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4 Gaussian Elimination 4.1 Lower and Upper Triangular Matrices ∗ ∗∗∗∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗∗∗∗ ∗ Figure 4.1: schematic representation of lower (left) and upper (right) matrices (n = 4); blank spaces represent a 0 4 Gaussian elimination goal: solve, for given A ∈ Rn×n and b ∈ Rn, the linear system of equations Ax = b. (4.1) In the sequel, we will often denote the entries of matrices by lower case letters, e.g., the entries of A are (A)ij = aij. Likewise for vectors, we sometimes write xi = xi. Remark 4.1 In matlab, the solution of (4.1) is realized by x = A\b. In python, the function numpy.linalg.solve performs this. In both cases, a routine from lapack1 realizes the actual computation. The matlab realization of the backslash operator \ is in fact very complex. A very good discussion of many aspects of the realization of \ can be found in [1]. 4.1 lower and upper triangular matrices A matrix A ∈ Rn×n is • an upper triangular matrix if Aij = 0 for j < i; • a lower triangular matrix if Aij = 0 for j > i. • a normalized lower triangular matrix if, in addition to being lower triangular, it satisfies Aii = 1 for i =1,...,n. Linear systems where A is a lower or upper triangular matrix are easily solved by “forward substitution” or “back substitution”: Algorithm 4.2 (solve Lx = b using forward substitution) Input: L ∈ Rn×n lower triangular, invertible, b ∈ Rn Output: solution x ∈ Rn of Lx = b for j = 1:n do j−1 xj := bj − ljkxk ljj J convention: empty sum = 0 K k=1 end for P . 1linear algebra package, see en.wikipedia.org/wiki/LAPACK 49 Algorithm 4.3 (solution of Ux = b using back substitution) Input: U ∈ Rn×n upper triangular, invertible, b ∈ Rn Output: solution x ∈ Rn of Ux = b for j = n:-1:1 do n xj := bj − ujkxk ujj k=j+1 ! end for P . The cost of Algorithms 4.2 and 4.3 are O(n2): Exercise 4.4 Compute the number of multiplications and additions in Algorithms 4.2 and 4.3. The set of upper and lower triangular matrices are closed under addition and matrix multipli- cation2: n×n Exercise 4.5 Let L1, L2 ∈ R be two lower triangular matrices. Show: L1 + L2 and L1L2 −1 are lower triangular matrices. If L1 is additionally invertible, then its inverse L1 is also a lower triangular matrix. Analogous results hold for upper triangular matrices. Remark 4.6 (representation via scalar products) Alg. 4.2 (and analogously Alg. 4.3) can be writen using scalar products: for j = 1:n do x(j) := b(j) − L(j, 1 : j − 1) ∗ x(1 : j − 1) L(j, j) end for h i. The advantage of such a formulation is that efficient libraries are available such as BLAS level 13. More generally, rather than realizing dot-products, matrix-vector products, or matrix- matrix-products directly by loops, it is typically advantageous to employ optimized routines such as BLAS. Remark 4.7 In Remark 4.6, the matrix L is accessed in row-oriented fashion. One can reorganize the two loops so as to access L in a column-oriented way. The following algorithm overwrites b with the solution x of Lx = b: for j = 1:n-1 do b(j) = b(j)/L(j, j) b(j +1: n) := b(j +1: n) − b(j)L(j +1: n, j) end for finis 6.DS 2That is, they have the mathematical structure of a ring 3Basic Linear Algebra Subprograms, see en.wikipedia.org/wiki/Basic Linear Algebra Subprograms 50 4.2 classical Gaussian elimination slide 23 The classical Gaussian elimination process transforms the linear system (4.1) into upper trian- gular form, which can then be solved by back substitution. We illustrate the procedure: a11x1 + a12x2 + ··· + a1nxn = b1 a21x1 + a22x2 + ··· + a2nxn = b2 . (4.2) . an1x1 + an2x2 + ··· + annxn = bn Multiplying the 1st equation by a21 l21 := a11 and subtracting this from the 2nd equation produces: a21 a21 a21 a21 (a21 − a11) x1 +(a22 − a12) x2 + ··· +(a2n − a1n) xn = b2 − b1 (4.3) a11 a11 a11 a11 0 (2) (2) (2) =:a22 =:a2n =:b2 Multiplying| the{z 1st equation} | by {z } | {z } | {z } a31 l31 := a11 and subtracting this from the 3rd equation produces: a31 a31 a31 a31 (a31 − a11) x1 +(a32 − a12) x2 + ··· +(a3n − a1n) ·xn = b3 − b1 (4.4) a11 a11 a11 a11 0 (2) (2) (2) =:a32 a3n b3 Generally,| multiplying{z } for |i = 2,...,n{z ,} the 1st equation| by{z li1 :=} ai1/a11 and| subtracting{z } this from the ith equation yields the following equivalent