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p1034 - SECTION 4: INTRODUCTION TO BASIC NUMERICAL LINEAR...

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SECTION 4: INTRODUCTION TO BASIC NUMERICAL LINEAR ALGEBRA IN R N ................. 46 BASIC LINEAR ALGEBRA FACTS ............................................................................................................ 47 GAUSS ELIMINATION ("PIVOTING"): ..................................................................................................... 56 INCOMPLETE GAUSS ELIMINATION (IGE): ........................................................................................... 64 MORE ON INCOMPLETE GAUSS ELIMINATION (IGE), AND LU MATRIX FACTORIZATION: ................ 68 POSITIVE DEFINITE MATRICES AND CHOLESKI'S FACTORIZATION: ................................................... 76
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EE103 Lecture Notes, Fall 2007, Prof S. Jacobsen Section 4 46 SECTION 4: INTRODUCTION TO BASIC NUMERICAL LINEAR ALGEBRA IN R n Recall an earlier example: 12 2 (1 10 ) 2 10 kk xx + = ++ = + The exact answer is * (1,1) ' x = . However, the vector (0,2)' x = results in residuals of 0 and 10 k for the first and second equations, respectively. Therefore, if one didn't know the exact solution it might appear that the vector x is not a bad solution for the system of equations. However, as we've already seen, the relative error is * * || 1.0 x = , a percentage error of 100%. Ex: Consider the system of equations .0003 1.566 1.569 .3454 2.436 1.018 + = −= The exact solution of this system of equations is * (10,1) ' x = . However, let's use (,) ( 1 0 , 4 ) n β = arithmetic to solve this system by Gauss elimination. That is we "pivot" on 11 .0003 a = . Upon doing so, in (10,4) arithmetic, the new 22 a coefficient is 22 1804 a =− and the new coefficient on the right hand side of the second equation becomes 2 1805 b . Of course, this implies that 2 1.001 x = and therefore 1 (1.569 1.566 1.001) /.0003 3.333 x = Hence, we have that (3.333,1.001)' x = . However, the relative error is enormous; in fact, the relative error for 1 x is 66.67%. Exercise: Interchange the two rows and pivot as we did above, in (10,4) arithmetic, and show that you arrive at the exact solution (10,1). Therefore, a mere interchange of rows leads to a far superior answer. In fact, look above at the expression for 1 x and it becomes
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EE103 Lecture Notes, Fall 2007, Prof S. Jacobsen Section 4 47 clear that the relatively small error in the value of 2 x (1.001 instead of 1.000) is being greatly inflated because of the division by the number .0003. Ex: Consider the system of linear equations 11 1 12 2 1 1 11 nn n n n ax ax ax b ax + ++ = + += ... ................................. .............. or, in matrix-vector notation, A xb = Basic Linear Algebra Facts You've all learned that if A has an inverse, denoted by 1 A , then the unique solution is given by 1 x Ab = * . However, the latter is not meant to be a prescription for a numerical method for computing the solution x * . That is, we'll see that there are better methods, especially for large systems (many equations and variables) and, in particular, one should not generally try to find 1 A and then execute the multiplication 1 . In fact, suppose we have another method for solving the x system of equations Ax b = . We can use that method for computing the inverse matrix. To see this, let 12 n ee e , ,..., denote the unit vectors in n-space. That is, 0 0 0 1 0 0 i e = ( , ,. .., , , ,. .., )' , where the "1" is in the i th position. Then, we use the method to solve the n linear systems of equations 1 i Ax e i n == , Let 1 i x in = , denote the respective unique solutions. Then the matrix n x xx [ , ] is the inverse of A . To see this, note 1 1 n nxn Ax x x Ax e e I = [ , ] [ ] [ ,. .., ] Much of our subsequent discussion will involve the development of methods of solving systems of linear equations. However, those methods are based upon some important concepts of numerical linear algebra.
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p1034 - SECTION 4: INTRODUCTION TO BASIC NUMERICAL LINEAR...

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