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Unformatted text preview: Notes on linear algebra (Monday 17th October, 2016, 23:10) page 153 Statement 6: Each column of AT has exactly one entry equal to 1. Hence, it remains to prove that Statements 4, 5 and 6 are satisfied. The definition of AT shows that the entries of AT are precisely the entries of A, just moved to different cells. Thus, Statement 4 follows from Statement 1. Hence, Statement 4 is satisfied. Let 1' 6 {1,2, . . ., a}. The i-th column of A has exactly one entry equal to 1 (by Statement 3). In other words, there exists exactly one j 6 {1,2, . . .,a} satisfying AL,- = 1. In view of (133), this rewrites as follows: There exists exactly one j 6 {1,2,. . . , a} satisfying (AT) . = 1. In other words, the t-th row of AT has exactly 1, one entry equal to 1. I Now, forget that we fixed 1'. We thus have shown that for each 1' 6 {1,2, . . . ,n}, the i-th row of AT has exactly one entry equal to 1. In other words, each row of AT has exactly one entry equal to 1. This proves Statement 5. A similar argument (but with the roles of rows and columns switched, and also the roles of i and j switched, and using Statement 2 instead of Statement 3) can be used to prove Statement 6. We thus have shown that Statements 4, 5 and 6 are satisfied. As we have said, this completes the proof of (134).] Everyi E {1,2,...,rt} andj E {1,2,...,a} satisfy (AAT) . = 5,3,. (135) ...
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