08bAss4sol

# 08bAss4sol - AS3sol/MATH1111/YKL/08-09 THE UNIVERSITY OF...

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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 7 Ans for Qn 1 (a) i). (b) When rank( A ) = n , dim N ( A ) = n - rank( A ) = 0. i.e. N ( A ) = { 0 } . This implies ( * ): A x = 0 has trivial solution only. (i.e. If A x = 0 , then x = 0 .) Let u N ( AB ). Then A ( B u ) = AB u = 0 . By ( * ), B u = 0 . 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 , the two columns (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 0 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 ) = { 0 } . (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 ) = R 3 . No. There are matrices B such that R 3 6 = N ( B ) + c ( B ). For example, B = 0 0 1 0 0 0 0 0 0 , we have c ( B ) = Span( e 1 ) and N ( B ) = Span( e 1 , e 2 ).

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