Cme 302: Numerical Linear Algebra

Since only the right-hand side is different in each of these systems, we need only compute the LU decomposition of A once, which requires 2n3/3 operations. For simplicity, we assume that pivoting is not required, and note that the case where pivoting is required can be handled in a similar fashion. Given the LU decomposition, we compute A−1 by solving the systems Lyj = ej and Uxj = yj for j = 1, . . . , n. Computing each column xj of A−1 requires n2 operations, resulting in a total of 2n3/3 + n(n2) = 5n3/3 operations. We can compute A−1 more efficiently by noting that A−1 = U−1L−1 and computing U−1, L−1, and the product U−1L−1 directly. Since the inverse of an upper triangular matrix is also an upper triangular matrix, computing column j of U−1 requires only approximately j2/2 operations, since, in solving the system Uxj = ej , we can ignore the last n− j components of xj since we know that they are equal to zero. As a result, computing U−1 requires only n3/6 operations. A similar result holds for computing L−1, which is a lower triangular matrix. To compute the product A−1 = U−1L−1, we note that if we number the northeast-to-southwest diagonals of A−1 starting with 1 for the upper left diagonal (the (1,1) element) and n for the lower right diagonal (the (n, n) element), then elements along diagonal j require only n − j + 1 multiplications to compute. It follows that the total operation count to compute the product of U−1 and L−1 is (2n− 1) + 2(2n− 3) + 3(2n− 5) + · · ·+ (n− 1)2 + n ≈ 1 3 n.