MATH535_HW2 - of the subspace U where U is the hyperplane U...

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Serge Ballif MATH 535 Homework 2 September 7, 2007 Let the m × n matrix A reduce to the matrix R in reduced row-echelon form. Let R have columns C j 1 , . . . , C j s corresponding to the independent variables X j 1 , . . . , X j s of the linear system of equa- tions AX = 0 . 1) Show that the vectors e j 1 - C j 1 , . . . , e j s - C j s are linearly independent, where e j n denotes the j th standard basis element of n . There is an error in the question since the column vectors C j i are of dimension m × 1, but we are discussing vectors of dimension n × 1. As a counterexample, consider the matrix A = ( 1 2 3 4 0 0 0 0 0 1 ), which has 3 independent variables. However, any three 2 × 1 matrices are linearly dependent. 2) Show that in fact these elements form a basis for the null space of A , Null ( A ) . Once again, elements of Null( A ) must be n × 1 matrices. The counterexample above shows the set cannot be a basis of any space. 3) The matrix 1 2 3 0 1 0 3 2 1 defines a linear map L : v 7→ Av from 3
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Unformatted text preview: of the subspace U , where U is the hyperplane U = n u = ± u 1 u 2 u 3 ² : u 1 + u 2 + u 3 = 0 o , and b) the ordered basis e 3 ,e 2 ,e 1 of F 3 . a) We calculate the image of the basis elements to find the matrix of the trans-formation. L ± 1-1 ² = ± 1 2 3 0 1 0 3 2 1 ²± 1-1 ² = ±-2 2 ² and L ± 1-1 ² = ± 1 2 3 0 1 0 3 2 1 ²± 1-1 ² = ±-1-1 1 ² . Hence the matrix for L | U is given by ±-2-1-1 2 1 ² . We check with the calculation L ± a ± 1-1 ² + b ± 1-1 ²² = ± 1 2 3 0 1 0 3 2 1 ²± a + b-b-a ² = ±-2 a-b-b 2 a + b ² = ±-2-1-1 2 1 ² ( a b ) . b) Each column of the matrix A tells us where each basis element will be mapped. Thus the linear transformation given by the ordered basis e 3 ,e 2 ,e 1 is given by swapping the first and third columns of A : ± 3 2 1 0 1 0 1 2 3 ² . Page 1 of 1...
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