13_hwc1SolnsODDA

13_hwc1SolnsODDA - f ( e 1 + e 2 )-f ( e 1 ) = 1 1 - 3 3...

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But then by Theorem 5, we would have f ( x ) = A f x for all x in IR 2 . Is the third piece of information that we are given consistent with this? Let us check: ± 2 3 3 2 ²± 1 1 ² = 1 ± 2 3 ² + 1 ± 3 2 ² = ± 5 5 ² . Since the right hand side is indeed f ³± 1 1 ²´ , all three pieces of information are consistent with f being linear, and f is the transformation corresponding to the matrix ± 2 3 3 2 ² . 2.21 Determine whether there is a linear transformation from IR 2 to IR 2 such that f µh 1 0 = h 3 3 i f µh 1 2 = h 3 1 i and f µh 1 1 = h 1 1 i If so, find the matrix of such a transformation. If not, explain why not. SOLUTION This one is requires a bit more thinking than Exercise 2.19, since although we are given f ( e 1 ), we are not given f ( e 2 ) – at least not directly. However, we can rewrite the given information as f ( e 1 ) = ± 3 3 ² f ( e 1 + 2 e 2 ) = ± 3 1 ² and f ( e 1 + e 2 ) = ± 1 1 ² If f were linear, we wold have to have f ( e 2 ) =
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Unformatted text preview: f ( e 1 + e 2 )-f ( e 1 ) = 1 1 - 3 3 =-2 1 1 . Therefore, by Theorem 2, If there is such a linear transformation f , it must be given by the matrix A f = 3-2 3-2 . ( * ) So far, we have used the rst and third pieces of given information about f . Is the second piece of information consistent? By Theorem 5 and ( * ), if f were linear, we would have to have f 1 2 = 3 3 3 1 1 1 . This is not the case: The right side is -1-1 , and we are told that the left side is 3 1 . Hence the three pieces of information are inconsistent with f being linear, and so there is no such linear transformation. 21 /september/ 2005; 22:09 14...
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