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7.9 LU Factorization  237




                            7.9         LU Factorization

                                        Let A be an n × m matrix of numbers. We sometimes want to factor A into a product of an n × n
                                        lower triangular matrix L and an n × m upper triangular matrix U. We will see why this is useful
                                        shortly. First we will develop a procedure for doing this.



                                          Amatrixis upper triangular if its only nonzero elements lie on or above the main
                                          diagonal. Equivalently, all elements below the main diagonal are zero. A matrix is lower
                                          triangular if its only nonzero elements are on or below the main diagonal. In the case
                                          that the matrix is not square, main diagonal elements are the 1,1, 2,2, ···, n,n ele-
                                          ments. If m > n, there will be columns beyond the columns containing these diagonal
                                          elements.




                                           To see how to construct L and U, consider an example. Let
                                                                      ⎛                 ⎞
                                                                       211 −3         5
                                                                  A = 236        1    4 ⎠ .
                                                                      ⎝
                                                                       621 −1        −3
                                        We will construct U using the elementary row operation of adding a scalar multiple of one row
                                        to another. We will not interchange rows or multiply individual rows by scalars in forming U.
                                           Begin with the leading entry in A. This is 2 in the 1,1 position. For a reason that will become
                                        clear when we construct L, highlight column one of A in some way, such as boldface (or if you
                                        are writing the matrix on a piece of paper, you might circle these elements):

                                                                      ⎛                 ⎞
                                                                        2  11 −3      5
                                                                  A = 2   36     1    4 ⎠ .
                                                                      ⎝
                                                                        6  21 −1 −3
                                        Now add scalar multiples of row one to the other rows to obtain zeros below the leading entry of
                                        2. In the matrix B, highlight the elements in column two below column one.

                                                                      ⎛                     ⎞
                                                                        2  1    1   −3   5
                                                              A → B = 0    2    5   4    −1 ⎠  .
                                                                      ⎝
                                                                        0  −1 −2    8   −18
                                        Row two has leading element 2 also. Add a scalar multiple (in this example, multiply by 1/2) of
                                        row two to row three to obtain a zero in the 3,2 position. After doing this, highlight the element
                                        in the 3,3 position.
                                                                      ⎛                     ⎞
                                                                       21     1   −3     5
                                                              B → C = 02      5    4    −1 ⎠  .
                                                                      ⎝
                                                                       00 1/2     10   −37/2
                                        In this example n = 3 and m = 5, so the diagonal elements are the 1,1; 2,2; and 3,3 elements,
                                        and there are two columns to the right of the columns containing this main diagonal. Notice that
                                        C is upper triangular. This is U:
                                                                   ⎛                      ⎞
                                                                     21     1   −3     5
                                                                U = 02      5    4    −1 ⎠  .
                                                                   ⎝
                                                                     001/2      10   −37/2



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