10_hwc1SolnsODDA

10_hwc1SolnsODDA - SOLUTION Reflecting e1 about y = x...

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Unformatted text preview: SOLUTION Reflecting e1 about y = x gives us e2 . Reflecting this about x = 0, the y – axis, doesn’t change e2 , so f (e1 ) = e2 . Reflecting e2 about y = x gives us e1 , and reflecting this about the y –axis gives us −e1 , so f (e2 ) = e1 . Hence, for this transformation, Af = [f (e1 ), f (e2 )] = [e2 , −e1 ] = 0 −1 1 0 . 2.15 Let f be the linear transformation from IR2 to IR2 given reflection about the line through the origin with slope s. Let g be the linear transformation from IR2 to IR2 given reflection about the x axis. What is the matrix Ag◦f corresponding to the composition of g with f ? (Your answer should be a matrix with entries depending on s.) How does your answer change if instead g is reflection about the y –axis? SOLUTION By Theorem 5, Ag◦f = Ag Af . Also, Theorem 2 tells us that Af = [f (e1 ), f (e2 )] and Ag = [g (e1 ), g (e2 )] . (∗∗) (∗) Therefore, to write down the matrix, we work out f (e1 ) and f (e2 ), and then g (e1 ) and g (e2 ). We can then use (∗∗) to write down Af and Ag . Finally, we can use (∗) to write down Ag◦f . To work out f (e1 ) and f (e2 ), consider the following diagram: The vector e1 is the vector indicated by the horizontal arrow, so that the segment from 0 to B has unit length. The reflected vector, f (e1 ) is the other arrow, and the line with slope s is the line running between these vectors. Since the line has slope s, we see that the segment from B to D has length s since the segment from O to B has unit length. Hence tan(ϕ) = s. 21/september/2005; 22:09 11 ...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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