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08aAss4sol

# 08aAss4sol - AS3sol/MATH1111/YKL/08-09 THE UNIVERSITY OF...

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Unformatted text preview: AS3sol/MATH1111/YKL/08-09 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111: Linear Algebra Assignment 4 Suggested Solution 1. (a) It is because N ( B ) ⊂ N ( AB ). (See Tutorial 8) (b) When rank( A ) = n , dim N ( A ) = n- rank( A ) = 0. i.e. N ( A ) = { } . This implies ( * ): A x = has trivial solution only. (i.e. If A x = , then x = .) Let u ∈ N ( AB ). Then A ( B u ) = AB u = . By ( * ), B u = . i.e. u ∈ N ( B ), or in other words, N ( AB ) ⊂ N ( B ). Hence we conclude N ( B ) = N ( AB ), thus rank( AB ) = r- dim N ( AB ) = r- dim N ( B ) = rank( B ) . 2. (a) As A is row equivalent to   1 1 2 0 1 1 0 0 0   , we see that (1 0 1) T , (1 1 3) T form a basis for c ( A ). Also, N ( A ) = Span((- 1- 1 1) T . Direct checking shows that   1 1   ,   1 1 3   ,  - 1- 1 1   are linearly independent. So c ( A ) + N ( A ) = R 3 . It is also easy to see that c ( A ) ∩ N ( A ) = { } . (You may check it directly or use the formula in Assignment 3 Qn. 4, dim c ( A ) = 2, dim N ( A ) = 1 so dim( c ( A ) ∩ N ( A )) = 0.) Hence c ( A ) ⊕ N ( A ) =...
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08aAss4sol - AS3sol/MATH1111/YKL/08-09 THE UNIVERSITY OF...

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