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 transformation. L ± 11 ² = ± 1 2 3 0 1 0 3 2 1 ²± 11 ² = ±2 2 ² and L ± 11 ² = ± 1 2 3 0 1 0 3 2 1 ²± 11 ² = ±11 1 ² . Hence the matrix for L  U is given by ±211 2 1 ² . We check with the calculation L ± a ± 11 ² + b ± 11 ²² = ± 1 2 3 0 1 0 3 2 1 ²± a + bba ² = ±2 abb 2 a + b ² = ±211 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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 Fall '08
 BROWNAWELL,WOODRO
 Math, Linear Algebra, Algebra, Serge Ballif, basis element, jth standard basis, basis e3

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