CME 302: NUMERICAL LINEAR ALGEBRA FALL 2005/06 LECTURE 8

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1 CME 30: NUMERICAL LINEAR ALGEBRA FALL 005/06 LECTURE 8 GENE H GOLUB 1 Positive Defiite Matrices A matrix A is positive defiite if x Ax > 0 for all ozero x A positive defiite matrix has real ad positive eigevalues, ad its leadig pricipal submatrices all have positive determiats From the defiitio, it is easy to see that all diagoal elemets are positive To solve the system Ax = b where A is positive defiite, we ca compute the Cholesky decompositio A = F F where F is upper triagular This decompositio exists if ad oly if A is symmetric ad positive defiite I fact, attemptig to compute the Cholesky decompositio of A is a efficiet method for checkig whether A is symmetric positive defiite It is importat to distiguish the Cholesky decompositio from the square root factorizatio A square root of a matrix A is defied as a matrix S such that S = SS = A Note that the matrix F i A = F F is ot the square root of A, sice it does ot hold that F = A uless A is a diagoal matrix The square root of a symmetric positive defiite A ca be computed by usig the fact that A has a eigedecompositio A = UΛU where Λ is a diagoal matrix whose diagoal elemets are the positive eigevalues of A ad U is a orthogoal matrix whose colums are the eigevectors of A It follows that ad so S = UΛ 1/ U is a square root of A A = UΛU = (UΛ 1/ U )(UΛ 1/ U ) = SS The Cholesky Decompositio The Cholesky decompositio ca be computed directly from the matrix equatio A = F F Examiig this equatio o a elemet-by-elemet basis yields the equatios a 11 = f 11, a 1j = f 11 f 1j, a kk = f 1k + f k + + f kk, a kj = f 1k f 1j + + f kk f kj, ad the resultig algorithm that rus for k = 1,, : f kk = ( a kk k 1 j=1 f jk ) 1/ j =,, j = k + 1,, f kj = ( a kj k 1 l=1 f lkf lj ) / fkk, j = k + 1,, This algorithm requires roughly half as may operatios as Gaussia elimiatio So if A is symmetric positive defiite, the we could compute the decompositio Date: September 0, 011, versio 10 A = F F, 1

2 kow as the Cholesky decompostio I fact, there are several ways to write A = GG for some matrix G sice A = F F = F QQ F = (F Q)(F Q) = GG for ay orthogoal matrix Q, but for the Cholesky decompositio, we require that F is lower triagular, with positive diagoal elemets We ca compute F by examiig the matrix equatio A = F F o a elemet-by-elemet basis, writig a 11 a 1 f 11 f 11 f 1 f 1 a 1 a = f 1 f f a 1 a f 1 f f f From the above matrix multiplicatio we see that f 11 = a 11, from which it follows that f 11 = a 11 From the relatioship f 11 f i1 = a i1 ad the fact that we already kow f 11, we obtai f i1 = a i1 f 11, i =,, Proceedig to the secod colum of F, we see that f1 + f = a Sice we already kow f 1, we have f = a f1 Next, we use the relatio f 1 f i1 + f f i = a i to compute f i1 = a i f 1 f i1 f I geeral, we ca use the relatioship a ij = f i f j to compute f ij, where f i is the ith colum of F Aother method for computig the Cholesky decompositio is to compute f 1 = 1 a11 a 1 where a i is the ith colum of A The we set A (1) = A ad compute A () = A (1) f 1 f1 0 = A 0 Note that [ A (1) 1 0 = B B 0 A where B is the idetity matrix with its first colum replaced by f 1 Writig C = B 1, we see that A is positive defiite sice [ 1 0 = CAC 0 A is positive defiite So we may repeat the process o A [ We partitio the matrix A ito colums, writig A = f = 1 [ 0 a () a () We the compute A 3 = A () f f a () a () 3 a () ad the compute

