LU Factorization. A m x n matrix A admits an LU factorization if it can be written in the form of A = LU

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1 LU Factorization A m n matri A admits an LU factorization if it can be written in the form of Where, A = LU L : is a m m lower triangular matri with s on the diagonal. The matri L is invertible and is called a unit lower triangular matri. L * * * * * *

2 LU Factorization (Contd.) U : Upper triangular matri. Note all entries in U below the diagonal entries i.e. u,u, and so on are zero. This is true, even when U is not square. The reason for considering a factorization A = LU, is that it drastically speeds up a solution of A = b linear system that uses row reduction, this is particularly true when solving linear systems of large dimensions. U * * * * * * * * * * * * *

3 Eample: Find an LU Factorization of A Since, A has rows, L should be, whereas since A has columns, U should be L = U A

4 Eample: (Contd.) Matri U is obtained by simply reducing A to row echelon form using forward elimination process. Note if A is not reducible to row echelon form, it implies LU Factorization does not eists for given matri However there are two methods to populate L, both will be discussed here: Method : L is obtained using the multipliers with negative sign, that are used during forward elimination process i.e. while reducing given matri A to row echelon form.

5 Eample: (Contd.) * * * L 9 ; Now reducing A to row echelon form, while populating L using the multipliers with negative sign, that are used during forward elimination process R +R R +R R +R * L 7 ; R +R R +R Net pivot column Net pivot column multipliers with negative sign multipliers with negative sign

6 Eample: (Contd.) L ; R +R multipliers with negative sign U L ; So L and U are as follows:

7 Eample: (Method ) * * * L ; Now populating L by dividing all entries in a pivot column from pivot position variable, for eample in column dividing,, 6 from pivot position variable. * L 9 ; R +R R +R R +R Net pivot column Now performing elementary row operations along st pivot column i.e. Column

8 Eample: (Method ) Now performing elementary row operations along net pivot column i.e. Column L 7 ; R +R R +R Net pivot column L U; R +R Now performing elementary row operations along net pivot column i.e. Column

9 Solving Linear system using LU Factorization. Decompose A into L and U.. So matri equation, A = b can be written as LU = b; Let y = U, so above equation becomes L y = b; Now writing above matri in augmented matri form i.e. [L I b] to find y. Since, U = y, so now writing this equation in augmented matri form i.e. [U I y] to find

10 Eample : Solve the equation A = b using LU factorization 7 b 6 7 ; A Step : Decomposing A into L and U * L 7 ; Now reducing A to row echelon form, while populating L using the multipliers with negative sign, that are used during forward elimination process R +R R +R Net pivot column multipliers with negative sign

11 Eample : (Contd.) L 7 ; R +R multipliers with negative sign Method : Now populating L by dividing all entries in a pivot column from pivot position variable, for eample in column dividing, 6 from pivot position variable. * L 7 ; 6 Now performing elementary row operations along st pivot column i.e. Column

12 Eample : (Contd.) L 7 ; R +R R +R Net pivot column Now performing elementary row operations along nd pivot column i.e. Column L 7 ; U R +R

13 Eample : (Contd.) Step : Now solving Ly = b in augmented form to find y b L R +R R +R y R +R

14 Eample : (Contd.) Step : Now solving U = y in augmented form to find 6 7 y U 7 R +R R +R ½ R R R +R / R

15 Eample : Solve the equation A = b using LU factorization 7 9 b ; A 7 8 ; U L Step : Decomposing A into L and U

16 Eample : (Contd.) Step : Now solving Ly = b in augmented form to find y by reducing it to reduced row echelon form Step : Now solving U = y in augmented form to find by reducing it to reduced row echelon form 9 y b L y U

17 Eercise Solve the equation A = b using LU factorization 9 6 Answer

18 Eigenvalues and eigenvectors The eigenvectors (or characteristic vectors) of a square matri are the nonzero vectors which, after being multiplied by the matri, remain proportional to the original vector (i.e. change only in magnitude, not in direction). For each eigenvector, the corresponding eigenvalue (or characteristic value) is the factor by which the eigenvector changes when multiplied by the matri.

