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42      1 Linear algebra



                   For further discussion of LU factorization with pivoting, consult Quateroni et al.
                   (2000).


                   Cholesky decomposition
                   Since the transpose of a lower triangular matrix L is an upper triangular matrix, we might
                   wonder if it is possible for some matrix A, that in the LU decomposition, we can set U = L T
                   to obtain
                                                   A = LL  T                         (1.207)

                   First, by taking the transpose of this equation, we see that any A that can be written in this
                                               T
                   manner must be symmetric, A = A ,
                                                            T
                                                    T T
                                             T
                                            A = (LL ) = LL = A                       (1.208)
                                                                                  N
                                                T
                   Also, for A to be nonsingular, L and L must be nonsingular, so that for any v ∈  , v  = 0,
                                                  T       T
                                                L v · L v > 0                        (1.209)
                   From
                                                    T
                                     T
                                                               T
                              T
                                                 T
                                                                      T
                                                       T
                                         T






                            v Av = v LL v = L v        L v = L v · L v               (1.210)
                                                           T
                   we see that a matrix A could be written as A = LL only if it were symmetric and positive-
                   definite; i.e.,
                                                            N
                                           T
                                          v Av > 0    ∀v ∈  , v  = 0                 (1.211)
                   While only a subset of all matrices are symmetric, positive-definite, they in fact form
                                                                               T
                   an important subset, especially in numerical optimization. Thus A = LL , known as a
                   Cholesky decomposition, is used frequently.
                     The special structure of a positive-definite matrix allows us to perform Cholesky factor-
                                                                              T
                   ization more quickly than LU decomposition. We start by writing A = LL explicitly,
                                                                             
                         a 11  a 21  a 31  ...  a N1    L 11
                        a 21  a 22  a 32  ...  a N2     L 21  L 22           
                                                                             
                         a 31  a 32  a 33               L 31  L 32  L 33
                                      ...                                    
                                           a N3   =                          
                        .     .    .   .    .        .      .    .   .       
                        . .   . .  . .  . .  . .     . .    . .  . .  . .    
                        a N1  a N2  a N3  ... a NN      L N1  L N2  L N3  ...  L NN
                                                                              
                                                        L 11  L 21  L 31  ...  L N1
                                                            L 22  L 32  ...  L N2  
                                                                              
                                                                      ...
                                                                  L 33     L N3     (1.212)
                                                                              
                                                   × 
                                                                       .    .
                                                                      .       
                                                                       .   . . 
                                                                           L NN
                   From the (1, 1) element we obtain
                                                         L 11 = (a 11 ) 1/2          (1.213)
                                        a 11 = L 11 × L 11
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