3 ad so o Note that which implies that a kk = f k1 + f k + + f kk, f ki a kk I other words, the elemets of F are bouded We also have the relatioship det A = det F det F = (det F ) = f 11f f Is the Cholesky decomposito uique? Employig a similar approach to the oe used to prove the uiquess of the LU decompositio, we assume that A has two Cholesky decompositios A = F 1 F 1 = F F The F 1 F 1 = F F1, but sice F 1 ad F are lower triagular, both matrices must be diagoal Let F 1 F 1 = D = F F 1 So F 1 = F D ad thus F 1 = DF ad we get D 1 = F F 1 I other words, D 1 = D or D = I Hece D must have diagoal elemets equal to ±1 Sice we require that the diagoal elemets be positive, it follows that the decompositio is uique I computig the Cholesky decompositio, o row iterchages are ecessary because A is positive defiite, so the umber of operatios required to compute F is approximately 3 /3 A variat of the Cholesky decompositio is kow as the square-root-free Cholesky decompositio, ad has the form A = LDL where L is a uit lower triagular matrix, ad D is a diagoal matrix with positive diagoal elemets This is a special case of the A = LDM factorizatio previously discussed The LDL ad Cholesky decompositios are related by F = LD 1/ 3 Baded Matrices A baded matrix has all of its ozero elemets cotaied withi a bad cosistig of select diagoals Specifically, a matrix A that has upper badwidth p ad lower badwidth q has the form a 11 a 1,p+1 a 1 a,p+1 a,p+ A = a q+1,1 a q+1,q+1 a q+1, Matrices of this form arise frequetly from discretizatio of partial differetial equatios The simplest baded matrix is a tridiagoal matrix, which has upper badwidth 1 ad lower badwidth 1 Such a matrix ca be stored usig oly three vectors istead of a two-dimesioal array Computig the LU decompositio of a tridiagoal matrix without pivotig requires oly O() operatios, ad produces bidiagoal L ad U Whe pivotig is used, this desirable structure is lost, ad the process as a whole is more expesive i terms of computatio time ad storage space 3

4 Various applicatios, such as the solutio of partial differetial equatios i two or more space dimesios, yield symmetric block tridiagoal matrices, which have a block Cholesky decompositio: A 1 B F 1 F1 G B B = G G B A G F F From the above matrix equatio, we determie that A 1 = F 1 F 1, B = G F 1 from which it follows that we ca compute the Cholesky decompositio of A 1 to obtai F 1, ad the compute G = B (F1 ) 1 Next, we use the relatioship A = G G + F F to obtai F F = A G G = A B (F 1 ) 1 F 1 1 B = A B A 1 1 B It is iterestig to ote that i the case of =, the matrix A B A 1 1 B is kow as the Schur complemet of A 1 Cotiuig with the block tridiagoal case with =, suppose that we wish to compute the factorizatio [ A B B 0 [ F [F = G G + [ X It is easy to see that X = B A 1 B, but this matrix is egative defiite Therefore, we caot compute a block Cholesky decompositio, but we ca achieve the factorizatio [ [ [ A B F 0 F G B = 0 G K 0 K where K is the Cholesky factor of the positive defiite matrix B A 1 B 4 Parallelism of Gaussia Elimiatio Suppose that we wish to perform Gaussia elimiatio o the matrix A = [ a 1 a Durig the first step of the elimiatio, we compute P (1) Π 1 A = [ P (1) Π 1 a 1 P (1) Π 1 a Clearly we ca work o each colum idepedetly, leadig to a parallel algorithm As the elimiatio proceeds, we obtai less beefit from parallelism sice fewer colums are beig modified at each step 5 Error Aalysis of Gaussia Elimiatio Suppose that we wish to solve the system Ax = b Our computed solutio x satisfies a perturbed system (A + ) x = b It ca be show that x x x A 1 1 A 1 A A 1 A 1 A A 1 A κ(a)r 1 κ(a)r where κ(a) = A A 1 is the coditio umber of A ad r = / A The coditio umber has the followig properties: κ(αa) = κ(a) where α is a ozero scalar 4

5 κ(i) = 1 κ(q) = 1 whe Q Q = I The perturbatio matrix is typically a fuctio of the algorithm used to solve Ax = b I this sectio, we will cosider the case of Gaussia elimiatio ad perform a detailed error aalysis, illustratig the aalysis origially carried out by JH Wilkiso The process of solvig Ax = b cosists of three stages: (1) Factorig A = LU, resultig i a approximate LU decompositio A + E = LŪ () Solvig Ly = b, or, umerically, computig y such that ( L + δ L)(y + δy) = b (3) Solvig Ux = y, or, umerically, computig x such that (Ū + δū)(x + δx) = y + δy Combiig these stages, we see that b = ( L + δ L)(Ū + δū)(x + δx) where = δ LŪ + LδŪ + δ LδŪ = ( LŪ + δ LŪ + LδŪ + δ LδŪ)(x + δx) = (A + E + δ LŪ + LδŪ + δ LδŪ)(x + δx) = (A + )(x + δx) Departmet of Computer Sciece, Gates Buildig B, Room 80, Staford, CA address: 5

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