08_hwc1SolnsODDA

08_hwc1SolnsODDA - and for those that are linear, write...

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of degree two”). In any case, we don’t have h ( a x ) = ah ( x ) for all x and a , so h is not homogeneous and therefore not linear. We now check to see that f is additive. Let x = ± x y ² and u = ± u v ² . Then f ( x + u ) = f ³± x + u y + v ²´ = ± x + u + y + v y + v - x - u ² = ± x + y y - x ² + ± u + v v - u ² = f ( x ) + f ( u ) . Hence f is both additive and homogeneous, so it is linear. Remark: To show that a transformation is not additive, or not homogeneous, you don’t need a general calculation; you just need one example. To see that g and h are not homogenous, consider x = ± 1 0 ² . Then g ( - x ) 6 = g ( - x ) and h ( - x ) 6 = h ( - x ). If we let u = ± - 1 0 ² , then x + u = 0, and so g ( x + u ) = 0 while g ( x ) + g ( u ) = 2 x 6 = 0. Likewise, h ( x + u ) = 0 while h ( x ) + h ( u ) = ± 2 2 ² 6 = 0. You can also see that neither is additive in the sense of (2.3). The problem with g comes from the absolute value and the fact that | a + b | < | a | + | b | when a and b have opposite signs. The problem with h comes from the squares. 2.5 Consider the following transformations from IR 2 to IR 3 . Which ones are linear? Explain your answers,
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Unformatted text preview: and for those that are linear, write down the corresponding matrix. f h x y i = y 1 x g h x y i = y x h h x y i = SOLUTION Note that f is not additive: f ( u + v ) 6 = f ( u ) + f ( v ), since the second entry of f ( u + v ) is 1, while the second entry of f ( u )+ f ( v ) is 2. Thus it is not linear. (It is also not homogeneous, as you can see from almost any example you might try.) This disposes of f . The transformations g and h are both additive and homogeneous, so they are linear. By Theorem 2, the corresponding matrices are: A g = 1 1 A h = . 21 /september/ 2005; 22:09 9...
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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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