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Unformatted text preview: Notes on linear algebra (Monday 17th October, 2016, 23:10) page 65 (because B, too, is upper-triangular). Now, fix two elements 1' and j of {1,2,. . . ,a} satisfying 1' > j. We shall prove that for every k 6 {1,2, . . . ,a}, we have AakBka = 0. (50) [Proof of (50): Let k E {1, 2,. . . , n}. Then, we are in one of the following two cases: Case 1: We have i g k. Case 2: We have i > k. We shall prove (50) in each of these two cases separately: 1. Let us first consider Case 1. In this case, we have i g k. Thus, I: 2 i, so that k 2 i > j. Hence, we can apply (49) to k instead of i. As a result, we obtain Bk”,- 2 0. Hence, AiJC Bk”,- 2 14,;ka = 0. Thus, (50) is proven in Case 1. ‘v" =0 2. Let us now consider Case 2. In this case, we have i > k. Hence, we can apply (48) to k instead of j. As a result, we obtain Ag, = 0. Hence, ALI: Bk”,- = v =0 0B,“- : 0. Thus, (50) is proven in Case 2. ...
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