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Elimination computes LU Alex Townsend February 18, 2015 In class we did elimination by hand on the 2 2 −1 0 −1 2 −1 2 + 21 1 −→ 0 0 −1 2 0 2 −→ 0 0 following matrix: −1 0 3 −1 2 −1 2 3 + 32 2 −1 0 3 −1 . 2 4 0 3 We then looked at the LU decomposition and saw: 1 0 0 2 2 −1 0 −1 2 −1 = − 1 1 0 0 2 0 −1 2 0 − 23 1 0 | {z } | {z }| A L −1 3 2 0 −1 . 4 3 0 {z } U The upper-triangular matrix U is the same as that from elimination and the unit lower-triangular matrix L contains the fractions we used during elimination (negated!). Why negated!? It’s a medium-length story. Think about what 2 −1 0 −1 2 −1 2 + 21 1 0 −1 2 is in terms of linear algebra. 1 1 2 0 Hence, elimination tells us 1 0 0 1 0 1 0 1 2 0 32 1 0 | {z L−1 It is exactly this matrix-matrix product: 0 0 2 −1 0 1 0 −1 2 −1 . 0 −1 2 0 1 that 0 0 2 1 0 −1 0 0 1 }| −1 2 −1 {z A 1 2 0 −1 = 0 2 0 } | −1 3 2 0 {z U 0 −1 . 4 3 } Thus, the fractions we used in elimination are telling us about L−1 , not L. When we invert we introduce minus signs. That is 1 1 2 0 −1 0 0 1 1 0 = − 12 0 1 0 0 0 1 0 , 0 1 1 0 0 0 1 2 3 −1 1 0 0 = 0 1 0 0 1 − 32 0 0 1 and 1 L = 0 0 0 1 2 3 0 1 0 12 1 0 −1 0 0 1 1 0 = 21 0 1 0 −1 1 0 0 1 0 0 0 1 0 0 1 2 3 −1 0 0 . 1 Hence, we have A = LU as follows: 2 −1 0 1 0 0 1 0 0 2 −1 0 −1 2 −1 = − 1 1 0 0 1 0 0 3 −1 2 2 4 0 −1 2 0 0 1 0 − 32 1 0 0 | {z } | {z }| {z 3 } A L U 1 0 0 2 −1 0 1 0 0 32 −1 . = − 21 4 0 − 23 1 0 0 {z }| {z 3 } | L U We all lived happily ever after. THE END 2