19 Definition If A is an n n matri, then a nonzero vector in R n is called an eigenvector of A, if A is a scalar multiple of ; that is, A=λ for some scalar λ. The scalar λ is called an eigenvalue of A, and is said to be an eigenvector of A corresponding to λ.

20 Eample: 6 6 Let A,u and v Are u and v eigenvectors of A? Au u Thus u is an eigenvector of matri A corresponding to an eigenvalue of.

21 Eample: (Contd.) v eigenvectors of A? Are u and and v 6,u 6 Let A 9 6 Av Thus v is not an eigenvector of matri A, because Av is not a multiple of v.

22 Eercise: Is the vector of A 8 an eigenvector A 8 6 Thus is an eigenvector of matri A corresponding to an eigenvalue of.

23 Eercise: A Where, is an eigenvectorsof A? Verify that ) ( A Thus is an eigenvector of matri A corresponding to an eigenvalue of.

24 Eercise: Determine which of the indicated column vectors (i.e.,, ) are eigenvector of given matri A. Give the corresponding eigenvalue. and, A i, ) ( Answer (i) : Only is an eigenvector of matri A corresponding to an eigenvalue of., 6 ) ( and, A ii Answer (ii) : Only is an eigenvector of matri A corresponding to an eigenvalue of.

25 Eercise: (Contd.) Determine which of the indicated column vectors (i.e.,, ) are eigenvector of given matri A. Give the corresponding eigenvalue., ) ( and, A iii Answer (iii) : and are eigenvectors of matri A corresponding to respective eigenvalues of and.

26 Eigenvalues and Eigenvectors of nn matrices To find the eigenvalues of an n n matri A, we can rewrite A=λ () using the properties of matri algebra as or equivalently, A=λI (A λi) = () Where, I is the multiplicative identity, whereas n

27 ( a a a Eigenvalues and Eigenvectors of nn matrices (A λi) = () For a linear system, Eq: can be rewritten as ) ( a a ) Eq : An obvious solution to Eq: is when a ( a ) = = = (i.e. trivial solution). However, we are seeking only non trivial solutions, as for λ to be an eigenvalue, there must be a nonzero solution of this equation a a

28 Recall!! A set of vectors is linearly dependent, if and only if we can find a set of scalars (i.e. weights), which are zero) such that (not all of,,, n v nn If such a set of scalars cannot be found i.e. the vector equation has only trivial solution (i.e. all scalars are zero), then set of vectors is said to be linearly independent. (vector equation of homogenoussystem) v, v,, v n

29 Eample:,, v, v v Let 6 a. Determine if the set {v, v, v } is linearly independent Solution Since, a set of vectors is linearly independent, if and only if the vector equation of homogenous system has only trivial solution. Therefore, we must determine if there is a nontrivial solution for given homogenous system using row operations on the associated augmented matri R +R R +R / R 6R +R

30 From row echelon form matri it can be seen that: Eample: (Contd.), : basic variables : free variable Therefore, each nonzero value of determines a nontrivial solution of giver linear system. Hence, v,v,v are linearly dependent (i.e. not linearly independent). b. Find a linear dependence relation among v,v,v To find a linear dependence relation among v,v,v reducing the row echelon matri to reduced row echelon form & Thus R +R

31 Eample: (Contd.) Now assigning any arbitrary value of, say = will yield:, : basic variables : free variable Substituting these values into vector equation will yield v v v Note this is one (out of infinitely many) possible linear dependence relations among v, v, v.

32 Characteristic equation Since, in case of linear dependence the determinant of the coefficient matri vanishes, therefore; to find a nonzero solution for Equation, we must have det (A λi)= This is called the characteristic equation of A Eigenvalues of square matri A are roots of the characteristic equation. Hence, an nn matri A has at least one eigenvalue and at most n numerically different eigenvalues

33 Terminologies Spectrum of A The set of the eigenvalues (or characteristic values) is called the spectrum of A. Spectral radius of A The largest of the absolute values of the eigenvalues of A is called the spectral radius of A. Eigenspace The set of all eigenvectors corresponding to an eigenvalue of A, together with, forms a vectorspace, called the eigenspace of A corresponding to this eigenvalue.

