Page 161 - Matrix Analysis & Applied Linear Algebra

3.10 The LU Factorization 155

T
entries. Conversely, if A = RR , where R is lower triangular with a positive
diagonal, then factoring the diagonal entries out of R as illustrated in Example
3.10.6 produces R = LD, where L is lower triangular with a unit diagonal and
D is the diagonal matrix whose diagonal entries are the r ii ’s. Consequently,
2
A = LD L T is the LDU factorization for A, and thus the pivots must be
2
positive because they are the diagonal entries in D . We have now proven that
A is positive definite if and only if it has a Cholesky factorization. To see why
T
such a factorization is unique, suppose A = R 1 R T 1 = R 2 R , and factor out
2
the diagonal entries as illustrated in Example 3.10.6 to write R 1 = L 1 D 1 and
R 2 = L 2 D 2 , where each R i is lower triangular with a unit diagonal and D i
2
2
T
T
contains the diagonal of R i so that A = L 1 D L = L 2 D L . The uniqueness
1
2
2
1
of the LDU factors insures that L 1 = L 2 and D 1 = D 2 , so R 1 = R 2 . Note:
More is said about the Cholesky factorization and positive definite matrices on
pp. 313, 345, and 559.
Exercises for section 3.10
 
1 4 5
3.10.1. Let A =  41826  .
31630
(a) Determine the LU factors of A.
(b) Use the LU factors to solve Ax 1 = b 1 as well as Ax 2 = b 2 ,
where
   
6 6
b 1 =  0  and b 2 =  6  .
−6 12
(c) Use the LU factors to determine A −1 .

3.10.2. Let A and b be the matrices
12 417 17
   
 36 −12 3   3 
23 −3 2 3
A =   and b =   .
02 −2 6 4
(a) Explain why A does not have an LU factorization.
(b) Use partial pivoting and find the permutation matrix P as well
as the LU factors such that PA = LU.
(c) Use the information in P, L, and U to solve Ax = b.

 
ξ 20
3.10.3. Determine all values of ξ for which A =  1 ξ 1  fails to have an
01 ξ
LU factorization.