S4 - Solution 4 December 2, 2008 Question 1 Do the row...

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Solution 4 December 2, 2008 Question 1 Do the row operations, we can obtain: 1 3 2 4 2 2 3 3 3 1 4 2 ---------------→ R 2 - 2 R 1 ,R 3 - 3 R 1 1 3 2 4 0 - 4 - 1 - 5 0 - 8 - 2 - 10 ------→ R 3 - 2 R 2 1 3 2 4 0 - 4 - 1 - 5 0 0 0 0 . Therefore, Rank ( A ) = 2. Consequently, Dim(Nullspace( A )) = column dimension of A - RankA = 4 - 2 = 2 , and Dim(Nullspace( A T )) = row dimension of A - RankA = 3 - 2 = 1 . From the above, we know that { (1 , 3 , 2 , 4) , (0 , - 4 , - 1 , - 5) } forms a group of vector bases of the Rowspace of A . For the Nullspace of A , we know it is in form ( - 5 x + y 4 , - x +5 y 4 ,x,y ), Therefore ( - 5 , - 1 , 4 , 0) and ( - 1 , - 5 , 0 , 4) forms a group of vector bases of the Nullspace(A). For orthogonality, we only need to verify that (1 , 3 , 2 , 4) T ( - 5 , - 1 , 4 , 0) = 0 (1 , 3 , 2 , 4) T ( - 1 , - 5 , 0 , 4) = 0 (0 , - 4 , - 1 , - 5) T ( - 5 , - 1 , 4 , 0) = 0 (0 , - 4 , - 1 , - 5) T ( - 1 , - 5 , 0 , 4) = 0 . 1
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Do column operations, we can obtain 1 3 2 4 2 2 3 3 3 1 4 2 -----------------------→ C 2 - 3 C 1 ,C 3 - 2 C 1 ,C 4 - 4 C 1 1 0 0 0 2 - 4 - 1 - 5 3 - 8 - 2 - 10 -------------------→ C 3 - 1 / 4 C 2 ,C 4 - 4 / 5 C 2 1 0 0 0 2 - 4 0 0 3 - 8 0 0 Therefore, (1 , 2 , 3) T , (0 , - 4 , - 8) T forms a group of vector bases of Columnspace(A), and (1 , - 2 , 1) T forms a group of vector bases of Nullspace( A T ). The orthogonality can be verified with ( (1 , 2 , 3) T (1 , - 2 , 1) = 0 (0 , - 4 , - 8) T (1 , - 2 , 1) = 0 . Ax = (0 , 0 , 0) T has infinitely number of solutions, since x in the Nullspace(A) is a solution. Ax
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This note was uploaded on 11/13/2010 for the course SEEM SEEM2018 taught by Professor Chan during the Spring '10 term at CUHK.

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S4 - Solution 4 December 2, 2008 Question 1 Do the row...

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