43_hwc1SolnsODDA

# 43_hwc1SolnsODDA - A h corresponding to h as in Theorem 2...

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(b) Find the vector a f such that f ( x ) = a f · x , which we know exists by Problem 5.23, and is unique by Problem 5.24. SOLUTION Computing the integral, Z 1 0 ± a + bx + cx 2 ² d x = a + b/ 2 + c/ 3 = a b c · 1 1 / 2 1 / 3 . Hence, if we deﬁne a f = 1 1 / 2 1 / 3 , we have f ( x ) = a f · x for all x in IR 3 . This solves both parts (a) and (b) at once, since we know that the transformation x 7→ a f · x is linear: This follows from the ﬁnal part of Theorem 8: Equation (4.9) in Theorem 8 says exactly this. 5.27 Let f and g be linear functionals on IR 3 . Deﬁne a function h from IR 3 to IR 2 by h ( x ) = h f ( x ) g ( x ) i . (a) Show that h is a linear transformation from IR 3 to IR 2 . (b) Let a f be the vector corresponding to f , as in Problem 5.23, and let a g be the vector corresponding to g . If a f = ³ 1 2 3 ´ and a g = ³ 3 2 1 ´ . Find the matrix
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Unformatted text preview: A h corresponding to h , as in Theorem 2. SOLUTION (a) We use the deﬁnition of linearity, given in Section 2.2: Since f and g are linear, for any numbers a and b , and any x and y in IR n , f ( a x + b y ) = af ( x ) = bf ( y ) and g ( a x + b y ) = ag ( x ) = bg ( y ) . Therefore, h ( a x ) = ³ f ( a x + b y )) g ( a x + b y )) ´ = ³ af ( x ) = bf ( y ) ag ( x ) = bg ( y ) ´ = a ³ f ( x ) g ( x ) ´ + b ³ f ( y ) g ( y ) ´ = ah ( x ) + bh ( y ) . Thus, h is linear. (b) The matrix is ³ 1 2 3 3 2 1 ´ . Section 6 21 /september/ 2005; 22:09 44...
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