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13_hwc1SolnsODDA

# 13_hwc1SolnsODDA - f e 1 e 2-f e 1 = ± 1 1 ²-± 3 3 ²...

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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 1 1 = 1 2 3 + 1 3 2 = 5 5 . Since the right hand side is indeed f 1 1 , all three pieces of information are consistent with f being linear, and f is the transformation corresponding to the matrix 2 3 3 2 . 2.21 Determine whether there is a linear transformation from IR 2 to IR 2 such that f 1 0 = 3 3 f 1 2 = 3 1 and f 1 1 = 1 1 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 f ( e 1 + 2 e 2 ) = 3 1 and f ( e 1 + e 2 ) = 1 1 If f were linear, we wold have to have
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Unformatted text preview: f ( e 1 + e 2 )-f ( e 1 ) = ± 1 1 ²-± 3 3 ² =-2 ± 1 1 ² . 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 ² . ( * ) So far, we have used the ﬁrst 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 ³± 1 2 ²´ = ± 3 3 3 1 ²± 1 1 ² . This is not the case: The right side is ±-1-1 ² , and we are told that the left side is ± 3 1 ² . 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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