation of A to E consists of the following steps : T 4 T 3 T 2 T 1 A = E Apply to the equation above or T 4 T 3 T 2 T 1 AA 1 = T 4 T 3 T 2 T 1 E EA 1 = A 1 = T 4 T 3 T 2 T 1 E Winter Semester 2006/7 Computational Physics I Lecture 8 19

20 Gauss-Jordan Elimination Let s try it on the previous example: After T 1 After T 2 After T After T Winter Semester 2006/7 Computational Physics I Lecture 8 20

21 With A 2 : Gauss-Jordan Elimination Start to see rounding issues: Winter Semester 2006/7 Computational Physics I Lecture 8 21

22 Gauss-Jordan Elimination As a check, we calculate the result for x for different initial matrices. Use matrices of the form A 2 : a ij = 1 i + j 1 Try different size square matrices. Should give us our initial vector b. Find that start into mumerical problems with n=m=10. Single precision calculation. All values should be Double precision calculation. All values are =1. Winter Semester 2006/7 Computational Physics I Lecture 8 22

23 Gauss-Jordan Elimination Check 100x100 matrix double precision Looks pretty good. However, there are cases where this simple Gauss-Jordan technique does not work so well. Winter Semester 2006/7 Computational Physics I Lecture 8 23

24 Pivoting What if the diagonal element which we divide by is very small or zero? (The element we divide by is called the pivot). This obviously causes problems. Resolved by a technique called pivoting (exchanging rows and columns). Partial pivoting - exchange rows in the region which have not yet been transformed to the unit matrix. Full pivoting - exchange rows and columns. These tricks are allowed according to the rules we stated on pages 4,5. Winter Semester 2006/7 Computational Physics I Lecture 8 24

25 Partial Pivoting Algorithm looks like this: 1. Loop over the n columns. We want to turn the columns one at a time into the unit matrix. Use the index k to specify which column element we want to turn into a For each value of k, we loop over the rows ik and look for the maximum value A(i,k). The row i which gives this maximum is swapped with row k. Use an index vector to keep track of the permutations Ind(i)=row which is to be considered as i th row. 3. Find the linear transformation: a) Loop over the n rows and make the diagonal element a 1. For Ind(i)=k, do this by dividing every element in A(Ind(i),j) by A(Ind(i),k), where j is the column index and Ind(i) is the row index. Also need to divide b(ind(i)) by A(Ind(i),k) b) For Ind(i)k, make the element in column k a 0. We do this by subtracting A(Ind(i),k)*A(Ind(k),j)/A(Ind(k),k) from A(Ind(i),j). For b(i), we subtract A(Ind(i),k)*b(Ind(k))/A(Ind(k),k) Winter Semester 2006/7 Computational Physics I Lecture 8 25

26 Partial Pivoting Example: Winter Semester 2006/7 Computational Physics I Lecture 8 26

27 Exercizes 1. Solve for x in Ax=b using the LU decomposition method a ij = 1 i + j +1 b j = j 2. Do it again with the Gauss-Jordan method 3. (more difficult). Now try a ij = i j i, j b j = j You will need to use pivoting. Winter Semester 2006/7 Computational Physics I Lecture 8 27