15_hwc1SolnsODDA

# 15_hwc1SolnsODDA - g ◦ f is not linear or can it be that...

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Two equations in one variable are often overdetermined, but in this case, both equations are solved by a = 2. Thus, a = 2 is the answer we seek. 2.25 Find all values of a , if any, for which there is a linear transformation from IR 2 to IR 2 such that f ±h 0 1 = h 3 3 i f ±h 2 0 = h 3 2 i and f ±h 1 1 = h a 3 i For each such a , ﬁnd the corresponding matrix. SOLUTION The ﬁrst two pieces of given information say that f ( e 2 ) = ³ 3 3 ´ and f (2 e 1 ) = ³ 3 2 ´ . If f is linear, we would have to have f ( e 2 ) = ³ 3 3 ´ and f ( e 1 ) = ³ 3 / 2 1 ´ , and then by Theorem 2, the matrix corresponding to f would be A f = ³ 3 3 / 2 3 1 ´ . By Theorem 5, we would then have to have f µ³ 1 1 ´¶ = ³ 3 3 / 2 3 1 ´³ 1 1 ´ = 1 ³ 3 / 2 1 ´ + 1 ³ 3 3 ´ = ³ 9 / 2 4 ´ . However, this is inconsistent with f µ³ 1 1 ´¶ = ³ a 3 ´ . (Even if we choose a = 9 / 2, we can only get agreement in the ﬁrst entries.) 2.27 Suppose that f is a transformation from IR n to IR n , and that g is a transformation from IR n to IR n , and that neither f nor g is linear, but both are invertible. Does it follow that
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Unformatted text preview: g ◦ f is not linear, or can it be that g ◦ f is a linear transformation from IR n to IR n ? Explain your answer. SOLUTION The composition can be linear. One can give simple examples using the fact that the identity transformation is linear. Therefore, if f is any invertible transformation from IR n to IR n , and g = f-1 , then g ◦ f will be the identity map, and therefore linear, whether or not f is linear. Indeed, take n = 1, and f ( x ) = tanh( x ), the hyperbolic tangent. This is an invertible map from IR onto (-1 , 1). Deﬁne g ( y ) to be given by g ( x ) = arctanh( y ) for-1 < y < 1, and by g ( y ) = 0 otherwise. Then g ( f ( x )) = x for all 21 /september/ 2005; 22:09 16...
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