34 Eigenvalue Problem The process of determining the eigenvalues and eigenvectors of a matri is called an eigenvalue problem.

35 Eample: Find the e igenvaluesand eigenvectors of A 7 8 From the characteristic equation det (A λi) λ 7 8 λ (det(a I) ) ( 6) ( 6) ( )( 6) Hence, the eigenvalues are λ, and λ 6 λ

36 Eample: (Contd.) Hence, the eigenvalues are λ and λ 6 Spectrum of A The set of the eigenvalues (or characteristic values) is called the spectrum of A. Therefore, spectrum of A is given by {, 6} Spectral radius of A The largest of the absolute values of the eigenvalues of A is called the spectral radius of A. Therefore, spectral radius of A is 6.

37 Eample: (Contd.) Now in case of matri there is no need to use GaussJordan elimination. To find the eigenvectors corresponding to we resort to the system (A λi) in equivalent form : 7 7 & (A I) It is apparent from above system that =. Thus, if we choose =, the eigenvector corresponding to eigenvalue of is

38 Eample: (Contd.) Similarly, to find the eigenvectors corresponding to 6 resort to the system (A λi) in equivalent form : we 7 7 & (A 6I) It is apparent from above system that = /7. Thus, if we choose =7, the eigenvector corresponding to eigenvalue of 6 is 7

39 Eercise: Find the e igenvaluesand eigenvectors of A From the characteristic equation λ det (A λi) λ λ (det(a I) ) ( 6) ( 6) ( )( 6) Hence, the eigenvalues are λ, and λ 6 Corresponding eigenvectors are λ & λ

40 Eercise: Find the e igenvaluesand eigenvectors of From the characteristic equation det (A λi) (det(a I) ) 8 ( ) ( ) ( )( ) λ, and λ λ Hence, the eigenvalues are λ 8 λ A & 8 Corresponding eigenvectors are λ 8 9 λ

41 Eercise: Find the eigenvaluesand eigenvectors of A 8 6 The eigenvalue λ λ isan eigenvalueof multipicit y. Corresponding singleeigenvector is λ λ

42 Eercise: Find the eigenvaluesand eigenvectors of A 7 The eigenvalue λ λ isan eigenvalueof multipicit y. Corresponding singleeigenvector is λ λ

43 Comple Eigenvalues Let A be a square matri with real entries. If iy, y of A, then its conjugate eigenvalue of A. is a comple eigenvalue is also an If is an eigenvector corresponding to λ, then its iy conjugate is an eigenvector to

44 Eample: Find the e igenvaluesand eigenvectors of A 6 From the characteristic equation 6 6 det (A λi) λ λ λ 6 From the quadratic formula, we obtain i 9 and i ( det(a I) ) Hence, the eigenvalues are λ i, and λ i

45 Quadratic formula: Revision In mathematics, a quadratic equation is a polynomial equation of the second degree. The general form is b b a ac ; indicates that both b b ac b b ac & a a are solutions of the Quadratic equation. Since, we have 9 (a ;b ;c 9), ( ) ( ) () ()(9) 6, ( i i, i; and i; & i )

46 Eample: (Contd.) To find the eigenvectors corresponding to we resort to the system in equivalent form : i ) λi (A 6 i) ( i) ( i i i) ( Now resotingto equivalent form I) (A It is apparent from above system that = ( i). Thus, if we choose =, the eigenvector corresponding to eigenvalue of (+i) is i

47 Eample: (Contd.) Since, if is an eigenvector corresponding to λ, then its conjugate is an eigenvector to Therefore; an eigenvector corresponding to i is: i

48 Eercise: Find the eigenvaluesand eigenvectors of A The compleeigenvalues are λ i and λ i Corresponding λ i and eigenvectorsare λ λ i

49 Eercise: A Find the eigenvaluesand eigenvectors of and compleeigenvalues are The i λ i λ i and i λ λ λ eigenvectorsare Corresponding