10aAss4sol

# 10aAss4sol - AS4sol/MATH1111/YKL/10-11 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111 Linear Algebra Assignment 4 Suggested

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AS4sol/MATH1111/YKL/10-11 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111: Linear Algebra Assignment 4 Suggested Solution 1. (a) No. As b = 1 2 a + 1 2 c , L ( b ) = 1 2 L ( a ) + 1 2 L ( c ) = 1 2 b + 1 2 a = ( 3 2 5 2 ) T if L is a linear transformation. (b) i. T ( a ) = ± 10 12 ² = 13 2 b + ( ± 1) c , T ( b ) = ± ± 4 ± 4 ² = ± 2 b . The matrix representation is ± 13 2 ± 2 ± 1 0 ² . ii. The standard matrix representation is ± ± 8 6 ± 9 7 ² . So the eigenvalues are ± 2 and 1, with eigenvectors ± 1 1 ² and ± 2 3 ² . Set G = ³± 1 1 ² ; ± 2 3 ²´ . 2. (a) i. As A is row equivalent to 0 @ 1 1 2 0 1 1 0 0 0 1 A , 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 0 @ 1 0 1 1 A , 0 @ 1 1 3 1 A , 0 @ ± 1 ± 1 1 1 A are linearly independent. So c ( A ) + N ( A ) = R 3 . It is also easy to see that c ( A ) \ N ( A ) = f 0 g . (One way is to apply the dimension formula: dim c ( A ) = 2, dim N ( A ) = 1 ) dim( c ( A ) \ N ( A )) = 0.) Hence c ( A ) ² N ( A ) = R 3 . ii. No. There are matrices B such that R 3 6 = N ( B ) + c ( B ).

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## This note was uploaded on 05/04/2011 for the course MATH 1111 taught by Professor Dr,li during the Spring '10 term at HKU.

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10aAss4sol - AS4sol/MATH1111/YKL/10-11 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111 Linear Algebra Assignment 4 Suggested

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