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GAUSSIAN ELIMINATION AND LU -FACTORIZATION 59 Consequently, we see that Ak+1 = Ek Pk Ak , and then Ak = Ek−1 Pk−1 · · · E1 P1 A. This justifies the claim made earlier, that Ak = Mk A for some invertible matrix Mk ; we can pick Mk = Ek−1 Pk−1 · · · E1 P1 , a product of invertible matrices. The fact that det(P (i, k)) = −1 and that det(Ei,j;β ) = 1 implies immediately the fact claimed above: We always have det(Ak ) = ± det(A). Furthermore, since Ak = Ek−1 Pk−1 · · · E1 P1 A and since Gaussian elimination stops for k = n, the matrix An = En−1 Pn−1 · · · E2 P2 E1 P1 A is upper-triangular.

Actually, it is possible to do better: This is the Cholesky factorization. 2. GAUSSIAN ELIMINATION AND LU -FACTORIZATION 63 The following easy proposition shows that, in principle, A can be premultiplied by some permutation matrix, P , so that P A can be converted to upper-triangular form without using any pivoting. 2, but for now we just need their definition. A permutation matrix is a square matrix that has a single 1 in every row and every column and zeros everywhere else. 2 that every permutation matrix is a product of transposition matrices (the P (i, k)s), and that P is invertible with inverse P .

There is a certain asymmetry in the LU -decomposition A = LU of an invertible matrix A. Indeed, the diagonal entries of L are all 1, but this is generally false for U . This asymmetry can be eliminated as follows: if D = diag(u11 , u22 , . . , unn ) is the diagonal matrix consisting of the diagonal entries in U (the pivots), then we if let U = D−1 U , we can write A = LDU , where L is lower- triangular, U is upper-triangular, all diagonal entries of both L and U are 1, and D is a diagonal matrix of pivots.

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Applications of Scientific Computation [Lecture notes] by Jean Gallier

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