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Unformatted text preview: Math 110 Spring 2008 Solutions to HW10 Section 6.1 6. Part c.: h x, i = h x, 0 + 0 i = 2 h x, i ⇒ h x, i = 0. It follows that h ,x i = h x, i = 0. Part d.: By part c., if x = 0, then h , 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 CS used the following chain of inequalities, where c = h x,y i / h y,y i (again, we assume y 6 = 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 CS 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 ....
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This note was uploaded on 06/15/2009 for the course MATH 110 taught by Professor Gurevitch during the Spring '08 term at University of California, Berkeley.
 Spring '08
 GUREVITCH
 Math, Linear Algebra, Algebra

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