**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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- Fall '08
- BROWNAWELL,WOODRO
- Math, Linear Algebra, Algebra, Serge Ballif, basis element, jth standard basis, basis e3