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Homework 11 Solutions

# Homework 11 Solutions - Math 110 Spring 2008 Solutions to...

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Math 110 Spring 2008 Solutions to HW10 Section 6.1 6. Part c.: h x, 0 i = h x, 0 + 0 i = 2 h x, 0 i ⇒ h x, 0 i = 0. It follows that h 0 , x i = h x, 0 i = 0. Part d.: By part c., if x = 0, then h 0 , 0 i = 0. By Axiom IV of an inner product, if x 6 = 0, then h x, x i > 0. Part e.: If h x, y i = h x, z i for all x V , then h x, y - z i = 0 for all x V . Then in particular, taking x = y - z , we get h y - z, y - z i = 0. Thus y - z = 0 by part d.; i.e., y = z . 11. || x + y || 2 + || x - y || 2 = h x + y, x + y i + h x - y, x - y i = 2 h x, x i + 2 h y, y i + h x, y i + h y, x i - h x, y i - h y, x i = 2 || x || 2 + 2 || y || 2 15. a. Our proof of C-S used the following chain of inequalities, where c = h x, y i / h y, y i (again, we assume y 6 = 0): 0 ≤ h x - cy, x - cy i = h x, x i - |h x, y i| / h y, y i . We get an equality in the C-S if and only if the first inequality is an equality. This in turn is true if and only if x - cy = 0, or x = cy . b. Similarly, our proof of the triangle inequality used the following chain of inequalities: h x + y, x + y i = h x, x i + 2 Rh x, y i + h y, y i ≤ h x, x i + 2 |h x, y i| + h y, y i ≤ h x, x i + 2 || x || · || y || + h y, y i = ( h x, x i + h y, y i ) 2 . We get an equality iff these two inequalities are equalities. The second one is an equality iff x = cy for some c F , by part a. The first and second are equalities iff x = cy and Rh x, y i = R

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