system of equations: a11x1 + a12x2 + ··· + a1nxn = b1 (2) (2) (2) a22 x2 + ··· + a2n xn = b2 . (4.5) . (2) (2) (2) an2 x2 + ··· + ann xn = bn Repeating this process for the (n − 1) × (n − 1) subsystem (2) (2) (2) a22 x2 + ··· + a2n xn = b2 . (2) (2) (2) an2 x2 + ··· + annxn = bn of (4.5) yields a11x1 + a12x2 + a13x3 + ··· + a1nxn = b1 (2) (2) (2) (2) a22 x2 + a23 x3 + ··· + a2n xn = b2 (3) (3) (3) a33 x3 + ··· + a3n xn = b3 (4.6) . (3) (3) (3) an3 x3 ··· ann xn = bn 51 0 0 0 0 0 0 0 A 0 0 0 0 0 0 Figure 4.2: Gaussian elimination: reduction to upper triangular form One repeats this procedure until one has reached triangular form. The following Alg. 4.8 realizes the above procedure: It overwrites the matrix A so that its upper triangle contains the final triangular form, which we will denote U below; the entries lik computed on the way will be collected in a normalized lower triangular matrix L. Algorithm 4.8 (Gaussian elimination without pivoting) Input: A Output: non-trivial entries of L and U; A is overwritten by U for k =1:(n − 1) do for i =(k + 1) : n do aik lik := akk A(i, [k + 1 : n])+ = −lik · A(k, [k + 1 : n]) end for end for Remark 4.9 In (4.8) below, we will see that A = LU, where U and L are computed in Alg. 4.8. Typically, the off-diagonal entries of L are stored in the lower triangular part of A so that effectively, A is overwritten by its LU-factorization. Exercise 4.10 Expand Alg. 4.8 so as to include the modifications of the right-hand side b. 4.2.1 Interpretation of Gaussian elimination as an LU-factorization With the coefficients lij computed above (e.g., in Alg. 4.8) one can define the matrices 1 .. 0 . . .. .. . . . 0 1 L(k) := , k =1,...,n − 1 . .. . lk ,k . +1 . . 0 .. . . .. .. 0 ··· 0 l 0 ··· 0 1 n,k 52 Exercise 4.11 Check that the inverse of L(k) is given by 1 .. 0 . . . .. .. . . 0 1 (L(k))−1 = . (4.7) . .. . −lk ,k . +1 . . 0 .. . . .. .. 0 ··· 0 −l 0 ··· 0 1 n,k Each step of the above Gaussian elimination process sketched in Fig. 4.2 is realized by a multiplication from the left by a matrix, in fact, the matrix (L(k))−1 (cf. (4.7)). That is, the Gaussian elimination process can be described as A = A(1) → A(2) =(L(1))−1A(1) → A(3) =(L(2))−1A(2) =(L(2))−1(L(1))−1A(1) → ... → A(n) =(L(n−1))−1A(n−1) = ... =(L(n−1))−1(L(n−2))−1 ... (L(2))−1(L(1))−1A(1) =:U upper triangular Rewriting this|{z yields,} the LU-factorization A = L(1) ··· L(n−1) U =:L The matrix L is a lower triangular matrix| as{z the product} of lower triangular matrices (cf. Exercise 4.5). In fact, due to the special structure of the matrices L(k), it is given by 1 . l .. L = 21 . . .. .. ln ··· ln,n 1 1 −1 as the following exercise shows: Exercise 4.12 For each k the product L(k)L(k+1) ··· L(n−1) is given by 1 .. 0 . . . .. .. . . 0 1 L(k)L(k+1) ··· L(n−1) = . . . lk ,k 1 +1 . . l .. k+2,k+1 . . .. .. 0 ··· 0 l l ··· l 1 n,k n,k+1 n,n−1 53 Thus, we have shown that Gaussian elimination produces a factorization A = LU, (4.8) where L and U are lower and upper triangular matrices determined by Alg. 4.8. 4.3 LU-factorization In numerical practice, linear systems are solved by computing the factors L and U in (4.8) and the system is then solved with one forward and one back substitution: 1. compute L, U such that A = LU 2. solve Ly = b using forward substitution 3. solve Ux = y using back substitution Remark 4.13 In matlab, the LU-factorization is realized by lu(A). In python one can use scipy.linalg.lu. As we have seen in Section 4.2.1, the LU-factorization can be computed with Gaussian elimi- nation. An alternative way of computing the factors L, U is given in the following section4 4.3.1 Crout’s algorithm for computing LU-factorization We seek L, U such that 1 u11 ··· ··· u1n a11 ··· ··· a1n .. .. l21 . . ! . = . . .. .. .. . l ··· l 1 u a ··· ··· a n1 n,n−1 nn n1 nn 2 2 This represents n equations for n unknowns, i.e., we are looking for lij,uij, such that n ! aik = lij ujk, ∀i, k =1, . , n. j=1 X L is lower triangular, U is upper triangular =⇒ min(i,k) ! aik = lij ujk ∀i, k =1,...,n (4.9) j=1 X 4One reason for studying different algorithms is that the entries of L and U are computed in a different order so that these algorithms differ in their memory access and thus potentially in actual timings.
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